undefined variable in php while giving sql values

up vote
-3
down vote

favorite

while inserting values into table its showing error

the error is like this

Notice: Undefined variable: pname in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: price in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: tax in /opt/lampp/htdocs/zubair/index.php
on line 24

Notice: Undefined variable: image in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: manufacturer in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: address in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: city in /opt/lampp/htdocs/zubair/index.php
on line 24

Notice: Undefined variable: state in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: country in
/opt/lampp/htdocs/zubair/index.php on line 24 Duplicate entry '0' for
key 'PRIMARY

<?

   

if(isset($_POST) & !empty($_POST)){

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);

  

}





$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }



         











?>

edited Nov 9 at 8:26

Sfili_81

383314

asked Nov 9 at 8:04

Zubair Nazer Oliyat

you are not passing any data in POST
– guradio
Nov 9 at 8:05

1

Possible duplicate of "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset" using PHP
– Tomasz
Nov 9 at 8:06

@guradio can u show me as an answer
– Zubair Nazer Oliyat
Nov 9 at 8:07

You have to set an else condition when you have empty fields. P.s. your code is open to SQL Injection
– Sfili_81
Nov 9 at 8:11

You are using & operator. Should be if(isset($_POST) && !empty($_POST)) otherwise $_POST is empty
– ivion
Nov 9 at 8:11

|
show 3 more comments

up vote
-3
down vote

favorite

while inserting values into table its showing error

the error is like this

Notice: Undefined variable: pname in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: price in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: tax in /opt/lampp/htdocs/zubair/index.php
on line 24

Notice: Undefined variable: image in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: manufacturer in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: address in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: city in /opt/lampp/htdocs/zubair/index.php
on line 24

Notice: Undefined variable: state in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: country in
/opt/lampp/htdocs/zubair/index.php on line 24 Duplicate entry '0' for
key 'PRIMARY

<?

   

if(isset($_POST) & !empty($_POST)){

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);

  

}





$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }



         











?>

edited Nov 9 at 8:26

Sfili_81

383314

asked Nov 9 at 8:04

Zubair Nazer Oliyat

you are not passing any data in POST
– guradio
Nov 9 at 8:05

1

Possible duplicate of "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset" using PHP
– Tomasz
Nov 9 at 8:06

@guradio can u show me as an answer
– Zubair Nazer Oliyat
Nov 9 at 8:07

You have to set an else condition when you have empty fields. P.s. your code is open to SQL Injection
– Sfili_81
Nov 9 at 8:11

You are using & operator. Should be if(isset($_POST) && !empty($_POST)) otherwise $_POST is empty
– ivion
Nov 9 at 8:11

|
show 3 more comments

up vote
-3
down vote

favorite

while inserting values into table its showing error

the error is like this

Notice: Undefined variable: pname in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: price in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: tax in /opt/lampp/htdocs/zubair/index.php
on line 24

Notice: Undefined variable: image in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: manufacturer in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: address in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: city in /opt/lampp/htdocs/zubair/index.php
on line 24

Notice: Undefined variable: state in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: country in
/opt/lampp/htdocs/zubair/index.php on line 24 Duplicate entry '0' for
key 'PRIMARY

<?

   

if(isset($_POST) & !empty($_POST)){

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);

  

}





$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }



         











?>

edited Nov 9 at 8:26

Sfili_81

383314

asked Nov 9 at 8:04

Zubair Nazer Oliyat

while inserting values into table its showing error

the error is like this

Notice: Undefined variable: pname in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: price in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: tax in /opt/lampp/htdocs/zubair/index.php
on line 24

Notice: Undefined variable: image in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: manufacturer in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: address in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: city in /opt/lampp/htdocs/zubair/index.php
on line 24

Notice: Undefined variable: state in
/opt/lampp/htdocs/zubair/index.php on line 24

Notice: Undefined variable: country in
/opt/lampp/htdocs/zubair/index.php on line 24 Duplicate entry '0' for
key 'PRIMARY

<?

