mysql_fetch_array() error

[Mung]

Can anyone help me to sort this out. I've been trying to get it to work for 2 days and my hair is starting to turn white.

$result = 'mysql_query(SELECT * FROM news);';
if ($result){
    echo "success!";
    }else{
die ('<br>Error'.mysql_error());
}

while ($row = mysql_fetch_array($result)) 
{
echo "<p> ".$row[content]." </p>";
} 

It generates this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/spaceon/public_html/chickens/test.php on line 71

Line 71 is the mysql_fetch_array.

Thanks

[Mike]

Use this code:

$result = mysql_query("SELECT * FROM news") or die("Error: " . mysql_error()); 

while ($row = mysql_fetch_array($result)) 
{ 
echo '<p> ' . $row['content'] . ' </p>'; 
} 

You had the query function itself inside of quotes, which just made it into a string. That caused the query to not exists, and therefor the fetch to fail.

[Icheb]

One starts to wonder why he didn't put mysql_fetch_array() in quotes, too .

[Mung]

thank you

Icheb? Would it have made it go faster?

[Icheb]

Definitely .

[Mung]

I'll add it now then

[Icheb]

But don't blame me if it doesn't work .

[Mung]

It's that fast it finishes before it's even started, what a great piece of custom code

[seanf]

Use this code:

echo '<p> ' . $row['content'] . ' </p>'; 

Come on Mike, go full throttle! (Umm, Angels)

echo '<p> ', $row['content'], ' </p>'; 

Sean

[Mike]

:sigh:, I always forget

Thanks for the tip, Sean