remove empty results with a 2nd query from array mysqli












0














I have 2 tables, 1 with companies(Costcenters) and one with clients (employees of those companies)



i need to make a form to edit or delete malformed ( like john smit and j. Smit) from those companies employees grouped by company (Costcenter), when i make a list with all those companies i get a lot of companies that has no clients. So I made a array of the companies (Costcenters) and check first if they have employees, this with the goal to remove the Costcenters without employees from the array ($bedrijven).



The form is no problem, but i cant find a way to get those companies removed from the bedrijven array.



require_once('conn.php');
$query = "SELECT bedrijfID, Houder, Costcenter, Actief FROM bedrijven
WHERE Actief = 'actief' ORDER BY Costcenter";
$results = mysqli_query($conn, $query);
if (!$results) printf("Query failed: %sn", $conn->error);

$bedrijven = ;
while($row = mysqli_fetch_assoc($results)) {
$bedrijven = $row['Costcenter'];
}
foreach ($bedrijven as $key => $item) {

$query1 = "SELECT * from customer where Costcenter = '$item' ORDER by
Client";
$customerresult = mysqli_query($conn, $query1) or
die(mysqli_error($conn));
if (!$customerresult) printf("Query failed: %sn", $conn->error);
if($customerresult->num_rows === 0) {
unset($bedrijven[$key]);
}
}


I am not familiar with PDO or funtions so tried it this way that does not work as i expected, the unset is not working.



the code is editted as it is working now, i hope it might help others as well. If any has a better solution please post.










share|improve this question
























  • customerresult is treated as a constant; it's missing the $ sign for it. Enable error reporting. So, is that your real code?
    – Funk Forty Niner
    Nov 12 '18 at 19:08












  • Its not a duplicate, I dont get any errors, just not getting the empty results removed from the bedrijven array..
    – aton Graaff
    Nov 12 '18 at 19:14












  • Why am I getting a downvote on a normal question ?
    – aton Graaff
    Nov 12 '18 at 19:29










  • Can you add the result of var_export($bedrijven); to your question?
    – The fourth bird
    Nov 12 '18 at 19:48
















0














I have 2 tables, 1 with companies(Costcenters) and one with clients (employees of those companies)



i need to make a form to edit or delete malformed ( like john smit and j. Smit) from those companies employees grouped by company (Costcenter), when i make a list with all those companies i get a lot of companies that has no clients. So I made a array of the companies (Costcenters) and check first if they have employees, this with the goal to remove the Costcenters without employees from the array ($bedrijven).



The form is no problem, but i cant find a way to get those companies removed from the bedrijven array.



require_once('conn.php');
$query = "SELECT bedrijfID, Houder, Costcenter, Actief FROM bedrijven
WHERE Actief = 'actief' ORDER BY Costcenter";
$results = mysqli_query($conn, $query);
if (!$results) printf("Query failed: %sn", $conn->error);

$bedrijven = ;
while($row = mysqli_fetch_assoc($results)) {
$bedrijven = $row['Costcenter'];
}
foreach ($bedrijven as $key => $item) {

$query1 = "SELECT * from customer where Costcenter = '$item' ORDER by
Client";
$customerresult = mysqli_query($conn, $query1) or
die(mysqli_error($conn));
if (!$customerresult) printf("Query failed: %sn", $conn->error);
if($customerresult->num_rows === 0) {
unset($bedrijven[$key]);
}
}


I am not familiar with PDO or funtions so tried it this way that does not work as i expected, the unset is not working.



the code is editted as it is working now, i hope it might help others as well. If any has a better solution please post.










share|improve this question
























  • customerresult is treated as a constant; it's missing the $ sign for it. Enable error reporting. So, is that your real code?
    – Funk Forty Niner
    Nov 12 '18 at 19:08












  • Its not a duplicate, I dont get any errors, just not getting the empty results removed from the bedrijven array..
    – aton Graaff
    Nov 12 '18 at 19:14












  • Why am I getting a downvote on a normal question ?
    – aton Graaff
    Nov 12 '18 at 19:29










  • Can you add the result of var_export($bedrijven); to your question?
    – The fourth bird
    Nov 12 '18 at 19:48














0












0








0







I have 2 tables, 1 with companies(Costcenters) and one with clients (employees of those companies)



i need to make a form to edit or delete malformed ( like john smit and j. Smit) from those companies employees grouped by company (Costcenter), when i make a list with all those companies i get a lot of companies that has no clients. So I made a array of the companies (Costcenters) and check first if they have employees, this with the goal to remove the Costcenters without employees from the array ($bedrijven).



