Morning,

 

I am getting the following error message:

 

Notice: Undefined variable: result in /Applications/MAMP/htdocs/website.com/admin/index.php on line 71

 

Here is my PHP code:

 

<?php

include("config.php");

if (isset($_POST["submit"])) {
	$txt = $_POST['txt'];
	$txt = mysql_real_escape_string($_POST['txt']);

	$result = mysql_query("UPDATE content SET txt='$txt' WHERE id='1'", $db);
}

if ($result) {
echo "<meta http-equiv=Refresh content=0;url=index.php>";
} 

else {

$result = mysql_query("SELECT * from content WHERE id='1'", $db);	
while ($row=mysql_fetch_assoc($result)){

require('fckeditor/fckeditor.php');

				$oFCKeditor = new FCKeditor('txt');
				$oFCKeditor->BasePath	= 'fckeditor/';
				$oFCKeditor->Value		= $row['txt'];
				$oFCKeditor->Height	 = '300';
				$oFCKeditor->Create();

}

}

?>

 

Line 71 is the following:

 

if ($result) {

 

any ideas, what I've done wrong?

 

Cheers

DFT

Hi again DFT,

The problem stems from the location of the first $result variable.

It is inside the first if() clause and so will only exist IF the form has been submitted. You then call the variable outside of the if() before the form has been submitted and so the variable hasnt been created yet - giving you the error.

 

I would guess that moving the { on line 10 down to where the else is on line 16 would do the trick.

 

<?php

include("config.php");

if (isset($_POST["submit"])) {
	$txt = $_POST['txt'];
	$txt = mysql_real_escape_string($_POST['txt']);

	$result = mysql_query("UPDATE content SET txt='$txt' WHERE id='1'", $db);


if ($result) {
echo "<meta http-equiv=Refresh content=0;url=index.php>";
}

} else {

$result = mysql_query("SELECT * from content WHERE id='1'", $db);	
while ($row=mysql_fetch_assoc($result)){

require('fckeditor/fckeditor.php');

				$oFCKeditor = new FCKeditor('txt');
				$oFCKeditor->BasePath	= 'fckeditor/';
				$oFCKeditor->Value		= $row['txt'];
				$oFCKeditor->Height	 = '300';
				$oFCKeditor->Create();

}

}
?>

Hi again DFT,

The problem stems from the location of the first $result variable.

It is inside the first if() clause and so will only exist IF the form has been submitted. You then call the variable outside of the if() before the form has been submitted and so the variable hasnt been created yet - giving you the error.

 

I would guess that moving the { on line 10 down to where the else is on line 16 would do the trick.

 

<?php

include("config.php");

if (isset($_POST["submit"])) {
	$txt = $_POST['txt'];
	$txt = mysql_real_escape_string($_POST['txt']);

	$result = mysql_query("UPDATE content SET txt='$txt' WHERE id='1'", $db);


if ($result) {
echo "<meta http-equiv=Refresh content=0;url=index.php>";
}

} else {

$result = mysql_query("SELECT * from content WHERE id='1'", $db);	
while ($row=mysql_fetch_assoc($result)){

require('fckeditor/fckeditor.php');

				$oFCKeditor = new FCKeditor('txt');
				$oFCKeditor->BasePath	= 'fckeditor/';
				$oFCKeditor->Value		= $row['txt'];
				$oFCKeditor->Height	 = '300';
				$oFCKeditor->Create();

}

}
?>

Morning Spike,

 

Thanks again. Works fine now.

 

DFT