函数的变量不在函数外部显示

Question

I am having trouble getting the value of a variable which is in a function outside the function...

Example:

function myfunction() {
    $name = 'myname';
    echo $name;
}

Then I call it...

if (something) {
    myfunction();
}

This is echoing $name, but if I try to echo $name inside the input value it won't show:

<input type="text" name="name" value="<?php echo $name ?>"

How can I get access to this variable value?

Answer 1

The $name variable is local until you explicitly define it as global:

function myfunction() {
    global $name;
    $name = 'myname';
    echo $name;
}

But this doesn't seem like a good use of globals here. Did you mean to return the value and assign it? (Or just use it once?)

function myfunction() {
    $name = 'myname';
    return $name;
}

...
$name = myfunction();