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PODoes PHP evaluate variables when a function is defined?
 

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  1. PODoes PHP evaluate variables when a function is defined?
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    <p>I am creating a multipage form in PHP, using a session. The <code>$stage</code> variable tracks the user's progress in filling out the form, <strong>(UPDATE) and is normally set in $_POST at each stage of the form.</strong></p> <p>On the second page (stage 2), the form's submit button gets its value like this:</p> <pre><code>echo '&lt;input type="hidden" name="stage" value="'; echo $stage + 1; echo '" /&gt;; </code></pre> <p>That works fine - <code>$stage + 1</code> evaluates to 3 if I'm on page 2. But since I'm doing this more than once, I decided to pull this code out into a function, which I defined at the top of my code, before <code>$stage</code> is mentioned.</p> <p>In the same spot where I previously used the code above, I call the function. I have verified that the function's code is the same, but now <code>$stage + 1</code> evaluates to 1.</p> <p><strong>Is PHP evaluating my variable when the function is defined, rather than when it's called? If so, how can I prevent this?</strong></p> <h3>Update 1</h3> <p>To test this theory, I set <code>$stage = 2</code> before defining my function, but it still evaluates to 1 when I call the function. What's going on?</p> <h3>Problem Solved</h3> <p>Thanks to everyone who suggested variable scope as the culprit - I'm slapping my forehead now. <strong><code>$stage</code> was a global variable, and I didn't call it <code>$GLOBAL_stage</code></strong>, like I usually do, to prevent this sort of problem.</p> <p>I added <code>global $stage;</code> to the function definition and it works fine. Thanks!</p>
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    1. USCharles
    1. USNathan Long
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    1. COWe really need to know how $stage is set in the first instance - is it a posted variable, a get variable, a session variable?
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      1. USBrynJ
    2. COPost your function and the call as a code snippet please.
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      1. USphilistyne
 

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