   

if(isset($_POST) & !empty($_POST)){

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);

  

}





$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }



         











?>

<?

   

if(isset($_POST) & !empty($_POST)){

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);

  

}





$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }



         











?>

<?

   

if(isset($_POST) & !empty($_POST)){

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);

  

}





$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }



         











?>

php sql

edited Nov 9 at 8:26

Sfili_81

383314

asked Nov 9 at 8:04

Zubair Nazer Oliyat

edited Nov 9 at 8:26

Sfili_81

383314

asked Nov 9 at 8:04

Zubair Nazer Oliyat

edited Nov 9 at 8:26

Sfili_81

383314

edited Nov 9 at 8:26

Sfili_81

383314

edited Nov 9 at 8:26

Sfili_81

383314

asked Nov 9 at 8:04

Zubair Nazer Oliyat

asked Nov 9 at 8:04

Zubair Nazer Oliyat

asked Nov 9 at 8:04

Zubair Nazer Oliyat

you are not passing any data in POST
– guradio
Nov 9 at 8:05

1

Possible duplicate of "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset" using PHP
– Tomasz
Nov 9 at 8:06

@guradio can u show me as an answer
– Zubair Nazer Oliyat
Nov 9 at 8:07

You have to set an else condition when you have empty fields. P.s. your code is open to SQL Injection
– Sfili_81
Nov 9 at 8:11

You are using & operator. Should be if(isset($_POST) && !empty($_POST)) otherwise $_POST is empty
– ivion
Nov 9 at 8:11

|
show 3 more comments

you are not passing any data in POST
– guradio
Nov 9 at 8:05

1

Possible duplicate of "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset" using PHP
– Tomasz
Nov 9 at 8:06

@guradio can u show me as an answer
– Zubair Nazer Oliyat
Nov 9 at 8:07

You have to set an else condition when you have empty fields. P.s. your code is open to SQL Injection
– Sfili_81
Nov 9 at 8:11

You are using & operator. Should be if(isset($_POST) && !empty($_POST)) otherwise $_POST is empty
– ivion
Nov 9 at 8:11

you are not passing any data in POST
– guradio
Nov 9 at 8:05

Possible duplicate of "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset" using PHP
– Tomasz
Nov 9 at 8:06

@guradio can u show me as an answer
– Zubair Nazer Oliyat
Nov 9 at 8:07

You have to set an else condition when you have empty fields. P.s. your code is open to SQL Injection
– Sfili_81
Nov 9 at 8:11

You are using & operator. Should be if(isset($_POST) && !empty($_POST)) otherwise $_POST is empty
– ivion
Nov 9 at 8:11

|
show 3 more comments

2 Answers
2

active

oldest

votes

up vote
0
down vote

According to your script variables defined if is set and not empty the $_POST variable. if you not pass your data to the script via POST method that section will not run. Then you get all the variables as undefined.
Probably your form method should change to POST. check your html form method if you using html form to test this.

answered Nov 9 at 8:13

Thushara Saman

1316

add a comment |

up vote
0
down vote

You are using the & operator. Use &&

<?

   

if(isset($_POST) && !empty($_POST)){  //<-- use the && operator

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);



$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }

  

}



?>

otherwise your $_POST variable is empty and be aware of mysql injection

edited Nov 9 at 8:17

answered Nov 9 at 8:13

ivion

17325

its not working
– Zubair Nazer Oliyat
Nov 9 at 8:16

have you check if you receive the value ?
– Sfili_81
Nov 9 at 8:17

What is the error?
– ivion
Nov 9 at 8:17

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
0
down vote

answered Nov 9 at 8:13

Thushara Saman

1316

add a comment |

up vote
0
down vote

answered Nov 9 at 8:13

Thushara Saman

1316

add a comment |

up vote
0
down vote

answered Nov 9 at 8:13

Thushara Saman

1316

answered Nov 9 at 8:13

Thushara Saman

1316

answered Nov 9 at 8:13

Thushara Saman

1316

answered Nov 9 at 8:13

Thushara Saman

1316

answered Nov 9 at 8:13

Thushara Saman

1316

add a comment |

up vote
0
down vote

You are using the & operator. Use &&

<?