The form is no problem, but i cant find a way to get those companies removed from the bedrijven array.



require_once('conn.php');
$query = "SELECT bedrijfID, Houder, Costcenter, Actief FROM bedrijven
WHERE Actief = 'actief' ORDER BY Costcenter";
$results = mysqli_query($conn, $query);
if (!$results) printf("Query failed: %sn", $conn->error);

$bedrijven = ;
while($row = mysqli_fetch_assoc($results)) {
$bedrijven = $row['Costcenter'];
}
foreach ($bedrijven as $key => $item) {

$query1 = "SELECT * from customer where Costcenter = '$item' ORDER by
Client";
$customerresult = mysqli_query($conn, $query1) or
die(mysqli_error($conn));
if (!$customerresult) printf("Query failed: %sn", $conn->error);
if($customerresult->num_rows === 0) {
unset($bedrijven[$key]);
}
}


I am not familiar with PDO or funtions so tried it this way that does not work as i expected, the unset is not working.



the code is editted as it is working now, i hope it might help others as well. If any has a better solution please post.










share|improve this question















I have 2 tables, 1 with companies(Costcenters) and one with clients (employees of those companies)



i need to make a form to edit or delete malformed ( like john smit and j. Smit) from those companies employees grouped by company (Costcenter), when i make a list with all those companies i get a lot of companies that has no clients. So I made a array of the companies (Costcenters) and check first if they have employees, this with the goal to remove the Costcenters without employees from the array ($bedrijven).



The form is no problem, but i cant find a way to get those companies removed from the bedrijven array.



require_once('conn.php');
$query = "SELECT bedrijfID, Houder, Costcenter, Actief FROM bedrijven
WHERE Actief = 'actief' ORDER BY Costcenter";
$results = mysqli_query($conn, $query);
if (!$results) printf("Query failed: %sn", $conn->error);

$bedrijven = ;
while($row = mysqli_fetch_assoc($results)) {
$bedrijven = $row['Costcenter'];
}
foreach ($bedrijven as $key => $item) {

$query1 = "SELECT * from customer where Costcenter = '$item' ORDER by
Client";
$customerresult = mysqli_query($conn, $query1) or
die(mysqli_error($conn));
if (!$customerresult) printf("Query failed: %sn", $conn->error);
if($customerresult->num_rows === 0) {
unset($bedrijven[$key]);
}
}


I am not familiar with PDO or funtions so tried it this way that does not work as i expected, the unset is not working.



the code is editted as it is working now, i hope it might help others as well. If any has a better solution please post.







php arrays mysqli






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 '18 at 20:40

























asked Nov 12 '18 at 19:05









aton Graaff

266




266












  • customerresult is treated as a constant; it's missing the $ sign for it. Enable error reporting. So, is that your real code?
    – Funk Forty Niner
    Nov 12 '18 at 19:08












  • Its not a duplicate, I dont get any errors, just not getting the empty results removed from the bedrijven array..
    – aton Graaff
    Nov 12 '18 at 19:14












  • Why am I getting a downvote on a normal question ?
    – aton Graaff
    Nov 12 '18 at 19:29










  • Can you add the result of var_export($bedrijven); to your question?
    – The fourth bird
    Nov 12 '18 at 19:48


















  • customerresult is treated as a constant; it's missing the $ sign for it. Enable error reporting. So, is that your real code?
    – Funk Forty Niner
    Nov 12 '18 at 19:08












  • Its not a duplicate, I dont get any errors, just not getting the empty results removed from the bedrijven array..
    – aton Graaff
    Nov 12 '18 at 19:14












