Undefined variable

[mmm104104]

I'm getting "Undefined variable: cat in..." error message in a following line

$query2 = mysql_query("SELECT * FROM Photos_Cat WHERE PhotosCat_ID='$cat'");

$cat is a variable from the url generated by previous page

Please advice. Thanks

[leatherback]

so..

if(isset($_GET['cat']))
  {
  $cat = mysql_real_escape_string($_GET['cat']);
  }
else
  {
  $cat = '';
  }

$query2 = mysql_query("SELECT * FROM Photos_Cat WHERE PhotosCat_ID='$cat'");

[bretticus]

Is $cat available to the current page? In other words, is $cat supposed to come from the URL on the current page? How are you passing the value for $cat from the previous page?

[mmm104104]

I pass it thru the url [url]http://...php?cat=29[/url]

after i made above change now i get

Undefined index: Photos_ID in...

[leatherback]

in which part? In the code you posted you do not have Photos_ID

[mmm104104]

my url is [url]http://...php?cat=29[/url]

and i use "cat" to define the query

	$query2  = mysql_query("SELECT * FROM Photos_Cat WHERE PhotosCat_ID='$cat'");

let thing i just added is (see above)
if(isset($GET['cat'])
{$cat = mysql_real_escape_string($GET['cat']); }

	else
		{$cat = '';}

note,
my original code works on different host, why?

[bretticus]

You may be misunderstanding variable persistence. The reason you got a cat is because you passed it in your URL first but fundamentally you initialized the variable $cat. In other words, you checked to see if the $GET['cat'] from your url was set. If so, you set the variable $cat to the value of $GET['cat'], else you set it to an empty string. The point is, the $cat variable is now defined. Maybe you get that because your new error is probably caused by attempting to use your SQL output...

You didn't show 'Photos_ID' in your post, but the "Index" variation of the error is usually a bad key in an associative or zero-based array. You probably need quotes like thus:

$row['Photos_ID']

as opposed to

$row[Photos_ID]

If that is NOT the issue, please post your code.

[bretticus]

To reiterate, there is nothing wrong with Leatherback's code at all. It did not cause the error. It fixed it and the next error in your code bubbled out.

mmm104104 wrote:

my original code works on different host, why?

Without seeign your code my assumption is that the other host (wrongly) has register_globals on. Unless you want your scripts vulnerable to cross site scripting, make a habit of keeping it off and substituting your $cat-like variables with super globals like $_GET['cat'] (since cat comes from the URL.)

[mmm104104]

here is the entire code, I'm new to this so I appreciate all your help

I marked errored lines with "***"

	if(isset($_GET['cat']))
		{$cat = mysql_real_escape_string($_GET['cat']);	}

*** else
{$cat = '';}

	$query2  = mysql_query("SELECT * FROM Photos_Cat WHERE PhotosCat_ID='$cat'");

*** while ( $imgcat = mysql_fetch_array($query2))
{
$id = $imgcat['PhotosCat_ID'];
$category = $imgcat['PhotosCat_Category'];
echo("<h1><b>" . $category . " Gallery</h1></b>");
}

	if ($cat == '1') {

*** $query = mysql_query("SELECT DATE_FORMAT(Photos_Date, '%b %Y') as Photos_NewDate, Photos_Title, Photos_FileName, Photos_FileName_sm FROM Photos");
while ( $img = mysql_fetch_array($query))
{
$id = $img['Photos_ID'];
$title = $img['Photos_Title'];
$filename = $img['Photos_FileName'];
$filename_sm = $img['Photos_FileName_sm'];
$category = $imgcat['Photos_Category'];
$date = $img['Photos_Date'];
$newdate = $img['Photos_NewDate'];

	if ($date == "0000-00-00"){
	echo("<a href='media/images/" . $filename . "' rel='lyteshow[" . $cat . "]' title='" . $title . "'><img class='images' src='media/images/" . $filename_sm . "'></a> &nbsp;");
		} 
	else {
	echo("<a href='media/images/" . $filename . "' rel='lyteshow[" . $cat . "]' title='" . $title . ", " . $newdate . "'><img class='images' src='media/images/" . $filename_sm . "'></a> &nbsp;");
		}
		}			
	echo("<p> </p>");}

	else{
	$query  = mysql_query("SELECT DATE_FORMAT(Photos_Date, '%b %Y') as Photos_NewDate, Photos_Title, Photos_FileName, Photos_FileName_sm FROM Photos WHERE PhotosCat_ID='$cat'");

*** while ( $img = mysql_fetch_array($query))
{
$id = $img['Photos_ID'];
$title = $img['Photos_Title'];
$filename = $img['Photos_FileName'];
$filename_sm = $img['Photos_FileName_sm'];
$category = $img['Photos_Category'];
$date = $img['Photos_Date'];
$newdate = $img['Photos_NewDate'];

	if ($date == "0000-00-00"){
	echo("<a href='media/images/" . $filename . "' rel='lyteshow[" . $cat . "]' title='" . $title . "'><img class='images' src='media/images/" . $filename_sm . "'></a> &nbsp;");
		} 
	else {
	echo("<a href='media/images/" . $filename . "' rel='lyteshow[" . $cat . "]' title='" . $title . ", " . $newdate . "'><img class='images' src='media/images/" . $filename_sm . "'></a> &nbsp;");
		}
		}
                    }

[bretticus]

Well, right off the bat, I see...

$query = mysql_query("SELECT DATE_FORMAT(Photos_Date, '%b %Y') as Photos_NewDate, Photos_Title, Photos_FileName, Photos_FileName_sm FROM Photos WHERE PhotosCat_ID='$cat'");
while ( $img = mysql_fetch_array($query))
{
$id = $img['Photos_ID'];//...

You have used mysql_fetch_array which will give you an associative array IF you pass the MYSQL_ASSOC constant as the second variable...

while ( $img = mysql_fetch_array($query, MYSQL_ASSOC))

Or you could just use...

while ( $img = mysql_fetch_assoc($query))

Now, you will actually make $img an associative array (can refer to array values via column names.)

Unfortunately, you never ask for the Photos_ID column in your SQL query, so that associative key will never exist in your $img array! Thus "undefined index." Fix those errors and see that they are not repeated in the rest of your code (I didn't look it all over.)

Good luck.

[leatherback]

*** else
-> What is the error for this line? Should NOT have an error

$query2 = mysql_query("SELECT FROM Photos_Cat WHERE PhotosCat_ID='$cat'");
** while

-> You most likely have an error in your query. I presume $cat is not parsed; Look at my sig and do as it sais:
$query2 = mysql_query("SELECT * FROM Photos_Cat WHERE PhotosCat_ID='$cat'") or die(mysql_error());

fix that one, and do the same for the other queries. Then tell us what the errors are, if you cannot fix them

[mmm104104]

here is an error I'm getting now.

mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in

in "***" line

$query = mysql_query("SELECT DATE_FORMAT(Photos_Date, '%b %Y') as Photos_NewDate, Photos_ID, Photos_Title, Photos_FileName, Photos_FileName_sm, Photos_Category FROM Photos WHERE PhotosCat_ID='$cat'" or die(mysql_error());
**** while ( $img = mysql_fetch_array($query))
{
$id = $img['Photos_ID'];
$title = $img['Photos_Title'];
$filename = $img['Photos_FileName'];
$filename_sm = $img['Photos_FileName_sm'];
$category = $img['Photos_Category'];
$date = $img['Photos_Date'];
$newdate = $img['Photos_NewDate'];

[mmm104104]

I think i got it, thanks