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I am encountering an "Undefined array key" error message in php

New here and to php. I wrote a code to toggle between two images and changing a letter to two colors when two buttons are pressed. It is working but I keep getting the following error messages:
Warning: Undefined array key "button2" in C:xampphtdocsphpcoursepractice3.php on line 11 (Shows when button1 is clicked)
Warning: Undefined array key "button1" in C:xampphtdocsphpcoursepractice3.php on line 10 (Shows when button2 is clicked)

Any help on how to resolve this?? Below is my code:

   <?php
   $btn1='';
   $btn2='';
   $result = "img.png";
   $style= "blue";

   if($_SERVER['REQUEST_METHOD']==='POST'){
      $btn1 = $_POST['button1'];
      $btn2 = $_POST['button2'];


       if(isset($btn1)){
           $result = "img.png";
           $style = "blue";
       }
       else {
           $result = "img2.png";
           $style = "red";
       }

   }



   ?>
   <?php include "head.php"
   ?>
   <body>

   <form action="" method="post">
   <button type="submit" name="button1">Button 1</button>
   <button type="submit" name="button2">Button 2</button>
   </form>
   <img src="<?php echo $result ?>">
   <h2 style="color:<?php echo $style?>">B</h2>
   </body>
   </html>

Tried everything to no avail.

Tags:

2

Answers


  1. 0

    You have two submit buttons in your form. When you click a button the browser will post a value for that button only. No value will be posted for the other button. Thus, your $_POST array doesn’t include a key for the second button. It makes no difference which button you click: the other button will be omitted.

    The way to test for a specific button is to use if isset().

    For example:

    
If (isset($_POST['button1'])){
  // Do something
}

If (isset($_POST['button2'])){
  // Do something else
}

    For your code try

    <?php
   $btn1='';
   $btn2='';
   $result = "img.png";
   $style= "blue";

   if($_SERVER['REQUEST_METHOD']==='POST'){
      
      // Don't need these
      //$btn1 = $_POST['button1'];
      //$btn2 = $_POST['button2'];


       if(isset($_POST['button1'])){
           $result = "img.png";
           $style = "blue";
       }
       else {
           $result = "img2.png";
           $style = "red";
       }

   }
    Login or Signup to reply.
  2. 0

    Only the button that’s used to submit the form gets sent to the server. So when you use button1, $_POST['button2'] is undefined.

    Use the same name for your submit buttons, and distinguish them by value instead of by name.

    <?php
$btn1='';
$btn2='';
$result = "img.png";
$style= "blue";

if($_SERVER['REQUEST_METHOD']==='POST'){
    $button_value = $_POST['button'];
    switch($button_value) {
    case "1":
        $result = "img.png";
        $style = "blue";
        break;
   case "2":
        $result = "img2.png";
        $style = "red";
        break;
   }
}

?>
<?php include "head.php"
?>
<body>

<form action="" method="post">
<button type="submit" name="button" value="1">Button 1</button>
<button type="submit" name="button" value="2">Button 2</button>
</form>
<img src="<?php echo $result ?>">
<h2 style="color:<?php echo $style?>">B</h2>
</body>
</html>
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