exit for loop with echo - php

Question

The following PHP code is working as expected. I need to echo "copy error" message when the code is unable to execute the command in 4 attempts.
for($num_tries = 0 ; $num_tries < 4 ; $num_tries++)
{
$cloudcmd = "cp abc xyz ";
system($cloudcmd,$status);
if($status != 0)
{
sleep(3) ;
continue ;
}
break ;
}
I tried to add echo command after break; but it does not seem to work.

Answer 1

Strictly, you don't need to introduce additional variables and checks to do this.
Keeping it simple, you can inspect the value of the for loop variable $num_tries once outside the for loop:
if($num_tries==4)....
You also have the $status variable available after the for loop.
if($status!=0)....

Answer 2

$success = false;
for($num_tries = 0 ; $num_tries < 4 ; $num_tries++)
{
$cloudcmd = "cp abc xyz ";
system($cloudcmd,$status);
if($status != 0)
{
sleep(3) ;
continue ;
}
$success=true;
break ;
}
if($success)
{
//do your thing
}

Answer 3

Try doing this:
for ($num_tries=0;$num_tries < 4;$num_tries++)
{
system('cp abc xyz',$status);
if ($status === 0)
{
break;
}
sleep(3);
}
if ($status !== 0)
{
echo 'error';
}
This loop will break as soon as $status is 0, meaning the system() call was a success, after the loop, the $status value will be accessible, still (PHP doesn't have block scope), so check it's value and echo if needed. You can easily replace the echo statement using an exit(), to stop the rest of the script.

Answer 4

<?php
for($num_tries = 1 ; $num_tries <=4 ; $num_tries++){
$cloudcmd = "cp abc xyz ";
system($cloudcmd,$status);
if($status != 0) {
if($num_tries!=4){
sleep(3);
continue;
} else {
echo "error";
}
}
break;
}
?>

Answer 5

A break statement causes the loop to break, and any code that occurs after the break, within the loop is skipped over.
Perhaps, you could do something like this:
$success = false;
for($num_tries = 0 ; $num_tries < 4 ; $num_tries++) {
if (success()) {
// do something
$success = true;
}
}
echo ($success) ? 'Success' : 'Failure';