solon

Add the following lines to your 'status.php' file: $month = $_GET['month']; $day_num = $_GET['day']; $year = $_GET['year']; this should work

Hey guys i know this might have been posted before but i couldnt find anything. My problem is with fade in fade out jquery effect. Here is the code: <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.min.js"></script> <script type="text/javascript"> function change(x) { if(x == '1') { $('#show').fadeOut(1000, function() { $('#show').innerHTML = '<div id="righttexttext"><p>We come at your place and discuss your ideas in order to </p><p>understand what you want to do. And it&#8217;s <strong>FREE!</strong> </p> <p>After all, it&#8217;s your website you should be happy with it.</p></div><div id="mangraphic"></div><div id="righttext2text"> <p> What is the <strong>Purpose </strong>of your website?</p> <p> What are the <strong>Goals </strong>you want your website to achive?</p> <p> What is your <strong>Target Audience</strong>?</p> <p> What kind of <strong>Content </strong>your clients want to find in your</p> <p> website?</p></div>'; document.getElementById("selectedbggraphic").style.top = '541px'; document.getElementById("1").style.color = "#292828"; document.getElementById("4").style.color = "#ffffff"; document.getElementById("3").style.color = "#ffffff"; document.getElementById("2").style.color = "#ffffff"; document.getElementById("5").style.color = "#ffffff"; }); $('#show').fadeIn(100); } else if(x == '2') { $('#show').fadeOut('slow', function() { $('#show').innerHTML = '<div id="righttexttext"><p>We come at your place and discuss your ideas in order to </p><p>understand what you want to do. And it&#8217;s <strong>FREE!</strong> </p> <p>After all, it&#8217;s your website you should be happy with it.</p></div><div id="mangraphic"></div><div id="righttext2text"> <p> What is the <strong>Purpose </strong>of your website?</p> <p> What are the <strong>Goals </strong>you want your website to achive?</p> <p> What is your <strong>Target Audience</strong>?</p> <p> What kind of <strong>Content </strong>your clients want to find in your</p> <p> Website?</p></div>'; document.getElementById("selectedbggraphic").style.top = '580px'; document.getElementById("2").style.color = "#292828"; document.getElementById("4").style.color = "#ffffff"; document.getElementById("3").style.color = "#ffffff"; document.getElementById("5").style.color = "#ffffff"; document.getElementById("1").style.color = "#ffffff"; $('#show').fadeIn('slow'); }); } } </script> When i call the function change() , fadeIn() and fadeOut() works exactly like hide() , show(). (no animation of fading) It might be something stupid but i cant find it Thanks for your help...

Thank you, i will try to fix it based on your inputs!

Thanks Mchl, I think i got the idea and actually found the problem... The table chatset is latin for some reason! Thanks again.

i use utf-8 as the encoding but it still stores it like that. I have this in all my files. <meta http-equiv="content-type" content="text/html; charset=utf-8" /> Am i doing something wrong?

The greek characters are stored like that(htlml entities) from a php mysql script. Is there a way to do it differently? Solon

Using php this would go like this: $original= "New York, NY 10011"; $after = explode(",",$original); //seperate by comma echo $after[0]; //will give you New York echo $after[1]; //will give you NY 10011 Since you want to get the second one seperated as well you should do: $again = explode(" ",$after[2]); //seperate by space For your second set use the same as the first one you see above.

Hey guys, I have a problem with searching greek characters in mysql using php script. In the database everything(in greek) is stored as html entities(i.e. α). Now, when i try to search from php like this $term = $_POST['search']; $fetch = mysql_query("SELECT `text`FROM pages INNER JOIN content ON (pages.page_id = content.page_id) WHERE frontpage = '1' and `pages`.`page_title_gr` like '%".$term."%' or `content`.`text` like '%".$term."%'"); the characters are displayed as normal greek letters (i.e. γεια) and i get no results. What might be the problem? Thanks in advance

Thanks for your reply PFMaBiSmAd, That was actually my problem... i fixed it using $_SERVER['HTTP_HOST'] instead of the actual address so it gets the url used by the user on the URL bar. Thank you very much i would never guessed that was the problem..

Hey guys... i have a login page and i use session id to keep track of what the user did before he logged in so i can carry his actions after he log's in. When i login the 1st time the session_id changes automatically... the second time it doesnt change and everything works. Does anyone had this problem before? if yes what might be the solution... Thanks in advance Solon

can you post your code so we can figure out where the problem is?

Where exactly do you request the change of the content?

Hey guys... this is the script i have: <script type="text/javascript"> function remove(x) { $.blockUI({ message: '<h1>Item is being removed from your Cart...</h1>' }); $('#s_cart').load('remove.php',{ pr : x, success: setTimeout("$('#shop_cart').load('thecart.php')",2000) //this one give me the wrong result }); setTimeout($.unblockUI,3000); } </script> As i understand from Firebug it actually executes after the first is done but i have a problem... The thing is that i am just loading the result of a php file that when it runs directly(not through jquery call) it gives me correct result ( i ask for the items in cart at that time). If i go through the jquery call it gives me an empty result. I am not sure if it's the timing the function is called or something else. If you have any advise please let me know Thanks in advanced, Solon

could you explain that a bit more please isedeasy?

