PHP: Looping

a_gajic posted this at 15:24 — 14th December 2001.

Joined: Aug 2001

Okay basically I'm getting an error and I don't know why. Maybe you guys can help me?

here is my code:

<?php
while($row = mysql_fetch_array($news))  {

// stuff goes here

}
?>

So what's wrong with it? It's the while line that is causing the error but I can't see a thing wrong with it, can you? I have also tried using the do while loop but it prints the same error for the while bit on that too.

Yes my database connections are correct

Thanks in advance,

Adam.

Mark Hensler posted this at 17:04 — 14th December 2001.

He has: 4,048 posts

Joined: Aug 2000

What is the error?
Is $news a valid mySQL result?

a_gajic posted this at 21:05 — 14th December 2001.

They have: 71 posts

Joined: Aug 2001

Yes news is valid.

Warning: Supplied argument is not a valid MySQL result resource in C:\apache\server\htdocs/agNews/read.php on line 8

that is the error

here is the line above the while command..

<?php
$news = mysql_query(\"SELECT * FROM news ORDER BY date DESC LIMIT 8;\") || die(\"Cannot select records\");
?>

Is it my query? It's the only other thing that I can think of..

Mark Hensler posted this at 05:36 — 15th December 2001.

He has: 4,048 posts

Joined: Aug 2000

No, that error tells you that $news is not valid.

This should give you more debuging help:

<?php
$query = \"SELECT * FROM news ORDER BY date DESC LIMIT 8\";
$result = mysql_query($query);
if (!$result) {
    echo \"\n\n\";
    echo \"<!--\n\";
    echo \"FILE: \" . __FILE__ . \"\n\";
    echo \"LINE: \" . __LINE__ . \"\n\";
    echo \"\n\";
    echo \"QUERY ERROR:\n\";
    echo mysql_error() . \"\n\";
    echo \"\n\";
    echo \"QUERY:\n\";
    echo $query . \"\n\";
    echo \"-->\n\";
    echo \"\n\";
}
while ($result && $row=mysql_fetch_array($result)) {
    // do something
}
?>

This should print the file and approximate line number where the faulty query was executed. It will also print the mySQL error and faulty query.

Mark Hensler
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