Reputation: 129
I'm trying to create an update function in PHP but the records don't seem to be changing as per the update. I've created a JSON object to hold the values being passed over to this file and according to the Firebug Lite console I've running these values are outputted just fine so it's prob something wrong with the sql side. Can anyone spot a problem? I'd appreciate the help!
<?php
$var1 = $_REQUEST['action']; // We dont need action for this tutorial, but in a complex code you need a way to determine ajax action nature
$jsonObject = json_decode($_REQUEST['outputJSON']); // Decode JSON object into readable PHP object
$name = $jsonObject->{'name'}; // Get name from object
$desc = $jsonObject->{'desc'}; // Get desc from object
$did = $jsonObject->{'did'};// Get id object
mysql_connect("localhost","root",""); // Conect to mysql, first parameter is location, second is mysql username and a third one is a mysql password
@mysql_select_db("findadeal") or die( "Unable to select database"); // Connect to database called test
$query = "UPDATE deal SET dname = {'$name'}, desc={'$desc'} WHERE dealid = {'$did'}";
$add = mysql_query($query);
$num = mysql_num_rows($add);
if($num != 0) {
echo "true";
} else {
echo "false";
}
?>
Upvotes: 0
Views: 823
Reputation: 4449
I believe you are misusing the curly braces. The single quote should go on the outside of them.:
"UPDATE deal SET dname = {'$name'}, desc={'$desc'} WHERE dealid = {'$did'}"
Becomes
"UPDATE deal SET dname = '{$name}', desc='{$desc}' WHERE dealid = '{$did}'"
On a side note, using any mysql_* functions isn't really good security-wise. I would recommend looking into php's mysqli or pdo extensions.
Upvotes: 1
Reputation:
you need to use mysql_affected_rows()
after update not mysql_num_rows
Upvotes: 1
Reputation: 204894
You need to escape reserved words in MySQL like desc
with backticks
UPDATE deal
SET dname = {'$name'}, `desc`= {'$desc'} ....
^----^--------------------------here
Upvotes: 1