How to solve Notice: Undefined index: id in C:\xampp\htdocs\invmgt\manufactured_goods\change.php on line 21

Question

I have a problem with my PHP code saying that "Notice: Undefined index" I am sure its very simple, since I am a beginner i am not getting well what is wrong exactly so please help me.

Here's the code

<?php require_once('../Connections/itemconn.php'); ?>
    <?php   

    $id=$_GET['id']; 

        $query=mysql_query("select * from manuf where id='$id' ")or die(mysql_error());
        $row=mysql_fetch_array($query);

        ?>

<form action="updateprice.php" method="post" enctype="multipart/form-data">
  <table align="center">
   <tr>
   <td> <label><strong>Item Name</strong></label></td>
     <td> <label> <?php echo $row['itemname']; ?></label><input type="hidden" name="id" value="<?php echo $id; ?> " />
     <br /></td>
    </tr>
    <tr>

     <td><label><strong>Unit price </strong></label></td>
  <td> <input type="text" name="pass" value="<?php echo $row['unitprice']; ?> " /><br /></td>
  </tr>

    <tr>
  <td> 
          <input type="reset" name="Reset" value="CANCEL" />
      <br></td>


     <td> 
          <input type="submit" name="Submit2" value="Update" />      </td>
   </tr>
</table>
</form>
</body>
</html>

You are using $_GET['id'] but there is no ID in your URL. Are you sure you're not mixing up $_POST and $_GET ? — Matheno, Oct 31 '14 at 08:42

Pupil · Accepted Answer · 2017-01-16T03:49:56.867

15

You are not getting value of $id=$_GET['id'];

And you are using it (before it gets initialised).

Use php's in built isset() function to check whether the variable is defied or not.

So, please update the line to:

$id = isset($_GET['id']) ? $_GET['id'] : '';

edited Jan 16 '17 at 03:49

answered Oct 31 '14 at 08:41

Pupil

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for me removing form action did the thing, its changed the link before query ran so i removed the action from form. – Abubakar Soomro Feb 18 '21 at 17:58

Rakesh Sharma · Answer 2 · 2014-10-31T08:52:16.917

5

if you are getting id from url try

$id = (isset($_GET['id']) ? $_GET['id'] : '');

if getting from form you need to use POST method cause your form has method="post"

 $id = (isset($_POST['id']) ? $_POST['id'] : '');

For php notices use isset() or empty() to check values exist or not or initialize variable first with blank or a value

$id= '';

edited Oct 31 '14 at 08:52

answered Oct 31 '14 at 08:45

Rakesh Sharma

13,680
5
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44

yeah updated with isset() also – Rakesh Sharma Oct 31 '14 at 08:49
No, remove the `empty` from your answer (see link at bottom of my answer as to why you shouldn't use `empty`) – Elias Van Ootegem Oct 31 '14 at 08:51

score 3 · Answer 3 · answered Oct 31 '14 at 08:48

3

Simply add this

$id = ''; 
if( isset( $_GET['id'])) {
    $id = $_GET['id']; 
}

answered Oct 31 '14 at 08:48

ajitpawarink

436
6
14

How to solve Notice: Undefined index: id in C:\xampp\htdocs\invmgt\manufactured_goods\change.php on line 21

3 Answers3

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