I'm creating a app that requires me to run a second php script while the first script is still running.
I'm new to php programing so I'm sure there's a simple function I can use that I'm just not aware of.
Looking forward to any help...
Shane
2 Answers
Since you are new to PHP I'm guessing you're looking for the include/require (and include_once/require_once) language constructs which will execute another PHP script as if it is part of the current script.
Otherwise if you want it to run as a separate process look into exec, shell_exec, or backticks. If you need the other PHP script to run as a background process make sure to redirect stdout somewhere (a file or maybe /dev/null if you don't need it) so that your currently executing script doesn't have to wait for it to finish to continue executing.
8 Comments
Bert
absolutely what i think he wants.
Shane E. Bryan
I think the exec operator will work great, but if I need to pass var. how would I do that? Also will the first script continue or will it wait for a response?
Paul
@ShaneE.Bryan Are you familiar with a linux shell (and running PHP ON Linux)? exec takes in a command if you want to pass arguments to that command you can just append them to the command string separated by whitespace like
'/usr/bin/php ' . $filename to run a php script whose name is stored in the variable $filename (If your php interpreter is located in /usr/bin/php). You probably also want to escape your arguments with escapeshellarg. The script will wait for a response unless you redirect stdout somewhere like: exec('/usr/bin/php ' . escapeshellarg($filename) . ' > /dev/null');Shane E. Bryan
So if I have multiple var. i would code it like this <code> exec('/usr/bin/php . escapeshellarg($filename?var=$var&var2=$var2) . ' > /dev/null'); </code>
Shane E. Bryan
Also if I run <?php phpinfo() ?> which field would include the root dir for php?
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This will actually require us to use some Javascript for an ajax call to execute our PHP and return it's data.
I prefer Jquery, which will look similar to this:
function callPHP(){
$.post('./filetocall.php', {variableid: 'id'}, function (response) {
$("#div_for_return_data").val(response);
});
}
filetocall.php can look like anything. It's output will populate the #div_for_return_data
eg:
<?php echo $_GET['variableid']; ?>
Then just call the Jquery function from anywhere.
exec('php path/to/the/script');