current community

your communities

more stack exchange communities

Stack Exchange
tour help
stack overflow careers
Take the 2-minute tour ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.
up vote 0 down vote favorite
1

I am running a script to update a MySQL table, and I have been echoing the update statement to help troubleshoot why the update statement was not working. When I copy and paste the echoed query for the mysql update into the SQL menu in phpmyadmin, it works perfectly, but when I run the script the update query does not execute. I know I have successfully connected to the database. Does anyone know why an echoed version of the update statement copy and pasted into the sql tab of phpmyadmin would execute successfully, but not the actual query that is supposed to be executed the line before the echoed statement? Here is my php: 

<!DOCTYPE html>
<html>
<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type">
<title>Untitled 1</title>
</head>
<body>
<?php
$user = 'tunemashercom';
$pass = 'password';
$host = 'tunemashercom.ipagemysql.com';
$db_name = 'user_library';
@ $db = mysqli_connect($host, $user, $pass, $db_name);  

if (mysqli_connect_errno()) {
    echo 'Error: Could not connect to database.';
    exit;
}

$oldsongs = $_POST['oldsongs'];
$newsongs = $_POST['newsongs'];
$count = count($newsongs);
$count2 = count($oldsongs);
echo "COUNT 1: ".$count." - COUNT 2: ".$count2."<br />";
for($i=0;$i<=$count;$i++){
    $newtitle = $newsongs[$i]["song"];
    echo $newtitle." - ";
    if($newtitle == ""){
        echo "NO NEW TAGS<br />";
    }
    else{
        $title = $oldsongs[$i]["title"];
        echo $title."<br />";
        $artist = $oldsongs[$i]["artist"];
        $album = $oldsongs[$i]["album"];
        $newartist = $newsongs[$i]["artist"];
        $newalbum = $newsongs[$i]["album"];
        $mbid = $newsongs[$i]["mbid"];          
        $insert = "UPDATE collection SET song='".$newtitle."', artist='".$newartist."', album='".$newalbum."', gille_id='".$mbid."' WHERE song='".$title."' AND artist='".$artist."' AND album='".$album."'";
        $results = $db->query($insert);         
        echo $insert."<br />";      
    }
}

mysqli_close($db);

?>
</body>
</html>
share|improve this question
3  
What error does it show when you try the update from the PHP side of things? Need a little more info here. –  AreYouSure Feb 8 '13 at 1:27
    
There where no errors from the PHP side of things. The script runs through perfectly and echoes the update statement for every iteration of the for loop, but when I check the database, none of the queries were actually executed. But when I copy and paste the statement that was echoed (the same update query) into the SQL form in phpmyadmin, the update query executes exactly as I wanted. –  jgille07 Feb 8 '13 at 15:17

2 Answers 2

up vote 0 down vote

Echo out every variable just before you run the query, for example in your code:

 else{
    $title = $oldsongs[$i]["title"];
    echo $title."<br />";
    $artist = $oldsongs[$i]["artist"];
    $album = $oldsongs[$i]["album"];
    $newartist = $newsongs[$i]["artist"];
    $newalbum = $newsongs[$i]["album"];
    $mbid = $newsongs[$i]["mbid"];          

    //echo all element after this:-
    echo '<br><br>';
    echo $newtitle ;  //echoing elements
     echo '<br><br>'; // in order to keep things clear
    echo $newartist;  //echoing elements
       echo '<br><br>';
    $insert = "UPDATE collection SET song='".$newtitle."', artist='".$newartist."', album='".$newalbum."', gille_id='".$mbid."' WHERE song='".$title."' AND artist='".$artist."' AND album='".$album."'";
    $results = $db->query($insert);         
    echo $insert."<br />";      
}
share|improve this answer
up vote 0 down vote

Without seeing the error it's difficult to know what the problem it, however, from my personal experience when that's happened to me on UPDATE queries it was always because I wrapped an ID with quotes.

Hopefully that helps.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.