PHP Submit on Select - php

Question

I am trying to not have a submit button, is there any way that the form can submit when the user selects a value?
<form method="post" action="<?php echo $_SERVER['PHP_SELF'] ?>" >
<table class="form">
<select name="category" class="formfield" id="category">
<option value="-1"> Category </option>
<?php
$sql_contry = "SELECT * FROM category";
$rs_c = mysql_query($sql_contry);
while ($row_c = mysql_fetch_array($rs_c)) {
echo '<option value="'.$row_c['category'].'">'.$row_c['category'].'</option>';
}
?>
</select>
</table>
</form>

Answer 1

Here an example
<form>
<select name='myfield' onchange='this.form.submit()'>
<option selected>Milk</option>
<option>Coffee</option>
<option>Tea</option>
</select>
<noscript><input type="submit" value="Submit"></noscript>
</form>
Using noscript tag it allows browsers that are not JavaScript-enabled to still function.

Answer 2

Yes - use javascript:
Html:
<form id="frm">
<select onchange="onSelectChange();">
<option>1</option>
<option>1</option>
<select>
</form>
js:
function onSelectChange(){
document.getElementById('frm').submit();
}

Answer 3

<form action="product.php" method="POST">
<select onchange="this.form.submit();" name="prod">
<option value="">Select product</option>
<option value="1">abc</option>
<option value="2">def</option>
<option value="3">ghi</option>
<option value="4">jkl</option>
<option value="5">mno</option>
</select>

Answer 4

Just change code as below, I haven't check code so there may be some error like capital/small letter I am not sure like, it's submit() or Submit() and so on..
<script language="javascript">
function submitForm(){
var val = document.myform.category.value;
if(val!=-1){
document.myform.submit();
}
}
</script>
<form method="post" name="myform" action="<?php echo $_SERVER['PHP_SELF'] ?>" >
<table class="form">
<select name="category" class="formfield" id="category" onchange="submitForm();">
<option value="-1"> Category </option>
<?php
$sql_contry = "SELECT * FROM category";
$rs_c = mysql_query($sql_contry);
while ($row_c = mysql_fetch_array($rs_c)) {
echo '<option value="'.$row_c['category'].'">'.$row_c['category'].'</option>';
}
?>
</select>
</table>
</form>