PHP add Value to Variable in Variable before echo












-1















i am looking for a solution:



$txt = "Welcome to ".$place." City!";

$place = "New York";

echo $txt; // should be "Welcome to New York City!"


The variable "$place" can only be declared after the variable "$txt".
Is that possible?



Thanks






php variables echo







share|improve this question























  • When $txt is set, it is concatenating "Welcome to ", the value of $place, and " City!" to form a new string. So no, you cannot. Why do you want to do this?

    – Jordan S
    Nov 21 '18 at 19:45








  • 2





    Can you explain the goal here, your example isn't possible, but I bet whatever you actually trying to achieve is

    – Dan
    Nov 21 '18 at 19:48











  • You can't define a variable before using it.

    – Funk Forty Niner
    Nov 21 '18 at 21:10
















-1















i am looking for a solution:



$txt = "Welcome to ".$place." City!";

$place = "New York";

echo $txt; // should be "Welcome to New York City!"


The variable "$place" can only be declared after the variable "$txt".
Is that possible?



Thanks






php variables echo







share|improve this question























  • When $txt is set, it is concatenating "Welcome to ", the value of $place, and " City!" to form a new string. So no, you cannot. Why do you want to do this?

    – Jordan S
    Nov 21 '18 at 19:45








  • 2





    Can you explain the goal here, your example isn't possible, but I bet whatever you actually trying to achieve is

    – Dan
    Nov 21 '18 at 19:48











  • You can't define a variable before using it.

    – Funk Forty Niner
    Nov 21 '18 at 21:10














-1












-1








-1








i am looking for a solution:



$txt = "Welcome to ".$place." City!";

$place = "New York";

echo $txt; // should be "Welcome to New York City!"


The variable "$place" can only be declared after the variable "$txt".
Is that possible?



Thanks






php variables echo







share|improve this question














i am looking for a solution:



$txt = "Welcome to ".$place." City!";

$place = "New York";

echo $txt; // should be "Welcome to New York City!"


The variable "$place" can only be declared after the variable "$txt".
Is that possible?



Thanks






php variables echo




php variables echo






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 21 '18 at 19:42









SteveO24SteveO24

11




11













  • When $txt is set, it is concatenating "Welcome to ", the value of $place, and " City!" to form a new string. So no, you cannot. Why do you want to do this?

    – Jordan S
    Nov 21 '18 at 19:45








  • 2





    Can you explain the goal here, your example isn't possible, but I bet whatever you actually trying to achieve is

    – Dan
    Nov 21 '18 at 19:48











  • You can't define a variable before using it.

    – Funk Forty Niner
    Nov 21 '18 at 21:10



















  • When $txt is set, it is concatenating "Welcome to ", the value of $place, and " City!" to form a new string. So no, you cannot. Why do you want to do this?

    – Jordan S
    Nov 21 '18 at 19:45








  • 2





    Can you explain the goal here, your example isn't possible, but I bet whatever you actually trying to achieve is

    – Dan
    Nov 21 '18 at 19:48











  • You can't define a variable before using it.

    – Funk Forty Niner
    Nov 21 '18 at 21:10

















When $txt is set, it is concatenating "Welcome to ", the value of $place, and " City!" to form a new string. So no, you cannot. Why do you want to do this?

– Jordan S
Nov 21 '18 at 19:45







When $txt is set, it is concatenating "Welcome to ", the value of $place, and " City!" to form a new string. So no, you cannot. Why do you want to do this?

– Jordan S
Nov 21 '18 at 19:45






2




2





Can you explain the goal here, your example isn't possible, but I bet whatever you actually trying to achieve is

– Dan
Nov 21 '18 at 19:48





Can you explain the goal here, your example isn't possible, but I bet whatever you actually trying to achieve is

– Dan
Nov 21 '18 at 19:48













You can't define a variable before using it.

– Funk Forty Niner
Nov 21 '18 at 21:10





You can't define a variable before using it.

