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();">
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.
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" >
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></div>
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();"
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'];?>">