How to solve Notice Undefined index id in C xampp htdocs invmgt manufactured goods change php on line 21

0 votes

I am having an error in the PHP code saying "Notice: Undefined index".

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>

As I am a beginner I cant solve this issue. Can someone help me with this?

Apr 30 in Other DevOps Questions by Kichu
• 19,040 points
2,077 views

1 answer to this question.

0 votes

The issue is that you are using the git id before it gets initialized. To solve this, use PHP's inbuilt isset() function to check whether the variable is defied or not.

So, update the line to:

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

This will remove the notice that you are facing. 

answered May 1 by narikkadan
• 37,660 points

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