   

if(isset($_POST) && !empty($_POST)){  //<-- use the && operator

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);



$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }

  

}



?>

otherwise your $_POST variable is empty and be aware of mysql injection

edited Nov 9 at 8:17

answered Nov 9 at 8:13

ivion

17325

its not working
– Zubair Nazer Oliyat
Nov 9 at 8:16

have you check if you receive the value ?
– Sfili_81
Nov 9 at 8:17

What is the error?
– ivion
Nov 9 at 8:17

add a comment |

up vote
0
down vote

You are using the & operator. Use &&

<?

   

if(isset($_POST) && !empty($_POST)){  //<-- use the && operator

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);



$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }

  

}



?>

otherwise your $_POST variable is empty and be aware of mysql injection

edited Nov 9 at 8:17

answered Nov 9 at 8:13

ivion

17325

its not working
– Zubair Nazer Oliyat
Nov 9 at 8:16

have you check if you receive the value ?
– Sfili_81
Nov 9 at 8:17

What is the error?
– ivion
Nov 9 at 8:17

add a comment |

up vote
0
down vote

You are using the & operator. Use &&

<?

   

if(isset($_POST) && !empty($_POST)){  //<-- use the && operator

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);



$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }

  

}



?>

otherwise your $_POST variable is empty and be aware of mysql injection

edited Nov 9 at 8:17

answered Nov 9 at 8:13

ivion

17325

You are using the & operator. Use &&

<?

   

if(isset($_POST) && !empty($_POST)){  //<-- use the && operator

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);



$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }

  

}



?>

otherwise your $_POST variable is empty and be aware of mysql injection

<?

   

if(isset($_POST) && !empty($_POST)){  //<-- use the && operator

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);



$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }

  

}



?>

<?

   

if(isset($_POST) && !empty($_POST)){  //<-- use the && operator

  $pname = mysql_real_escape_string($_POST['pname']);

  $price = mysql_real_escape_string($_POST['price']);

  $tax = mysql_real_escape_string($_POST['tax']);

  $image = mysql_real_escape_string($_POST['image']);

  $manufacturer = $_POST['manufacturer'];

  $address = mysql_real_escape_string($_POST['address']);

  $city = mysql_real_escape_string($_POST['city']);

  $state = mysql_real_escape_string($_POST['state']);

  $country = mysql_real_escape_string($_POST['country']);



$CreateSql = "INSERT INTO `products` (product_name, price, tax, image, manufacturer, address, city, state, country) VALUES ('$pname', '$price', '$tax', '$image', '$manufacturer', '$address', '$city', '$state', '$country')";



$connection = mysqli_connect("localhost","root","","product");

$res = mysqli_query($connection, $CreateSql) or die(mysqli_error($connection));



  if($res){

    $smsg = "Successfully inserted data, Insert New data.";

  }else{

    $fmsg = "Data not inserted, please try again later.";

  }

  

}



?>

edited Nov 9 at 8:17

answered Nov 9 at 8:13

ivion

17325

edited Nov 9 at 8:17

answered Nov 9 at 8:13

ivion

17325

answered Nov 9 at 8:13

ivion

17325

answered Nov 9 at 8:13

ivion

17325

its not working
– Zubair Nazer Oliyat
Nov 9 at 8:16

have you check if you receive the value ?
– Sfili_81
Nov 9 at 8:17

What is the error?
– ivion
Nov 9 at 8:17

add a comment |

its not working
– Zubair Nazer Oliyat
Nov 9 at 8:16

have you check if you receive the value ?
– Sfili_81
Nov 9 at 8:17

What is the error?
– ivion
Nov 9 at 8:17

its not working
– Zubair Nazer Oliyat
Nov 9 at 8:16

have you check if you receive the value ?
– Sfili_81
Nov 9 at 8:17

What is the error?
– ivion
Nov 9 at 8:17

add a comment |

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