Inserting variable from input in form action

Question

I have a form which queries the user about which type of list he wants to see. As the submit button is pressed, I want to be able to have the value listtype displayed in the address so I can access it by using GET methods, for instance to change the page of the list that is being shown. In this form, I tried the following code:

<?php
    echo '<form method=post action="index.php?operacao=checkList&number=1&pagesize=10&listtype=', $_GET['listtype'] ,'" name="listForm">
    <ol>

        <li><label for="listtype">Type:</label>
        <select name="listtype">
        <option value="default">Select</option>
        <option value="doctor">Doctor</option>
        <option value="nurse">Nurse</option>
        <option value="patient">Patient</option>
        </select></li>

        <div id="form-button">
            <li><input type="submit" value="Submit" onclick="javapopup();"></li>
        </div>

    </ol>
    </form>';
?>

Although, as I open the form, I get an error that says the variable listtype is not defined. How can I solve this problem?

You have <form method=post and you're asking for a GET. Try using <form method="get" — Funk Forty Niner
– Funk Forty Niner, Commented Nov 20, 2013 at 15:02
Your undefined error message could be caused by an unnamed variable. For instance: $var=$_GET['var']; we need to see your PHP handler. — Funk Forty Niner
– Funk Forty Niner, Commented Nov 20, 2013 at 15:15

John Conde · Accepted Answer · 2013-11-20 15:01:54Z

2

$_GET['listtype'] is not going to be set the first time you load this page which is why you are getting this error. You need to check to see if it is set before you use it and, if it is not, not use it in that situation. In your case because your form is set to POST it will never populate the $_GET superglobal which will cause this error to never go away.

echo '<form method=post action="index.php?operacao=checkList&number=1&pagesize=10&listtype=', $_GET['listtype'] ,'" name="listForm">

probably should resemble something like this:

$listype = (isset($_POST['listtype'])) ? $_POST['listtype'] : '';
echo '<form method=post action="index.php?operacao=checkList&number=1&pagesize=10&listtype=', $listype ,'" name="listForm">

FYI, using a raw $_GET or $_POST variable in your code the way you are using it leaves you wide open to Cross Site Scripting attacks. So you should be sanitizing this variable anyway.

$listype = (isset($_POST['listtype'])) ? htmlspecialchars($_POST['listtype']) : '';
echo '<form method=post action="index.php?operacao=checkList&number=1&pagesize=10&listtype=', $listype ,'" name="listForm">

answered Nov 20, 2013 at 15:01

John Conde

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6 Comments

Funk Forty Niner Over a year ago

So using post as the form method doesn't matter?

John Conde Over a year ago

Actually, that's another issue they have. Let me address that.

Funk Forty Niner Over a year ago

I knew there was something more to just the form method, which is why I didn't want to make it as an actual answer. Questions like these it seems, always opens up a can of worms, as it were.

John Conde Over a year ago

Agreed. Once I see more than two issues in one code snippet I just vote to close as not knowing enough about what they're doing just yet.

Funk Forty Niner Over a year ago

Exactly. The OP has some JS happening javapopup() and probably querying a DB. Next thing we know, risks in coming back to say "It still doesn't work... my info doesn't show up now etc." Yep, see it all too often, and a vote to close is usually what's best to do until we know everything.

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