Skip parent if using PHP - php

Question

Not sure if this is even possible, but the basic idea of the code is as follows:
if(isset($_POST['submit'])) {
if($_POST['field'] == 0) {
// do not process the code in the if statement
}
// code to process if the above validation criteria is not met
}
I basically want to try and keep it as simple as possible without lots of if's everywhere, just the quick validation statements at the top, and if none of those if statements are triggered, it will process the code to update the database etc.
I have tried the continue statement function with no joy, I believe that only works with loops.
Thanks!

Answer 1

if(isset($_POST['submit']) && $_POST['field'] != 0) {
}
As a function
function testCondition() {
return isset($_POST['submit']) &&
$_POST['field'] != 0;
}
if (testCondition()) { ... }

Answer 2

You could put all of this in a function and then call 'return' from the inside 'if':
if(isset($_POST['submit']) && validate_form())
{
...
}
function validate_form()
{
if($_POST['field'] == 0) {
return false;
}
if(another check that fails) {
return false;
}
...
return true;
}

Answer 3

To accomplish this as you originally desired, you need to use the break statement. From the PHP manual:
$arr = array('one', 'two', 'three', 'four', 'stop', 'five');
while (list(, $val) = each($arr)) {
if ($val == 'stop') {
break; /* You could also write 'break 1;' here. */
}
echo "$val<br />\n";
}
/* Using the optional argument. */
$i = 0;
while (++$i) {
switch ($i) {
case 5:
echo "At 5<br />\n";
break 1; /* Exit only the switch. */
case 10:
echo "At 10; quitting<br />\n";
break 2; /* Exit the switch and the while. */
default:
break;
}
}