Reputation: 44312
I have a simple PHP file with the following:
<?php
echo 'catid=$_GET["catid"]';
<?>
When I run the page, the output is:
catid=$_GET["catid"]
I'm accessing the page as www.abc.com/temp.php?catid=3. I'd like $_GET to execute so I see:
catid=3
What am I doing wrong?
Upvotes: 1
Views: 197
Reputation: 2597
There are a few options to combine a variable with a string.
<?php
$var = "something";
// prints some something
echo 'some ' . $var; // I prefer to go for this one
// prints some something
echo "some $var";
// prints some $var
echo 'some $var';
// prints some something
echo "some {$var}";
?>
Upvotes: 0
Reputation: 3103
You have to use double quoted string to notify PHP that it might contains variables inside. $_GET is array, so you will need to put the variable statement in {}.
<?php
echo "catid={$_GET['catid']}";
?>
Upvotes: 0
Reputation: 7653
The best way to use variables in string is:
echo "catid={$_GET['catid']}";
Upvotes: 0
Reputation: 2344
$_Get
is a variable, and to echo a variable you do not need parenthesis around it.
<?php
echo 'catid='.$_GET["catid"];
?>
please see this : source
Upvotes: 1
Reputation: 1427
You can use non-array variables for that:
$getCatID = $_GET["catid"];
echo "catid=$getCatID";
Or you can use (recommended):
echo 'catid=' . $_GET["catid"];
Upvotes: 0
Reputation: 23580
You have to cancat the two:
echo 'catid=' . $_GET["catid"];
or you could use "
(double quotes):
echo "catId=$someVar";
Upvotes: 2