I've got a problem! ahh!

Anon

The following code JUST WON'T WORK! What i'm trying to do is if the ID #s in the database 'links' does not start with one and consecutively count through the last row # (start 1, 2, 3, 4, end 5), then it will check each row one by one, if the first row isn't 1, then it'll change it to 1 and so on. But this doesn't work! I don't know why either, i've tried every thing I, an amateur at this PHP, can come up with.

require ("includes.php");

$count = 0;

$result = mysql_query("SELECT * FROM links)
or exit ("Invalid query");

if ($row = mysql_fetch_array($result)) {
do {
$count += 1;
$oID = $row[id];
if ($oID != $count) {
$sql = "UPDATE links SET id='$count' WHERE id=$oID";
$result = mysql_query($sql);
print "oID: $oID <br>"; # used for test results
print "COUNT: $count <br>"; # used for test results
}
} while ($row = mysql_fetch_array($result));
}

I usually end up with the error that says that the MySQL fetch statement is wrong or won't work or something like that.
any help? any tips? any corrections? thanks for your help!

Anon

Here's one problem:

$result = mysql_query("SELECT * FROM links)

missing a quote, it should be:

$result = mysql_query("SELECT * FROM links")

And things like this:

$oID = $row[id];

are bad programming in PHP - you need to put quotes around the key or you'll throw warnings if you've got them turned on (which a lot of people do)... so:

$oID = $row['id'];

Also just as an observation... seems pointless to get a row, do the loop, then do another loop... you should just start with a while at the top, get rid of that first if, and don't do a do...while, just a regular while.

Anon

Hmm, I have fixed those mistakes but I still get the following error:

Warning: Supplied argument is not a valid MySQL result resource in /home/floorgaz/public_html/links/reID.php on line 17

Here's the updated script:

require ("includes.php");

$count = 0;

$result = mysql_query("SELECT * FROM links");

while ($row = mysql_fetch_array($result)) {
$count += 1;
$oID = $row['id'];
if ($oID != $count) {
$sql = "UPDATE links SET id='$count' WHERE id=$oID";
$result = mysql_query($sql);
}
}