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Invalid argument supplied for foreach()
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Join Date: May 2008
Posts: 74
Reputation:
Solved Threads: 0
hi friends,
my code is as follow,
when i use foreach loop in template, it give the inavlid argument "Invalid argument supplied for foreach() error"
i am not understanding where i comit mistake.
my code is as follow,
------sql query here------------
$result = mysql_query($sql):
while ($row = mysql_fetch_array($res))
{
$row[tableLegends];
$link[]=$row;
}foreach($link AS %links){
--------code here-------
}i am not understanding where i comit mistake.
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Join Date: May 2008
Posts: 74
Reputation:
Solved Threads: 0
I used following code as per your suggestion,
but the script is giving the same error.
------sql query here------------
$result = mysql_query($sql):
while ($row = mysql_fetch_array($res))
{
$row[tableLegends];
$link = array ();
$link =$row;
}but the script is giving the same error.
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Join Date: Aug 2009
Posts: 49
Reputation:
Solved Threads: 8
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foreach($link AS %links){ --------code here------- }
i am not understanding where i comit mistake.
foreach($link as %links) There's a percent sign (%) there instead of a dollar sign ($).
Last edited by EvolutionFallen; Sep 3rd, 2009 at 4:58 am.
The proper use of this would be declaring the $link variable as an array before you start looping through the array results of the query. Your code should look like this:
Also looking at your foreach code, what is
PHP Syntax (Toggle Plain Text)
$result = mysql_query($sql); //Make sure to use a semicolon at the end of a line, not a colon! $link = array(); //Here is where you declare the array while ($row = mysql_fetch_array($res)) { //$row[tableLegends]; <---Not sure what this is for, and since it doesn't appear to have a function I commented it out $link[] = $row; //You were right by approaching this with $link[] and not $link, if you used $link then link would only contain the value of row and won't be an array of the rows }
Also looking at your foreach code, what is
%links . Don't you mean $links : PHP Syntax (Toggle Plain Text)
foreach($link as $links) { ... }
My suggestion stated _before_ the query...
php Syntax (Toggle Plain Text)
$link = array (); $result = mysql_query($sql): while ($row = mysql_fetch_array($res)) { $link[]=$row; }
"If it is NOT source, it is NOT software."
-- NASA
-- NASA
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My suggestion stated _before_ the query...
php Syntax (Toggle Plain Text)
$link = array (); $result = mysql_query($sql): while ($row = mysql_fetch_array($res)) { $link[]=$row; }
PHP Syntax (Toggle Plain Text)
$link = array(); $result = mysql_query($sql); //<-- Semicolon while ($row = mysql_fetch_array($res)) { $link[] = $row; }
$links , %links as used, but your solution provides a safety if there are no rows returned (Which is a good thing to have!). •
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Glad you saw that typo too, FlashCreations. I was starting to worry that maybe %<varnamehere> was some PHP syntax I'd never heard of before =P
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Well observed... missed the colon.
I left the template alone, because I was unsure if it was supposed to be php, or a specific templating syntax.
If all your questions have been answered I ask that you mark the thread as solved to avoid further confusion!
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