Asked  7 Months ago    Answers:  5   Viewed   57 times

if i call a php file via jquery ajax, that contains a script to do some stuff that takes a while — for instance uploading a big video — and then I close the page: does the php script keep loading the video or not?

 Answers

80

See here: http://php.net/manual/en/function.ignore-user-abort.php

int ignore_user_abort ([ bool $value ] )

Sets whether a client disconnect should cause a script to be aborted.

When running PHP as a command line script, and the script's tty goes away without the script being terminated then the script will die the next time it tries to write anything, unless value is set to TRUE

There also is a PHP configuration option of the same name: http://php.net/manual/en/misc.configuration.php

By default, if you do nothing, according to the PHP manual the default is to abort the script. http://php.net/manual/en/features.connection-handling.php

NECESSARY UPDATE

It seems I (unknowingly) tricked my way to "reputation points", because I did NOT supply the (correct) answer, but here it is now thanks to testing and continued nudging from "mellamokb":

Quote: "Ok, I took a look at the PHP source code and, if I didn't miss anything, I now have the answer. The "ignore_user_abort" flag is only checked when PHP receive an error trying to output something to the user. So, in my understanding, there is no way to interrupt code which doesn't produce any output."

Okay, I wasn't totally off, but it is important to know that it all depends on whether or not your script produced any output!

If you read THIS, also DO check out the comments below.

Wednesday, March 31, 2021
 
medhybrid
answered 7 Months ago
77

No, it doesn't have one. For this reason most popular libraries come with one in their utility packages. Check out jQuery's inArray and Prototype's Array.indexOf for examples.

jQuery's implementation of it is as simple as you might expect:

function inArray(needle, haystack) {
    var length = haystack.length;
    for(var i = 0; i < length; i++) {
        if(haystack[i] == needle) return true;
    }
    return false;
}

If you are dealing with a sane amount of array elements the above will do the trick nicely.

EDIT: Whoops. I didn't even notice you wanted to see if an array was inside another. According to the PHP documentation this is the expected behavior of PHP's in_array:

$a = array(array('p', 'h'), array('p', 'r'), 'o');

if (in_array(array('p', 'h'), $a)) {
    echo "'ph' was foundn";
}

if (in_array(array('f', 'i'), $a)) {
    echo "'fi' was foundn";
}

if (in_array('o', $a)) {
    echo "'o' was foundn";
}

// Output:
//  'ph' was found
//  'o' was found

The code posted by Chris and Alex does not follow this behavior. Alex's is the official version of Prototype's indexOf, and Chris's is more like PHP's array_intersect. This does what you want:

function arrayCompare(a1, a2) {
    if (a1.length != a2.length) return false;
    var length = a2.length;
    for (var i = 0; i < length; i++) {
        if (a1[i] !== a2[i]) return false;
    }
    return true;
}

function inArray(needle, haystack) {
    var length = haystack.length;
    for(var i = 0; i < length; i++) {
        if(typeof haystack[i] == 'object') {
            if(arrayCompare(haystack[i], needle)) return true;
        } else {
            if(haystack[i] == needle) return true;
        }
    }
    return false;
}

And this my test of the above on it:

var a = [['p','h'],['p','r'],'o'];
if(inArray(['p','h'], a)) {
    alert('ph was found');
}
if(inArray(['f','i'], a)) {
    alert('fi was found');
}
if(inArray('o', a)) {
    alert('o was found');
}  
// Results:
//   alerts 'ph' was found
//   alerts 'o' was found

Note that I intentionally did not extend the Array prototype as it is generally a bad idea to do so.

Wednesday, March 31, 2021
 
radmen
answered 7 Months ago
37

To answer your question directly, see ignore_user_abort

More broadly, you probably have an architecture problem here.

If many users can initiate this stuff, you'll want the web application to add jobs to some kind of queue, and have a set number of background processes that chew through all the work.

Thursday, July 29, 2021
 
LoicTheAztec
answered 3 Months ago
61

You can switch the version here:

1. Press CTRL+ALT+S

2 Search for JavaScript & click the select field. and then select ECMAScript 6

View image.

Sunday, August 22, 2021
 
RajaReddy PolamReddy
answered 2 Months ago
24

Where are you displaying the content is more important. If you are writing it to a spreadsheet, then you have to change the settings of the spreadsheet to the correct language (French). In addition, you can use the advanced parameters to specify what charset the original content is in. For example,

var result = UrlFetchApp.fetch( url , {contentType : 'text/html; charset=utf-8'});
var wholePage = result.getContentText(); 
var scrap = wholePage.match( /<title>(.*?)</title>/ ); 
var title = scrap[1];
Friday, August 27, 2021
 
van_folmert
answered 2 Months ago
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