Error while using php variable in a query string

Question

I'm trying to pass the return value of a php function through query string but its not working. Please look into the code below:

Here I'm trying to replace $myvar in the query string.

<button onClick="parent.location='http://www.example.com&id=$myvar'">Design Your product</button>

The php function to retrieve $myvar is given below:

<?php
$myvar=Design();
function Design() 
 {
// my function code

return $result;
}

Andreas Eriksson · Accepted Answer · 2011-09-21 06:44:47Z

4

<button onClick="parent.location='http://www.example.com&id=<?php echo $myvar; ?>'">Design Your product</button>

answered Sep 21, 2011 at 6:44

Andreas Eriksson

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2 Comments

nkn Over a year ago

Thanx 4 ur immediate help but i'm still getting error. It says "Call to undefined funtion Design()" Have i made mistake in the function call ?? <?php $myvar=Design(); function Design() { // my function code return $result; }

Andreas Eriksson Over a year ago

See Trogvar's answer - You must tell the preprocessor what the definition of the function is before you call it. Setting $myvar to Design() before the actual function is defined results, predictably, in a Call to undefined function Design().

Trogvar · Accepted Answer · 2011-09-21 08:15:01Z

0

You try to call the function before it has been defined. This should work

<?php
    function Design()
    {
         return "something";
    }
    $myvar = Design();
?>
<button onClick="parent.location='http://www.example.com?id=<?php echo $myvar; ?>'">Design Your product</button>

answered Sep 21, 2011 at 8:15

Trogvar

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