[Resolved] mysql_fetch_array not a valid argument

Radon3k

I'm trying to display a news article based upon the ID of that article. However (and my code is below) I get this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/elite/public_html/index.php on line 92

The variables $title and $author are global variables in another function. Thanks for any help you can provide. 🙂

<?php function displayArticle($news) {
	$article = mysql_query("SELECT * FROM news WHERE ID = $news");
	if(!$article) {
		echo("Could not retrieve article.");
	}

while($articlerow = mysql_fetch_array($article)) {
	$article = $articlerow['message'];

	echo("<b>$title</b><br>");
	echo("<b>Posted by: $author<br>");
	echo("<b>On:</b> $date<br><br>");
	echo("$message<br><br>");
	echo("<a href='index.php'>Home</a>");
}
}

if($_GET['news']) {
	displayArticle($id);
} else {
	displayNewsTitles();
}
?>

matt_4013

It's not receiving a valid result set from mysql, so it's got nothing to turn into an array.

Change the line:

$article = mysql_query("SELECT * FROM news WHERE ID = $news");

to:

$article = mysql_query("SELECT * FROM news WHERE ID = $news") or die(mysql_error());

So that you can see exactly where the problem lies! 🙂

Radon3k

Hi, this is the error that is returned:

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

. But my line one is just this:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

laserlight

funny, I would think that the if(!$article) check would have revealed that, though the while loop should have been placed in an else comditional block to that check.

Radon3k

Well I suppose one of the problems might be here:

<?php function displayNewsTitles() {
	global $title,$author,$date;
	$news = mysql_query("SELECT title,author FROM news");
	if(!$news) {
		echo("Unable to retrieve the title and/or author.");
	}

echo("The latest news from Elite 504:<br><br>");

while($newsrow = mysql_fetch_array($news)) {
	$newsid = $newsrow['ID'];
	$title = $newsrow['title'];
	$author = $newsrow['author'];
	$date = $newsrow['date'];

	echo("<img src='arrow.gif' height='10' width='10'> ");
	echo("<a href=\"$_SERVER[PHP_SELF]?news=$newsid\"><u>$title</u></a><br>");
	echo("&nbsp;&nbsp;Posted by: $author<br>");
	echo("&nbsp;&nbsp;On: $date<br><br>");
}
}?>

The link does not display any ID # as it should. Any reason why?

laserlight

At runtime, what does the argument of displayArticle() contain?
You might want to place it in quotes when running the SQL query.

incidentally,
$date = $newsrow['date'];
isnt likely to be correct since you did not select date in the query.

matt_4013

What's it displaying instead?

Radon3k

displayArticle($news) is what is contained.

matt_4013

Is this line:

echo("<a href=\"$_SERVER[PHP_SELF]?news=$newsid\"><u>$title</u></a><br>");

outputting the correct link? If NOT, exactly what is it outputting.

Radon3k

It out puts http://www.something.com?news=

That's it.

matt_4013

Well there's a problem in itself 🙂

duh. I just reread the code, I can't believe I missed it.

In your sql statement, you aren't selecting the id or the date from your table.

SELECT id, title, author, date FROM news

That would probably be the better way to go about it 😉

That should fix your link up. Btw, it's usually a good idea to keep your case standard, ie, not have some things in upper-case, and some-lower, some with first letter uppper, that type of thing, because it just gets a little confusing 😉

Anyway, fix that up, and then give the other one ago, if it's still failing, make sure the link is good, if it is, we'll get back onto it 🙂

Radon3k

heh I suppose selecting the proper items would be helpful. OK so that problem is resolved but I still get an error when trying to view the actual article. Here is the modified code:

<?php function displayNewsTitles() {
	global $title,$author,$date;
	$news = mysql_query("SELECT ID,title,author,date FROM news");
	if(!$news) {
		echo("Unable to retrieve the title and/or author.");
	}

echo("The latest news from Elite 504:<br><br>");

while($newsrow = mysql_fetch_array($news)) {
	$newsid = $newsrow['ID'];
	$title = $newsrow['title'];
	$author = $newsrow['author'];
	$date = $newsrow['date'];

	echo("<img src='arrow.gif' height='10' width='10'> ");
	echo("<a href=\"$_SERVER[PHP_SELF]?news=$newsid\"><u>$title</u></a><br>");
	echo("&nbsp;&nbsp;Posted by: $author<br>");
	echo("&nbsp;&nbsp;On: $date<br><br>");
	}
}


function displayArticle($news) {
	$article = mysql_query("SELECT * FROM news WHERE ID = $news") or die(mysql_error());
	if(!$article) {
		echo("Could not retrieve article.");
	}

while($articlerow = mysql_fetch_array($article)) {
	$article = $articlerow['message'];

	echo("<b>$title</b><br>");
	echo("<b>Posted by: $author<br>");
	echo("<b>On:</b> $date<br><br>");
	echo("$message<br><br>");
	echo("<a href='index.php'>Home</a>");
}
}

if($_GET['news']) {
	displayArticle($id);
} else {
	displayNewsTitles();
}?>

matt_4013

if($_GET['news']) {
    displayArticle($id);
} else {
    displayNewsTitles();
}?>

Why don't you change this to

if($_GET['news']) {
    displayArticle($_GET['news']);
} else {
    displayNewsTitles();
}?>

and see what happens

Radon3k

Ok well I changed some code around and made it look all nice and pretty. I'm still getting the error, however, it does display the "Posted by:" and "On:". It doesn't display, however, the $author or $date. I don't understand what I'm doing wrong. Thanks for the help so far. 🙂

matt_4013

You keep missing out little things 😉 Notice you're referring to the author and date variables, but you haven't defined them?

function displayArticle($news) {
    $article = mysql_query("SELECT * FROM news WHERE ID = $news") or die(mysql_error());

// I took out the checking on $article, because a) it wasn't working, and b) that's what the die function does anyway

while($articlerow = mysql_fetch_array($article)) {
    $message = $articlerow['message'];

// I added these in, and changed the name of your variable above

    $title = $articlerow['title'];
    $author = $articlerow['author'];
    $date = $articlerow['date'];

    echo("<b>$title</b><br>");
    echo("<b>Posted by: $author<br>");
    echo("<b>On:</b> $date<br><br>");
    echo("$message<br><br>");
    echo("<a href='index.php'>Home</a>");
}
}

Radon3k

Those variables were delcared as global variables in the displayNewsTitles() function.

matt_4013

They certainly were, but they were also being rewritten everytime it looped through your while, and would as such only contain the last value, not all of them.

Radon3k

I'm still getting the same error, and according to the code, it has to do with that SQL statement. Am I trying too hard to understand this or what? I do appreicate the help thus far, however. 🙂

matt_4013

ok, cool, can you make the area around the query this, and see what it outputs? My thought is to put single quotes around the value in the query, but it shouldn't be required :\

$sql = "SELECT * FROM news WHERE ID = $news";

echo "<h2>$sql</h2>";

$article = mysql_query($sql) or die(mysql_error());

I have a meeting to go to, then I'm off home, so If you haven't figured it in about an hour, I'll be back on then 😉

Radon3k

Well I'm not sure what I did to fix it, but it works now. I changed some code here and there. I'll be it was my displayArticle($_GET['news']) because at one point it was set to, "name" instead of, "news". All the little mistakes are what get me caught up. Go figure :rolleyes:. Well thank you very much for the help, I do appreciate it! 🙂