Thank you for your great help and time you spent for me. But I found my mistake that in the style tag, I don't know exactly, there is one style iframe {display:none;}
That is why it is not showing FCKEditor. Because FCKEditor is using iframe.
I am using FCKEditor in my website. It's working all over in website fine.
But there is one page in which it does not showing properly. I tried all things but in that page is does not showing properly.
I have one hidden div tag. When user clicks on "Edit Description" link from Edit button, hidden div tag is shown but FCKEditor is not getting shown there. Please give me some tips.
Select mytext from tblhebrew where mytext='שאלות נפוצות'
When I run this query directly in phpMyAdmin then it's working perfectly. But when I run same query using php script then it does not working. Why? Please see this is my php script code...
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html dir="ltr" lang="he" xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Home Page</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8"/>
</head>
<body>
<?php
$link= mysql_connect("localhost","root","");
mysql_select_db("dbhebrew");
//mysql_query("insert into tblhebrew (test,description) values ('עדיין לא רשום ? הרשם עכשיו ','עדיין לא רשום ? הרשם עכשיו ')");
$txt="שאלות נפוצות";
$query="Select mytext from tblhebrew where mytext='".mysql_real_escape_string($txt)."'";
echo $query;
$result=mysql_query($query) or die(mysql_error());
$row=mysql_fetch_assoc($result);
echo "output:".$row['mytext'];
?>
</body>
</html>
I want help to generate www.bestuff.com like paging in my application because my client want same thing and in today's era this thing is found everywhere.
I searched a lot but still could not found help.
Here I am providing link to that paging so that you can check it.
I want to know that how can we compare two hebrew string in MySQL database?
I tried and search a lot on net but still could not found the answer.
My website is dealing with Hebrew(main) and English as well.
So please let me know How can I compare these two languages in SQL queries. Because this problem is became headech for me now. My client is becoming engry and engry and engry..... on me.
Is there any solution for this?
Please help me out.
I am waiting for your reply. Please do it fast as soon as possible.
amit if you have set the session time for 10 minute.so then it does't matter that user close the browser. session will automaticaly destroyed after 10 minute.
Dear Freddavis,
You all right that when browser get closed all session varialbes will be destroyed.
My main and serious problem is how to Update table when browser get closed?
I am finding this answer since long time and have visited most websites and forums but could not found solutions yet.
Is there anybody who can solve this? Is there any expert in daniweb who can solve my problem?
See my exact problem 1. visit http://bestuff.com 2. At the last there is section of "Active Members", in which it shows top 40 to 44 users who are online. 3. When I get logged in, I will be shown on that list because I just logged in last. 4. When I close browser without logging out force fully then my profile image will be removed from that section.
If you have any other solution for that please help me that how to do?
I am following the bestuff.com's active member session. I am doing the following thing... 1. when user logs in, I update my database's field with value 'O' (means user is logged in and fires query which has flag value 'O')
But the problem is how to check user's status when User closes browser?
I set 10 minutes session time out but it only works when user refreshes or visits the page.
Please help me out of this... hell...
Please reply me fast and please admin section also take part in this query/problem...
Thanks in advance
index.php ----------
session_start();
if (isset($_SESSION['LAST_ACTIVITY']) && (time() - $_SESSION['LAST_ACTIVITY'] > 600))
{
// last request was more than 30 minates ago
//session_destroy(); // destroy session data in storage
//session_unset(); // unset $_SESSION variable for the runtime
header("location:logout.php");
}
$_SESSION['LAST_ACTIVITY'] = time(); // update last activity time stamp
logout.php ----------
session_start();
include_once 'commoninclude.php';
$objLogin=new MySQLConnection();
$objLogin->Open();
$updateMemberOnlineStatus="Update tblMemberRegistration Set OnlineStatus='F' Where ID=".$_SESSION['memberid'];
$objLogin->UpdateQuery($updateMemberOnlineStatus);
$objLogin->Close();
session_unset();
session_destroy();
header("location:index.php");
Dear Friend, Please go to this link http://w3schools.com/php/php_ajax_rss_reader.asp It has explained how to read RSS feed with very easy example and working successfully. You can visit my site in which I have used that example here http://sabsyp.com/rss/simple.php It may take few seconds to load becaused it fetches RSS feed from most busy websites called Google and MSN.
Here I am showing you how to Redirect from Window Application to Web Application from VB.NET or C#.NET.
The below code is from my VB.NET application. Just copy and paste the code into your application and you will see the magic.
Private Sub sitelink_LinkClicked(ByVal sender As System.Object, ByVal e As System.Windows.Forms.LinkLabelLinkClickedEventArgs) Handles sitelink.LinkClicked
Dim openBrowser As New System.Diagnostics.Process
openBrowser.EnableRaisingEvents = False
openBrowser.StartInfo.FileName = "iexplore"
openBrowser.StartInfo.Arguments = http://www.themacrosoft.com"
openBrowser.Start()
End Sub
Can anybody tell me how to use Web Services using PHP? I want to do following things in my website: 1. I want to use RSS feed in my website. 2. Want to display latest news from other sites who are providing this facility. 3. Want to show temperature from other sites.
Please make this fast as soon as possible.
I work with OOPs. So if possible please make it with class. Thank you.
In the below HTML and JAVASCRIPT code, nice work has been done. In JavaScript first we are finding the "check_list" array and then assigning it to out array variable and then selecting multiple checkboxes.
<html>
<head>
<title>CheckAll and UnCheckAll</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<script LANGUAGE="JavaScript">
function CheckAll()
{
var a=new Array();
a=document.getElementsByName("check_list[]");
//alert("Length:"+a.length);
/*var p=0;
for(i=0;i<a.length;i++){
if(a[i].checked){
alert(a[i].value);
p=1;
}
}*/
for (i = 0; i < a.length; i++)
a[i].checked = true ;
}
function UnCheckAll(chk)
{
var a=new Array();
chk=document.getElementsByName("check_list[]");
for (i = 0; i < chk.length; i++)
chk[i].checked = false ;
}
</script>
</head>
<body>
<form name="myform" action="checkbox1.php" method="post">
<b>Scripts for Web design and programming</b><br>
<input type="checkbox" name="check_list[]" value="1">ASP<br>
<input type="checkbox" name="check_list[]" value="2">PHP<br>
<input type="checkbox" name="check_list[]" value="3">JavaScript<br>
<input type="checkbox" name="check_list[]" value="4">HTML<br>
<input type="checkbox" name="check_list[]" value="5">MySQL<br>
<input type="button" name="Check_All" value="Check All" onClick="CheckAll();">
<input type="button" name="Un_CheckAll" value="Uncheck All" onClick="UnCheckAll();">
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>
Just to take note that there is no Foreign key in MySQL unless your database's type is InnoDB. I assume you are using MyISAM.
Hope it helps.
Hi Friends,
I also have read in many websites that we cannot create foreign key in MyISM type database or tables. But It is possible to create foreign key in MyISM type database or tables. Here I am submitting two tables and the relationship between them using foreign key concept. So do practice in your maching also. Two tables 1. Country 2. State
SQL for Country: Create Table Country(Id int(2) Primary Key Not Null Auto_Increment, CName varchar(10))
SQL for State: Create Table State (Id int(2) Primary Key Not Null Auto_Increment,SName varchar(10),cid int(2),foreign key (cid) references country(id))
Try, it is working successfully in my machine (laptop). Cheers :)
