PHP update MYSQL Form not working

Question

Im trying to edit the lastname (lname) but its not working ,

im getting this error :

ERROR: Could not able to execute UPDATE tablename SET fname = '', lname = '' WHERE fname = . You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

<?php
$link = mysqli_connect("IP","DB","PASS (hiden ofc)", "DBN");

// Check connection
if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

// Attempt update query execution
$sql = "UPDATE tablename SET fname = '$nfname', lname = '$nlname' WHERE fname = $fname";
if(mysqli_query($link, $sql)){
    echo "Records were updated successfully.";
} else {
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// Close connection
mysqli_close($link);
?>

--- HTML CODE ---

<html>
<body>
<h1>Test editing </h1>
<form action="edit.php" method="post">
OrginalFirstname: <input type="text" name="fname" /><br><br>
Lastname: <input type="text" name="lname" /><br><br>
NewFirstname: <input type="text" name="nfname" /><br><br>
 
<input type="submit" />
</form>
</body>
</html>

Why are you updating all the fields if you only want to edit the last name? Where are those variables assigned? — chris85
– chris85, Commented Dec 25, 2015 at 22:47
Only update fields that need to be updated. You also should use parameterized queries. — chris85
– chris85, Commented Dec 26, 2015 at 0:23

Phiter · Accepted Answer · 2015-12-26 00:40:23Z

4

You forgot to put apostrophes on the last part of your query:

$sql = "UPDATE tablename SET fname = '$nfname', lname = '$nlname' WHERE fname = '$fname'";

This should work.

Make sure you escaped all the variables with mysqli_real_escape_string. If one of the variables has a non-escaped apostrophe, the query will fail again.

If the PHP code in your question is the entire code, then you are not getting the values from the $_POST[].

You can get the values into your variables using extract($_POST); on the beggining of your code.

edited Dec 26, 2015 at 0:40

answered Dec 25, 2015 at 22:16

Phiter

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2 Comments

Tᴀʀᴇǫ Mᴀʜᴍᴏᴏᴅ Over a year ago

I think that's the issue he isn't assigning the variables which he is using.

Jonas Over a year ago

Aaaah i didnt know i had to add it all seems to be working currently Thank you very much! and sorry for my mistakes

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