– Funk Forty Niner
Nov 21 '18 at 21:10












3 Answers
3






active

oldest

votes


















1














It's unclear what you're trying to accomplish. But this is pretty close to your example and works as expected:



$txt = "Welcome to %s City!";

$place = "New York";

printf($txt, $place); // should be "Welcome to New York City!"


https://3v4l.org/jEQvZ






share|improve this answer































    0














    You can't do that directly. The string is concatenated using the value of $place at the time of assignment so any changes to $place will not be reflected in the string. You can emulate something like this by writing a function that will generate the string once you know the place it should reference.



    function welcome($place) {
    
    return 'Welcome to '.$place.' City!';
    
}
    

    
$place = 'New York';
    

    
echo welcome($place);





    share|improve this answer































      0














      Yes it's possible.

      Just create a placeholder and str_replace it.

      Notice I changed " to '.

      If you use " it will read $place as a variable. With ' $place is string and can be replaced later in the code.



      $txt = 'Welcome to $place City!';
      
$place = "New York";
      
echo str_replace('$place', $place, $txt);


      https://3v4l.org/DVtQa






      share|improve this answer
























      • Thank you! It works fine!

        – SteveO24
        Nov 21 '18 at 19:57






      • 1





        @SteveO24 no problem. Feel free to accept the answer that you find the best solution. (Not implying that it's mine, they are all very similar)

        – Andreas
        Nov 21 '18 at 20:00











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      It's unclear what you're trying to accomplish. But this is pretty close to your example and works as expected:



      $txt = "Welcome to %s City!";
      
$place = "New York";
      
printf($txt, $place); // should be "Welcome to New York City!"


      https://3v4l.org/jEQvZ






      share|improve this answer




























        1














        It's unclear what you're trying to accomplish. But this is pretty close to your example and works as expected:



        $txt = "Welcome to %s City!";
        
$place = "New York";
        
printf($txt, $place); // should be "Welcome to New York City!"


        https://3v4l.org/jEQvZ






        share|improve this answer


























          1












          1








          1







          It's unclear what you're trying to accomplish. But this is pretty close to your example and works as expected:



          $txt = "Welcome to %s City!";
          
$place = "New York";
          
printf($txt, $place); // should be "Welcome to New York City!"


          https://3v4l.org/jEQvZ






          share|improve this answer













          It's unclear what you're trying to accomplish. But this is pretty close to your example and works as expected:



          $txt = "Welcome to %s City!";
          
$place = "New York";
          
printf($txt, $place); // should be "Welcome to New York City!"


          https://3v4l.org/jEQvZ







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 21 '18 at 19:55









          DanDan

          5,44531432




          5,44531432

























              0














              You can't do that directly. The string is concatenated using the value of $place at the time of assignment so any changes to $place will not be reflected in the string. You can emulate something like this by writing a function that will generate the string once you know the place it should reference.



              function welcome($place) {
              
    return 'Welcome to '.$place.' City!';
              
}
              

              
$place = 'New York';
              

              
echo welcome($place);





              share|improve this answer




























                0














                You can't do that directly. The string is concatenated using the value of $place at the time of assignment so any changes to $place will not be reflected in the string. You can emulate something like this by writing a function that will generate the string once you know the place it should reference.



                function welcome($place) {
                
    return 'Welcome to '.$place.' City!';
                
}
                

                
$place = 'New York';
                

                
echo welcome($place);





                share|improve this answer


























                  0












                  0








                  0







                  You can't do that directly. The string is concatenated using the value of $place at the time of assignment so any changes to $place will not be reflected in the string. You can emulate something like this by writing a function that will generate the string once you know the place it should reference.



                  function welcome($place) {
                  
    return 'Welcome to '.$place.' City!';
                  
}
                  

                  
$place = 'New York';
                  

                  
echo welcome($place);





                  share|improve this answer













                  You can't do that directly. The string is concatenated using the value of $place at the time of assignment so any changes to $place will not be reflected in the string. You can emulate something like this by writing a function that will generate the string once you know the place it should reference.



                  function welcome($place) {
                  
    return 'Welcome to '.$place.' City!';
                  
}
                  

                  
$place = 'New York';
                  

                  
echo welcome($place);






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 21 '18 at 19:49









                  jfadichjfadich

                  3,4381723




                  3,4381723























                      0














                      Yes it's possible.

