[Falgern]

Hello
I've got a bit of PHP problem.
I'm trying to make a page that updates a row in a database. and another page reads from that row and gets its text from it.. some kind of easy posting/updating system.
But i've run into some really big problem.

<?php 
$submit = $_POST['submit']; 
$text = $_POST['text']; 

$db = mysql_connect(localhost, ******, *****); 
mysql_select_db(******); 

if ($submit) { 
$result = mysql_query("UPDATE pages SET lol = $_POST[text] WHERE id = '1'" ); 


} else{ 
?> 
<form action="indexupdate.php" method="post"> 
Text <input type="text" name="text" size="30" style="width: 300px; height: 100px;"> 


<p><INPUT TYPE="Submit" name="submit" value="Change"> 

<?php 
$submit = $_POST['submit']; 
$lname = $_POST['text']; 
} // end if 
?>

Thats the code i've done so far. but it doesnt work, i cant however see anything wrong in it.

also, is there an easy way to make a page to read from database?


Thank you.

[Le Froid]

$ = mysql_query("UPDATE pages SET lol =  $ WHERE id = '1'" );

remove the result maby? if nothing is using "$result" it is not needed. See if it works

[XTZGZoReX]

Also, I doubt you can start an If-statement in one PHP enclosed block of code, and end it in another.. Besides, doing it the normal way is more efficient.

[siliconslayer]

I dont think you can put variables straight into a mysql quey like that

i tend to append it into another variabe first

$query = "UPDATE pages SET lol = ".$_POST['text']." WHERE id = '1'";
$result = mysql_query($query);

* not sure if you need lol = '".$_POST['text']."' <-- i added (') single quotations try both

i suggest you read about sql injection also

To read form a database do

$db = "database name here";
$query = "SELECT * FROM".$db." WHERE lol = '$lol' ORDER BY lol DESC"; // <---adjust the where and order by statements as desired.

$result = mysql_query($query);
while ($output = mysql_fetch_assoc($result)) //<-- you can use mysql_fetch_array() here instead
{
echo $output['lol'];
}