scootstah
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Everything posted by scootstah
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You can't. And even if you did, if an attacker could gain access to your filesystem all he has to do is include whatever method you used to encrypt it and he has full database access.
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Well, the code you provided isn't going to output that. So, assuming the object looks like this: $object = (object) array('id' => 5, 'stand_by' => 'test', 'problem' => 'test'); Then you would get "id" by doing $koid = $object->id;
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All you need is this: $lastyear = date('Y', strtotime('-1 year'));
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You need Javascript for this. Working example using jQuery here: http://jsfiddle.net/TxLxR/
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I would use an AJAX request that refreshes every 5 - 10 seconds or so and calls your script.
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I just use my mobile browser (SkyFire) to browse the forums.
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I think it should be $deleteIDs_ary = array_filter($deleteIDs_ary);
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Heh... Yeah, like you said, that's what IT people are for. I used a tutorial to setup my local Postfix, typing in a bunch of config commands in terminal going "no idea what that does, no idea what that does..."
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I absolutely hate setting up Postfix/dovecot. No matter what I do, something ALWAYS goes wrong Yeah, it took me way too long to configure Postfix to simply send an email to my Gmail for local development (sending activation emails and such). If I had to set one up for use as like an email service? F that.
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What is the URL you are using to try and get the profile?
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HUH???????????????? Why would you want to return to "add_comment.php" when you are on "add_comment.php"??? Debbie Because you go to add_comment.php, aren't logged in so you get redirected to login.php, and then get redirected back to add_comment.php after logging in. Making assumptions that the user wants to add a comment just because they logged in while reading an article is a little weird. Don't make assumptions, do things that seem natural and expected.
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Select form to display data from MySQL not working when using AJAX
scootstah replied to webguync's topic in PHP Coding Help
Change var id = $('select[name="category_id"] option:selected').text(); To var id = $('select[name="category_id"] option:selected').val(); -
On add_comment.php set $_SESSION['returnToPage'] to add_comment.php
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Why can't you just set the session when they go to add_comment.php?
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Select form to display data from MySQL not working when using AJAX
scootstah replied to webguync's topic in PHP Coding Help
He already is. ;P -
Select form to display data from MySQL not working when using AJAX
scootstah replied to webguync's topic in PHP Coding Help
Actually, you can ditch that entire function, as well as the onchange bit on the select with this: $(function(){ $('select[name="category_id"]').change(function(){ var id = $('select[name="category_id"] option:selected').text(); $.ajax({ url: 'Query.php?category_id=' + id, success: function(response){ $('#DataDisplay').html(response); } }); }); }); -
Readable Programming Language to Create Audio Software
scootstah replied to Glese's topic in Miscellaneous
http://www.cplusplus.com/doc/ http://www.cplusplus.com/reference/ http://docs.oracle.com/javase/6/docs/ http://docs.oracle.com/javase/6/docs/api/ -
Select form to display data from MySQL not working when using AJAX
scootstah replied to webguync's topic in PHP Coding Help
really, I would recommend that you switch to using the AJAX api with Jquery, much easier and less coding. Agreed. Here's an example: $.ajax({ url: "Query.php?category_id=" + str, success: function(response){ $('#DataDisplay').html(response); } }); -
Readable Programming Language to Create Audio Software
scootstah replied to Glese's topic in Miscellaneous
C++ and Java both have pretty good documentation. As well as huge communities with tons of resources, so you shouldn't ever feel abandoned with either platform. -
Select form to display data from MySQL not working when using AJAX
scootstah replied to webguync's topic in PHP Coding Help
Of course, because the id is supposed to be sent as a query string. What does the AJAX response look like? xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { alert(xmlhttp.responseText); -
Select form to display data from MySQL not working when using AJAX
scootstah replied to webguync's topic in PHP Coding Help
So how does your query look now? -
Select form to display data from MySQL not working when using AJAX
scootstah replied to webguync's topic in PHP Coding Help
You're missing a double quote. <select name="category_id" onchange=showPlayers(this.options[this.selectedIndex].value);"> Should be: <select name="category_id" onchange="showPlayers(this.options[this.selectedIndex].value);"> -
Select form to display data from MySQL not working when using AJAX
scootstah replied to webguync's topic in PHP Coding Help
function showPlayers(str) { alert(str); -
.text selects the text inside the option, IE <option>HERE</option> .value selects the value of the option, IE <option value="HERE"></option>