I have a php file:
<?php
$a = 1;
function test(){
echo $a;
}
test();
?>
And I get this error:
Notice: Undefined variable: a in X:\...\test.php on line 4
Using XAMPP @ 32bit W7.
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Its because of the function scope. you would need to write global $a; at the top of the function, but you should avoid global variables. Read a bit about scope in phpAndy Groff– Andy Groff2011-09-04 02:36:31 +00:00Commented Sep 4, 2011 at 2:36
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2php.net/manual/en/language.variables.scope.phpNullUserException– NullUserException2011-09-04 02:37:18 +00:00Commented Sep 4, 2011 at 2:37
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PHP does not have real global variables. You need to chain to the shared scope if you want this to work. (Example too abstract. But it's frequently misused by newcomers; function parameters are usually the advisable approach.)mario– mario2011-09-04 02:51:17 +00:00Commented Sep 4, 2011 at 2:51
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1Also NARQ: Please formulate your posts as actual questions. Add more details and real code if you want case-specific help. See higher quality question and sample code howtos.mario– mario2011-09-04 02:54:43 +00:00Commented Sep 4, 2011 at 2:54
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3 Answers
Variables have function scope. $a inside the function is not the same as $a outside the function. Inside the function you have not defined a variable $a, so it doesn't exist. Pass it into the function:
$a = 1;
function test($a) {
echo $a;
}
test($a);
Comments
You have trouble understanding variable scope. $a is defined in the global scope, but not in the function scope. If you want your function to know what $a contains, you have two choices :
- Make it global (usually a bad solution)
- Add a new argument to your function, and pass your variable to your function
Comments
You can use global as advised, but that is bad practice. If you need variables in a function from outside the function then pass them as parameters.
$a = 1;
function test($a) {
echo $a;
}
test($a);