  • Why am I getting a downvote on a normal question ?
    – aton Graaff
    Nov 12 '18 at 19:29










  • Can you add the result of var_export($bedrijven); to your question?
    – The fourth bird
    Nov 12 '18 at 19:48
















customerresult is treated as a constant; it's missing the $ sign for it. Enable error reporting. So, is that your real code?
– Funk Forty Niner
Nov 12 '18 at 19:08






customerresult is treated as a constant; it's missing the $ sign for it. Enable error reporting. So, is that your real code?
– Funk Forty Niner
Nov 12 '18 at 19:08














Its not a duplicate, I dont get any errors, just not getting the empty results removed from the bedrijven array..
– aton Graaff
Nov 12 '18 at 19:14






Its not a duplicate, I dont get any errors, just not getting the empty results removed from the bedrijven array..
– aton Graaff
Nov 12 '18 at 19:14














Why am I getting a downvote on a normal question ?
– aton Graaff
Nov 12 '18 at 19:29




Why am I getting a downvote on a normal question ?
– aton Graaff
Nov 12 '18 at 19:29












Can you add the result of var_export($bedrijven); to your question?
– The fourth bird
Nov 12 '18 at 19:48




Can you add the result of var_export($bedrijven); to your question?
– The fourth bird
Nov 12 '18 at 19:48












2 Answers
2






active

oldest

votes


















2














If I understand what you are going for, this is better done as a single Query. A JOIN can be used first to bind your tables, and then additional WHERE operators can be used if needed to refine your search. I'm not 100% sure if I'm reading right that this is is exactly how you wanted to join the data, but if you play with different JOIN operators you'll get it.



$query = "SELECT Costcenter.bedrijfID, Costcenter.Houder, Costcenter.Costcenter, Costcenter.Actief, customer.* FROM Costcenter
LEFT JOIN customer ON customer.Costcenter = Costcenter.Costcenter
WHERE Actief = 'Costcenter.actief' AND Costcenter.Costcenter != "" ORDER BY Costcenter.Costcenter";


The biggest reason for doing it this way this that a single SQL call processes WAY faster than trying to parse your data from multiple calls in PHP.






share|improve this answer























  • That would look for empty Costcenters in the bedrijven table. the goal was to look in if there were Costcenters that had no customers in the customers table and so remove those costcenters from the bedrijven array.
    – aton Graaff
    Nov 12 '18 at 20:43










  • Sorry, I got the array name and table name backwards, but the process is still the same. By selecting Costcenter and joining customer to it, you get a combined dataset. Then your WHERE statement filters out empty cost centers. Doing it this way can literally be thousands of times faster which will make a huge difference as your database grows.
    – Nosajimiki
    Nov 20 '18 at 18:13










  • the join way of the query is a perfect solution to make it less heavvy , thatnks for the advice.
    – aton Graaff
    Nov 21 '18 at 11:02



















0














thanks to the requestion of The fouth bird i discovered i have been wasting a lot of time on a simple solution. i should not have done :
unset($bedrijven['Costcenter']);
but



 unset($bedrijven[$key]);


you must unset the key in the array not the value....






share|improve this answer





















    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53268540%2fremove-empty-results-with-a-2nd-query-from-array-mysqli%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    If I understand what you are going for, this is better done as a single Query. A JOIN can be used first to bind your tables, and then additional WHERE operators can be used if needed to refine your search. I'm not 100% sure if I'm reading right that this is is exactly how you wanted to join the data, but if you play with different JOIN operators you'll get it.