                      Just create a placeholder and str_replace it.

                      Notice I changed " to '.

                      If you use " it will read $place as a variable. With ' $place is string and can be replaced later in the code.



                      $txt = 'Welcome to $place City!';
                      
$place = "New York";
                      
echo str_replace('$place', $place, $txt);


                      https://3v4l.org/DVtQa






                      share|improve this answer
























                      • Thank you! It works fine!

                        – SteveO24
                        Nov 21 '18 at 19:57






                      • 1





                        @SteveO24 no problem. Feel free to accept the answer that you find the best solution. (Not implying that it's mine, they are all very similar)

                        – Andreas
                        Nov 21 '18 at 20:00
















                      0














                      Yes it's possible.

                      Just create a placeholder and str_replace it.

                      Notice I changed " to '.

                      If you use " it will read $place as a variable. With ' $place is string and can be replaced later in the code.



                      $txt = 'Welcome to $place City!';
                      
$place = "New York";
                      
echo str_replace('$place', $place, $txt);


                      https://3v4l.org/DVtQa






                      share|improve this answer
























                      • Thank you! It works fine!

                        – SteveO24
                        Nov 21 '18 at 19:57






                      • 1





                        @SteveO24 no problem. Feel free to accept the answer that you find the best solution. (Not implying that it's mine, they are all very similar)

                        – Andreas
                        Nov 21 '18 at 20:00














                      0












                      0








                      0







                      Yes it's possible.

                      Just create a placeholder and str_replace it.

                      Notice I changed " to '.

                      If you use " it will read $place as a variable. With ' $place is string and can be replaced later in the code.



                      $txt = 'Welcome to $place City!';
                      
$place = "New York";
                      
echo str_replace('$place', $place, $txt);


                      https://3v4l.org/DVtQa






                      share|improve this answer













                      Yes it's possible.

                      Just create a placeholder and str_replace it.

                      Notice I changed " to '.

                      If you use " it will read $place as a variable. With ' $place is string and can be replaced later in the code.



                      $txt = 'Welcome to $place City!';
                      
$place = "New York";
                      
echo str_replace('$place', $place, $txt);


                      https://3v4l.org/DVtQa







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Nov 21 '18 at 19:49









                      AndreasAndreas

                      16.5k41744




                      16.5k41744













                      • Thank you! It works fine!

                        – SteveO24
                        Nov 21 '18 at 19:57






                      • 1





                        @SteveO24 no problem. Feel free to accept the answer that you find the best solution. (Not implying that it's mine, they are all very similar)

                        – Andreas
                        Nov 21 '18 at 20:00



















                      • Thank you! It works fine!

                        – SteveO24
                        Nov 21 '18 at 19:57






                      • 1





                        @SteveO24 no problem. Feel free to accept the answer that you find the best solution. (Not implying that it's mine, they are all very similar)

                        – Andreas
                        Nov 21 '18 at 20:00

















                      Thank you! It works fine!

                      – SteveO24
                      Nov 21 '18 at 19:57





                      Thank you! It works fine!

                      – SteveO24
                      Nov 21 '18 at 19:57




                      1




                      1





                      @SteveO24 no problem. Feel free to accept the answer that you find the best solution. (Not implying that it's mine, they are all very similar)

                      – Andreas
                      Nov 21 '18 at 20:00





                      @SteveO24 no problem. Feel free to accept the answer that you find the best solution. (Not implying that it's mine, they are all very similar)

                      – Andreas
                      Nov 21 '18 at 20:00


















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