Guys i found the solution... Sorry for taking your time.. The problem was when i was calling the function from <option onclick="function(x)"></option> instead of <select onchange="function(x)"></select>

You can create a jquery function that loads the result in the div you want. it would be something like the following: <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.2.6/jquery.min.js"></script> //Include the jQuery Library <script type="text/javascript"> function load_cart(x) // Create your function with a value to recognize who's cart is it... lets say the customer id value as a parameter { $('#cart').load('cart_cont.php',{ //#cart is the id of the div (or any element) you want to load the result s_id : x //s_id:x - s_id is the value you will get with $_POST['s_id']; in your php file }); //x is the parameter you send to the function when calling it i.e. load_cart(2) } //cart_cont.php should have your php code that returns the result </script>

Hey guys, I need some help cause i am really confused here... I have two AJAX functions (jQuery) the same structure(exact same structure) in two different files. I am using AJAX with PHP. when i use the one it works in all browsers(IE, Firefox, Chrome). The second AJAX function works only in Firefox. This is the first Function with the .php file that calls : <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.2.6/jquery.min.js"></script> <script type="text/javascript"> function ins(x) { $('#show').load('temp1.php',{ pr : x }); } </script> The temp1.php file: <?php include ("db.php"); $pr = $_POST["pr"]; $get_p = mysql_query("query") or die (mysql_error()); $gp = mysql_fetch_array($get_p); $qry = mysql_query("query2") or die(mysql_error()); $ch = mysql_query("query3") or die (mysql_error()); $x=0; while($ch1 = mysql_fetch_array($ch)) { $pr = $ch1[1]; $id = $ch1[0]; $x +=1; echo "<table cellspacing='2' width='350' bgcolor='#333333' border='0' > <tr> <td colspan = '2'>$pr</td> <input type='hidden' id='$x' value='$id'/> <td>"; ?> <input type='button' value='X' class='remove' onclick='rem_one(document.getElementById("<?php echo $x; ?>").value)' /></td></tr> <?php } ?> Now the second function with the .php file that works only in Firefox: <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.2.6/jquery.min.js"></script> <script type="text/javascript"> function show_cat(x) { $('#cat').load('get_cat.php',{ s_id : x }); } </script> <select name="section"> <option value="Select">Select Section</option> <?php $chs = mysql_query("query") or die (mysql_error()); while($chs1 = mysql_fetch_array($chs)) { echo '<option onclick="show_cat('.$chs1[0].')" value="'.$chs1[0].'">'.$chs1[1].'</option>'; } ?> </select> The get_cat.php file: <option value="Select">Select Cat</option> <?php include("db.php"); if(isset($_POST['s_id'])) { $s_id = $_POST['s_id']; $chc = mysql_query("query1") or die (mysql_error()); while($chc1 = mysql_fetch_array($chc)) { echo '<option onclick="show_br('.$chc1[0].')" value="'.$chc1[0].'">'.$chc1[1].'</option>'; } } else { echo '<option value="Select">Error</option>'; } ?> Thanks for any help Solon

thats the problem i am not. i checked every single line and there is nowhere "test" written.

The reason i did not post the code is because i dont know what part of the code produces the error. There are 4 files included in 1 and over 3000 lines of code. Although the only part that has to do with actually connecting to the db is <?php $conn=mysql_connect('localhost','test2','password'); if (!$conn = mysql_connect('localhost', 'test2', 'password') ) { die(mysql_error()); } if (!mysql_select_db('dbname',$conn) ) { die('Could not select database'); } ?>

Hey guys , i get the following error and i have no idea why and how to fix it! There is no 'test'@'%' registered I should say that "test" is the username i had previously on my database. Now i changed that to "test2" and on a different server. I check very carefully on every page that had these info and changed it from "test" to "test2". Any help would be great! Thanks in advance, Solon

Hey guys , i get the following error and i have no idea why and how to fix it! There is no 'test'@'%' registered I should say that "test" is the username i had previously on my database. Now i changed that to "test2" and on a different server. I check very carefully on every page that had these info and changed it from "test" to "test2". Any help would be great! Thanks in advance, Solon

Ok, if i got that right you want to do something like in google's first search page... In order to do that you need to use AJAX in combination with PHP. That is not so hard. Do you have any available code to see and help you on that?

yes when i said ssh i ment shell commands.And that helped 'thrope' thnx.

hey guys i recently got a new linux server and i need help to add a directory into open_basedir . I have no idea about ssh so i would really appreciate any help! Thanks in advance

ok i will check my code but i should tell you that nothing has changed in the code. Yesterday on the old server worked ok and with no changes on the code only the transfer on the new server caused the problem. The reason i am asking here and didnt go to my code is that i have a joomla website on as well and i get the same problem.