hi all! i have here a code to delete a record

<?php

    define ('DB_NAME', 'vincegac_vince');
    define ('DB_USER', 'root');
    define ('DB_PASSWORD', '');
    define ('DB_HOST', 'localhost');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!$link) {
    die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
    die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}
    $value1 = $_POST["emp_number"];
    $value2 = $_POST["fname"];
    $value3 = $_POST["mname"];
    $value4 = $_POST["lname"];
    $value5 = $_POST["homeadd"];
    $value6 = $_POST["emailadd"];
    $value7 = $_POST["ofcnum"];
    $value8 = $_POST["mobilenum"];
    $value9 = $_POST["homenum"];
    $value10 = $_POST["position"];
    $value11 = $_POST["practice"];
    $value12 = $_POST["projname"];
    $value13 = $_POST["projmgr"];
    $value14 = $_POST["teamlead"];

$sql = "DELETE FROM employee WHERE emp_number = '$value1' AND fname = '$value2' AND mname = '$value3' AND lname = '$value4' AND homeadd = '$value5' AND emailadd = '$value6' AND ofcnum = '$value7' AND mobilenum = '$value8' 
        AND homenum = '$value9' AND position = '$value10' AND practice = '$value11' AND projname = '$valu12' AND projmgr = '$value13' AND teamlead = '$value14'";

if (!mysql_query($sql)) {
    die('Error: ' . mysql_error());
    }
//echo '<p>One member added! <a href="demo-form.php">add another member</a></p>';
echo '<p>Back to <a href="main.php">Home page</a></p>';
mysql_close();

?>

but when i click my delete button, it says, undefined index. im using post method.

Thanks!

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Will u share click button HTML code too? Are you using JS also? If means, share that tooo.. :)

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Will u share click button HTML code too? Are you using JS also? If means, share that tooo.. :)

The query have projname = '$valu12'
it should be projname = '$value12'

You can use a unique id to delete a row

Yep, as code_rum has suggested the problem is likely to be caused by a mispelling of a variable.

In the future, a tip to debugging such errors is to either echo out the variables when you get an error (and see if it is null, or has the expected value), or to use the isset() function.

hi guys! yes i saw that one. but thats not the cause. my button was not inside the form. :)

THANKS! :)

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