MySQL/PHP update query not executing

Question

I tested the variables in the update statement and checked if a database connection is established, however the query doesn't run, can you please show me the error in my code.

for($i=0; $i <= $numcourses; $i++){
    echo '<div class="new'.$i.'" id="new'.$i.'"><label>'.$course_names[$i].'</label>
    <input name="edit'.$i.'" type="submit" value="Edit" /><input name="delete'.$i.'" type="submit" value="Delete" /><br /></div>';
    $name="edit".$i;    
    if (isset($_POST[$name])){
        echo '<input name="text" type="text" value="'.$course_names[$i].'" /><input name="save'.$i.'" type="submit" value="Save"/>';
    }   
    $name2="save".$i;
    if (isset($_POST[$name2])){
        include "includes/open.php";
            $newname=($_POST['text']);
            $int=$i+1;
            $query = "UPDATE course SET cname = '".$newname."' WHERE cid = '".$int."'";
            mysql_query($query) or die(mysql_error());
            include "includes/close.php";
    }
}

Update: Thanx Marc B, adding or die(mysql_error());showed me the error in my code, everything works again and I'm back on track.

Have you tried just running the query in the database (sans PHP)? What error message do you get...? — amaidment
– amaidment, Commented May 3, 2012 at 14:41
If you need any more info to help me solve this problem, please let me know. — grasshopper
– grasshopper, Commented May 3, 2012 at 14:42
Ok... so you are trying to update a record, while using an incremented value. Are you sure that those 'cid'-s match. — Mirko Akov
– Mirko Akov, Commented May 3, 2012 at 14:46
As you're just starting out, forget all that stuff about mysql_query and so on. Move to MySQLi and prepared statements etc. The mysql_* stuff is all deprecated. (Or will be soon) — Cylindric
– Cylindric, Commented May 3, 2012 at 14:46

Marc B · Accepted Answer · 2012-05-03 14:42:15Z

3

You have no error handling on your query calls:

mysql_query($query) or die(mysql_error());
                   ^^^^^^^^^^^^^^^^^^^^^^

which would tell you if there's any problems with the query execution. On a meta level, you're wide open to SQL injection attacks, so you'd better read up about that and fix the problem before you go any further with your code.

answered May 3, 2012 at 14:42

Marc B

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2 Comments

grasshopper Over a year ago

Aaaah, ok, didn't know. I added your code and it throws "No database selected", so a connection was probably not established, thanx. Going to see if I can get it to work, will update.

Marc B Over a year ago

That'd be a missing mysql_select_db() call, in that case. If you hadn't connected at all, it'd say "no connection" or something similar. Never assume a query call succeeded. Even if the SQL statement itself is 100% valid, there's way too many other reasons for things to blow up to NOT check for failure.

user1327073 · Accepted Answer · 2012-05-03 14:47:23Z

2

$query = "UPDATE course SET cname = '".$newname."' WHERE cid = '".$int."'";

is cID an integer ? in the update statement, looks to me like a string, try to echo every query and check the validity by executing it directly in your db

answered May 3, 2012 at 14:47

user1327073

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Comments

Sameer Anand · Accepted Answer · 2012-05-03 14:46:57Z

1

where do you connect to the database??

use mysql_connect(string hostname, string username, string password'); to connect to the database and then execute the query after selecting your database using mysql_select_db..

answered May 3, 2012 at 14:46

Sameer Anand

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1 Comment

grasshopper Over a year ago

I establish the connection in includes, it was missing from my code which caused the error.

Devator · Accepted Answer · 2012-05-03 14:49:35Z

1

First you should remove the extra ; on $name="edit".$i;;

Then, how do you post the values? I see no <form> attributes in your code, hence it cannot be posted.

Also, everything is in a for loop. $newname=($_POST['text']); is never being set.

answered May 3, 2012 at 14:49

Devator

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Comments

Tamas Szoke · Accepted Answer · 2012-05-03 14:51:26Z

1

Maybe instead of this:

if (isset($_POST[$name2]))

try this:

if ($name2!="")

answered May 3, 2012 at 14:51

Tamas Szoke

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