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php progress bar?

Answer 1

A:

The best approach is to have your javascript make ajax calls to the server and find out where it is at, so you can give some feedback to the user.

You can start with having a progress bar for uploading, and then you just continue to show the progress of the application.

James Black 2009-10-22 00:16:41

Answer 2

+2 A:

Use AJAX to ask PHP what it's doing at the moment, and have a queue system that defines what tasks are completed or not. Here's a very basic example:

When the task is started, make it set the following variables:

$_SESSION['isEbayLoaded'] = false;
$_SESSION['isAmazonLoaded'] = false;

Once PHP loads eBay, it would set isEbayLoaded to true, and then AJAX could simply ask PHP if ebay was loaded yet, then amazon, then whatever else you needed. All PHP would have to do is check if those variables were set as true, which would mean that the tasks had been completed.

Bruno De Barros 2009-10-22 00:23:01

Are session variables accessible from the main page without reloading that page? That's something you might want to think about.

iddqd 2009-10-22 18:51:11

The session variables are being read by the page being called through AJAX, so as they're set in the main script, they should become available for requests after they were set. If not, one could still use temp files, or the database, to store the information, which would then be read when the AJAX request is made. To avoid calling PHP, one could even store a file as .json and then use JavaScript to request the file and process the information on what tasks are complete and what tasks aren't.

Bruno De Barros 2009-10-22 19:11:35

Answer 3

A:

I've accomplished this before with a really simple solution:

ob_flush();

All it required was creating a single div with appropriate class names:

<div class="loading">
    Loading...
</div>

Then, my PHP was basically:

foreach ($sites as $site) {
   printAndFlush(<<<CSS
     <style>
       div.loading {
         background: url($site.png) no-repeat right center;
       }
     </style>
CSS;

   scrapeSite($site);
}

This would use a background style to show which site you would be scraping.

If you must use markup, you can simply print out a new "Loading" div each time (with a unique ID), along with a style to set the previous div's style to "display: none".

Of course, remember that some browsers require at least 1024 bytes (somethings 4k!) before rendering anything.

Eric 2009-10-22 04:11:32

Minor detail, but you never close the parenthesis in your PrintAndFlush function. Also, I don't see any reason to print the style tags more than once.

iddqd 2009-10-22 18:52:19

Not an actual function, just a pseudo-php function.The reason the style tags were printed more than once were to force their compilation by the browser, as each <style> declaration overwrote the previous style.

Eric 2009-11-03 16:58:02

This is one of the few approaches that doesn't require AJAX to function.

Eric 2010-01-12 05:24:01

Answer 4

A:

I would make a request for each site, and use the output from that page (XML, JSON, whatever) to let the other page know what's going on.

For example...psuedo code:

gather sites
for loop that loads ajax page
ajax page returns each request with a status message e.g. - "Loading Amazon...."

That's it!

iddqd 2009-10-22 18:54:54

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