Undefined variable row even on accessing it with index values

Question

I am getting the data from db and updating it here and trying to sending back to db. The error is undefined variable row.I am accessing with index values instead of their names.

<!DOCTYPE HTML>
<?php
    include_once('DBConnection.php');

    if( isset($_GET['edit']) )
    {
        $id = $_GET['edit'];
        $res= mysql_query("SELECT * FROM details WHERE id='$id'");
        $row= mysql_fetch_array($res);
    }

    if( !empty($_POST['newName'])&& isset($_POST['newName'])&& !empty($_POST['email'])&& isset($_POST['email'])&&!empty($_POST['phonenumber'])&& isset($_POST['phonenumber']))
    {
        $newName = $_POST['newName'];
        $EMail = $_POST['email'];
        $PhoneNumber = $_POST['phonenumber'];
        $id      = $_POST['id'];
        $sql     = "UPDATE details SET name='$newName' email='$EMail' phonenumber='$PhoneNumber' WHERE id='$id'";
        $res     = mysql_query($sql) 
                                    or die("Could not update".mysql_error());
        echo "<meta http-equiv='refresh' content='0;url=random.php'>";
    }

?>
<form action="edit.php" method="POST">
PhoneNumber: <input type="text" name="phonenumber" value="<?php echo $row[3];?>"/><br />
EMail: <input type="text" name="email" value="<?php echo $row[2];?>"/><br />
Name: <input type="text" name="newName" value="<?php echo $row[1];?>"/><br />
SlNo<input type="hidden" name="SlNo" value="<?php echo $row[0];?>"/>
<input type="submit" value=" Update "/>
</form>

If your GET does not contain `edit`, then no `$row` will be defined, that's where the error source is from — Justinas, Oct 29 '18 at 14:17
1. $row need to be set with default values, 2. do not use mysql_query use mysqli_query — talsibony, Oct 29 '18 at 14:19

score 0 · Answer 1 · answered Oct 29 '18 at 14:18

0

You haven't done a SELECT query and executed it in your second if. There is no $row variable defined, yet you talk about it in your form.

answered Oct 29 '18 at 14:18

delboy1978uk

12,118
2
21
39

score 0 · Accepted Answer · answered Oct 29 '18 at 14:22

In this instance, you need to be sure $row is defined before using it. Try this:

<!DOCTYPE HTML>
<?php
    include_once('DBConnection.php');

    if( isset($_GET['edit']) )
    {
        $id = $_GET['edit'];
        $res= mysql_query("SELECT * FROM details WHERE id='$id'");
        $row= mysql_fetch_array($res);
    }

    if( !empty($_POST['newName'])&& isset($_POST['newName'])&& !empty($_POST['email'])&& isset($_POST['email'])&&!empty($_POST['phonenumber'])&& isset($_POST['phonenumber']))
    {
        $newName = $_POST['newName'];
        $EMail = $_POST['email'];
        $PhoneNumber = $_POST['phonenumber'];
        $id      = $_POST['id'];
        $sql     = "UPDATE details SET name='$newName' email='$EMail' phonenumber='$PhoneNumber' WHERE id='$id'";
        $res     = mysql_query($sql) 
                                    or die("Could not update".mysql_error());
        echo "<meta http-equiv='refresh' content='0;url=random.php'>";
    }

?>
<form action="edit.php" method="POST">
PhoneNumber: <input type="text" name="phonenumber" value="<?php echo (isset($row)? $row[3]: "");?>"/><br />
EMail: <input type="text" name="email" value="<?php echo (isset($row)? $row[2]: "");?>"/><br />
Name: <input type="text" name="newName" value="<?php echo (isset($row)? $row[1]: "");?>"/><br />
SlNo<input type="hidden" name="SlNo" value="<?php echo (isset($row)? $row[0]: "");?>"/>
<input type="submit" value=" Update "/>
</form>

score 0 · Answer 3 · answered Oct 29 '18 at 14:24

The query() method shall return FALSE on fail, which is a boolean, which has no rows...

Before trying to access the result rows, check if it is an array :

$res = mysql_query($sql);
if (!is_array($res)) {
    return ('error');
}
$row = mysql_fetch_array($res);

and also make sure $row exists, as Elisha said.

Also to concatenate strings and variables : PHP - concatenate or directly insert variables in string (Note that you method perhaps works, I don't know)

Undefined variable row even on accessing it with index values

3 Answers3