Using link to submit form into php using $_PHP_SELF and $_POST

Question

I'm trying to get a form to submit from a link rather than an input submit button.

    <form id="formadd" name="formadd" method="post" action="<?php $_PHP_SELF ?>">

    ...form data 

     <a name="ADD" onclick="document.getElementById("formadd").submit();" href="" class="button primary">Add Camera to Database</a>
     <!-- <input type="submit" name="ADD" value="Add Camera"> -->

The PHP that gets called looks like so:

<?php 
            if(isset($_POST['ADD'])) {
                $SIGID = $_POST['SIGID'];
                $LOC = $_POST['LOC'];
                $URL = $_POST['URL'];
                $IMG = $_POST['IMG'];
                $LAT = $_POST['LAT'];
                $LON = $_POST['LON'];
                $CAMTYPE = $_POST['CAMTYPE'];



                $sql = "INSERT INTO cam_markers (SIGID, LOC, URL, LAT, LON, CAMTYPE, IMG) 
                   VALUES ('$SIGID', '$LOC', '$URL', '$IMG', '$LAT', '$LON', '$CAMTYPE')";
                $retval = mysql_query($sql);

                if(! $retval ) {
                   die('Could not add data: ' . mysql_error());
                }
                echo 'Added data successfully' . PHP_EOL;
                mysql_close($connection);
            }
?>

The link currently doesn't work, but it works using a normal submit input, so I know the script is okay.

My best guess was that the problem had to do with the fact that $_POST only creates an array from form elements, and since the hyperlink is not a form element, its name identifier is not getting passed.

But, I tried it without the isset() conditionals and just tried to echo the data that was being passed, which did not work. So, that seems to indicate something else is going on.

Any help/workaround is appreciated. Thanks.

EDIT: I got it working by doing this: <a href="javascript:document.formadd.submit();">

You don't need to specify action if form is being submitted on the same page itself but if you want to use $_SERVER['PHP_SELF'] ; — Just_Do_It, Nov 07 '16 at 21:38
If all you want is a variable to test with `isset()`, then you can add a `` — AbraCadaver, Nov 07 '16 at 21:50
However, you are in danger: http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php?s=1|38.9724 — AbraCadaver, Nov 07 '16 at 21:51
I changed to single quotes but it still isn't working. I'm not sure why not either. Seems like it should work. — Auz Beauxdine, Nov 07 '16 at 22:18

Just_Do_It · Answer 1 · 2016-11-07T21:47:25.990

1

action="<?php $_PHP_SELF ?>">

is wrong. You don't need to specify action if form is being submitted on the same page itself but if you want to use $_SERVER['PHP_SELF'] ; so now

action="<?php echo $_SERVER['PHP_SELF'];?>">

edited Nov 07 '16 at 21:47

answered Nov 07 '16 at 21:41

Just_Do_It

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score 0 · Answer 2 · answered Nov 07 '16 at 21:30

0

Use a submit button.

If you don't like the way it looks (the most likely reason for you to try to avoid using the right tool for the job), apply CSS.

answered Nov 07 '16 at 21:30

Quentin

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I realize that's most straightforward way. But my workflow involves using frameworks to quickly prototype, and I'd prefer to not have to go and edit the sass/css if I can help it. Thanks anyway. – Auz Beauxdine Nov 07 '16 at 21:36

score 0 · Answer 3 · answered Nov 07 '16 at 21:33

You should use a submit button.

button {
  background: none!important;
  border: none;
  padding: 0!important;
  font-family: arial, sans-serif;
  color: #069;
  text-decoration: underline;
  cursor: pointer;
}

<button type="submit">your button that looks like a link</button>

score -1 · Answer 4 · answered Nov 07 '16 at 21:31

A couple of things... If action is blank, the browser submits to the current page, so

action=""   fixes it.

You can use the PHP_SELF if you like, but it's not getting printed so either

action="<?php echo $_PHP_SELF; ?>"

or, the shorthand syntax

action="<?=$_PHP_SELF?>"

Finally your code looks for $_POST['ADD'] so create a hidden one within your form..

<input type="hidden" name="ADD" value="ADD" >

score -1 · Answer 5 · answered Nov 07 '16 at 21:35

-1

The technique you used is correct, you can use a link and .submit(). The only problem I see is the incorrect quote:

<a name="ADD" onclick="document.getElementById("formadd").submit();"
                                               ^       ^

Use single quotes or escape it.

onclick="document.getElementById('formadd').submit();"

answered Nov 07 '16 at 21:35

sidyll

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Keep in mind this implies that for this to work your final user must have javascript enabled. The styled standard submit button does not. – sidyll Nov 07 '16 at 21:37

Using link to submit form into php using $_PHP_SELF and $_POST

5 Answers5