    $query = "SELECT Costcenter.bedrijfID, Costcenter.Houder, Costcenter.Costcenter, Costcenter.Actief, customer.* FROM Costcenter
    LEFT JOIN customer ON customer.Costcenter = Costcenter.Costcenter
    WHERE Actief = 'Costcenter.actief' AND Costcenter.Costcenter != "" ORDER BY Costcenter.Costcenter";


    The biggest reason for doing it this way this that a single SQL call processes WAY faster than trying to parse your data from multiple calls in PHP.






    share|improve this answer























    • That would look for empty Costcenters in the bedrijven table. the goal was to look in if there were Costcenters that had no customers in the customers table and so remove those costcenters from the bedrijven array.
      – aton Graaff
      Nov 12 '18 at 20:43










    • Sorry, I got the array name and table name backwards, but the process is still the same. By selecting Costcenter and joining customer to it, you get a combined dataset. Then your WHERE statement filters out empty cost centers. Doing it this way can literally be thousands of times faster which will make a huge difference as your database grows.
      – Nosajimiki
      Nov 20 '18 at 18:13










    • the join way of the query is a perfect solution to make it less heavvy , thatnks for the advice.
      – aton Graaff
      Nov 21 '18 at 11:02
















    2














    If I understand what you are going for, this is better done as a single Query. A JOIN can be used first to bind your tables, and then additional WHERE operators can be used if needed to refine your search. I'm not 100% sure if I'm reading right that this is is exactly how you wanted to join the data, but if you play with different JOIN operators you'll get it.



    $query = "SELECT Costcenter.bedrijfID, Costcenter.Houder, Costcenter.Costcenter, Costcenter.Actief, customer.* FROM Costcenter
    LEFT JOIN customer ON customer.Costcenter = Costcenter.Costcenter
    WHERE Actief = 'Costcenter.actief' AND Costcenter.Costcenter != "" ORDER BY Costcenter.Costcenter";


    The biggest reason for doing it this way this that a single SQL call processes WAY faster than trying to parse your data from multiple calls in PHP.






    share|improve this answer























    • That would look for empty Costcenters in the bedrijven table. the goal was to look in if there were Costcenters that had no customers in the customers table and so remove those costcenters from the bedrijven array.
      – aton Graaff
      Nov 12 '18 at 20:43










    • Sorry, I got the array name and table name backwards, but the process is still the same. By selecting Costcenter and joining customer to it, you get a combined dataset. Then your WHERE statement filters out empty cost centers. Doing it this way can literally be thousands of times faster which will make a huge difference as your database grows.
      – Nosajimiki
      Nov 20 '18 at 18:13










    • the join way of the query is a perfect solution to make it less heavvy , thatnks for the advice.
      – aton Graaff
      Nov 21 '18 at 11:02














    2












    2








    2






    If I understand what you are going for, this is better done as a single Query. A JOIN can be used first to bind your tables, and then additional WHERE operators can be used if needed to refine your search. I'm not 100% sure if I'm reading right that this is is exactly how you wanted to join the data, but if you play with different JOIN operators you'll get it.



    $query = "SELECT Costcenter.bedrijfID, Costcenter.Houder, Costcenter.Costcenter, Costcenter.Actief, customer.* FROM Costcenter
    LEFT JOIN customer ON customer.Costcenter = Costcenter.Costcenter
    WHERE Actief = 'Costcenter.actief' AND Costcenter.Costcenter != "" ORDER BY Costcenter.Costcenter";


    The biggest reason for doing it this way this that a single SQL call processes WAY faster than trying to parse your data from multiple calls in PHP.






    share|improve this answer














    If I understand what you are going for, this is better done as a single Query. A JOIN can be used first to bind your tables, and then additional WHERE operators can be used if needed to refine your search. I'm not 100% sure if I'm reading right that this is is exactly how you wanted to join the data, but if you play with different JOIN operators you'll get it.



    $query = "SELECT Costcenter.bedrijfID, Costcenter.Houder, Costcenter.Costcenter, Costcenter.Actief, customer.* FROM Costcenter
    LEFT JOIN customer ON customer.Costcenter = Costcenter.Costcenter
    WHERE Actief = 'Costcenter.actief' AND Costcenter.Costcenter != "" ORDER BY Costcenter.Costcenter";


    The biggest reason for doing it this way this that a single SQL call processes WAY faster than trying to parse your data from multiple calls in PHP.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 20 '18 at 18:04

























    answered Nov 12 '18 at 20:05









    Nosajimiki

    6701411




    6701411












    • That would look for empty Costcenters in the bedrijven table. the goal was to look in if there were Costcenters that had no customers in the customers table and so remove those costcenters from the bedrijven array.
      – aton Graaff
      Nov 12 '18 at 20:43










    • Sorry, I got the array name and table name backwards, but the process is still the same. By selecting Costcenter and joining customer to it, you get a combined dataset. Then your WHERE statement filters out empty cost centers. Doing it this way can literally be thousands of times faster which will make a huge difference as your database grows.
      – Nosajimiki
      Nov 20 '18 at 18:13










    • the join way of the query is a perfect solution to make it less heavvy , thatnks for the advice.
      – aton Graaff
      Nov 21 '18 at 11:02


















    • That would look for empty Costcenters in the bedrijven table. the goal was to look in if there were Costcenters that had no customers in the customers table and so remove those costcenters from the bedrijven array.
      – aton Graaff
      Nov 12 '18 at 20:43










    • Sorry, I got the array name and table name backwards, but the process is still the same. By selecting Costcenter and joining customer to it, you get a combined dataset. Then your WHERE statement filters out empty cost centers. Doing it this way can literally be thousands of times faster which will make a huge difference as your database grows.
      – Nosajimiki
      Nov 20 '18 at 18:13










    • the join way of the query is a perfect solution to make it less heavvy , thatnks for the advice.
      – aton Graaff
      Nov 21 '18 at 11:02
















    That would look for empty Costcenters in the bedrijven table. the goal was to look in if there were Costcenters that had no customers in the customers table and so remove those costcenters from the bedrijven array.
    – aton Graaff
    Nov 12 '18 at 20:43




    That would look for empty Costcenters in the bedrijven table. the goal was to look in if there were Costcenters that had no customers in the customers table and so remove those costcenters from the bedrijven array.
    – aton Graaff
    Nov 12 '18 at 20:43












    Sorry, I got the array name and table name backwards, but the process is still the same. By selecting Costcenter and joining customer to it, you get a combined dataset. Then your WHERE statement filters out empty cost centers. Doing it this way can literally be thousands of times faster which will make a huge difference as your database grows.
    – Nosajimiki
    Nov 20 '18 at 18:13




    Sorry, I got the array name and table name backwards, but the process is still the same. By selecting Costcenter and joining customer to it, you get a combined dataset. Then your WHERE statement filters out empty cost centers. Doing it this way can literally be thousands of times faster which will make a huge difference as your database grows.
    – Nosajimiki
    Nov 20 '18 at 18:13












    the join way of the query is a perfect solution to make it less heavvy , thatnks for the advice.
    – aton Graaff
    Nov 21 '18 at 11:02




    the join way of the query is a perfect solution to make it less heavvy , thatnks for the advice.
    – aton Graaff
    Nov 21 '18 at 11:02













    0














    thanks to the requestion of The fouth bird i discovered i have been wasting a lot of time on a simple solution. i should not have done :
    unset($bedrijven['Costcenter']);
    but



     unset($bedrijven[$key]);


    you must unset the key in the array not the value....






    share|improve this answer


























      0














      thanks to the requestion of The fouth bird i discovered i have been wasting a lot of time on a simple solution. i should not have done :
      unset($bedrijven['Costcenter']);
      but



       unset($bedrijven[$key]);


      you must unset the key in the array not the value....






      share|improve this answer
























        0












        0








        0






        thanks to the requestion of The fouth bird i discovered i have been wasting a lot of time on a simple solution. i should not have done :
        unset($bedrijven['Costcenter']);
        but



         unset($bedrijven[$key]);


        you must unset the key in the array not the value....






        share|improve this answer












        thanks to the requestion of The fouth bird i discovered i have been wasting a lot of time on a simple solution. i should not have done :
        unset($bedrijven['Costcenter']);
        but



         unset($bedrijven[$key]);


        you must unset the key in the array not the value....







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 12 '18 at 20:38









        aton Graaff

        266




        266






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53268540%2fremove-empty-results-with-a-2nd-query-from-array-mysqli%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            這個網誌中的熱門文章

            Tangent Lines Diagram Along Smooth Curve

            Yusuf al-Mu'taman ibn Hud

            Zucchini