believeinsharing

any other option

@ Manixat: Thx for reply I tried your code bt its not working

@SocialCloud: thx for ur quick responce I did created .htaccess file and its has following code RewriteEngine on RewriteCond %(REQUEST_FILENAME) !-d RewriteCond %(REQUEST_FILENAME)\.php -f RewriteRule ^(.*)$ $1.php I got this form google.. I have .htaccess file in my page folder where i have all my php pages Directory of my website is: MainFolder thn it has subfolders for Images, Pages, Styles, and Lib Thx

Hi I am developing one website its in PHP and I am using Eclipse When I click on any page link its shows "www............com/home.php or www............com/blog.php " in address bar where home and blog are my php pages I wanted to hide those php extenstions and would like to see only "www............com/home or www............com/blog" Is it possible and if yes how? Thx in advance for help

I just need to add <br clear="left" /> it break thn seq n div is nw in next line thx for all help

thx for ur reply My code is <h4 class="blogTitle"> <span style="float: right; width:auto;height:auto;">7</span> <a href="blogPage.php?srNo='.$row['srNo'].'">Title</a> </h4> <div id="blogBody"> <span style="float: left; padding:0px 10px 5px 0px;"><img src="C:\Users\Desktop\New folder\img\2.jpg"/></span> <div id="blogText">This is info</div> </div> <div id="jump-link"><a href="blogPage.php?srNo='.$row['srNo'].'"> See more...</a></div> <div id="blogFooter"> <div id="blogFooterLine1"> <span class="postAuthor">POSTED BY:</span> </div> </div> In above code i wanted to display img and div "blogText" side by side and div "jump-link" should be below img but i am case everything is right hand side of img

Hi, I have one one span with id="imgPart" and div with id="infoPart". I am using span to show img and div to display text, If info is more (height is more than img height) page looking good but if text is short than other divs are coming up with img level. I would love to adjust the height of div with height of span(containing ) img. thx in advance

Hi, Any one knows simple way to validate captcha? Thanks

@Psycho: thx for reply, I am going that my contactus.php form has following code: if ($_SESSION['captchaError']==true) { echo "<p id='contErrorMsg' style='display: none;color:red;'>Please fill the required fields</p>"; } ?> and i m setting $_SESSION['captchaError'] value in send.php file when contactus page will reload it will check $_SESSION['captchaError'] value n base on tht will display or not above <p> my que is is their any way to check that page is loading first time or second time. Thanks in advance

I am trying to add captcha to my contact us form.... I did that but I do have prob while validating captcha. As i know captcha validation should be on server side. my code is: <?php session_start(); $captcha=$_SESSION['captcha_code']; $ecaptcha=$_POST['code']; echo "actual captcha :".$captcha ."</br>"; echo "entered captcha:".$ecaptcha; $_SESSION['captchaError']==false; if ($ecaptcha==$captcha) { echo "right captcha entered"; } else { $_SESSION['captchaError']==true; echo "wrong captcha entered"; header('Location:contactus.php') ; } ?> because, I m using header('Location:contactus.php') it showing contactus.php directly without error msg; is their any other way to do this? thanks in advance

I am trying to add online radios on my webpage.... I found some links which can play radios online , I embedded thm to my page but prob is some of them are start automatically once page load. importunately I have 2 radio links which start playing n because of that no is sounds good. I am trying to add following url http://player.streamtheworld.com/liveplayer.php?CALLSIGN=ARNCITY their is no option to pause it. thanks

thanks peipst9lker n beven

is their any way to open a popup window i php page. I want to click on a click and would like to open a page as a popup nt a new full page Thanks

Hi, I am using eclipse for php development, but its very hard to solve php prob. So anyone knows good php debugger for eclipse or anyway to debug php file. Thanks

Hi, I have seen some website pages where lets say; out of 35, first 10 records or forums are coming of first page. On same page they have link for next 10 records and so on. but in this case only some part of page is going to refresh not the whole page. is their any simple way to do this using php. Thanks in advance

thx for reply... I dont have any code right now, what i wanted is If some one add some blog with blog title n blog info. I want to display some text of updated blog on home page with "read more" link

is their any way to display part of article with read more option...

Hi All, I knw how to code basic php page, bt i need logical help... Lets c I have 3 pages n all has "Banner" as command part, So one option is to repeat 'banner' code on every page... but i wanted to avoid repeating code. I thought to write another PHP file only with banner code in it and 'include' that on every page. My que is, should that banner.php has whole html tags like 'html', 'head', 'title', n 'body'. or simply echo all html tags or just add the code. should banner. php file like <html> <head> <title></title></head> <body> <? echo "This is banner Info"; ?> </body></html> or <? echo "<html> <head> <title></title></head> <body> This is banner Info </body></html>"; ?> or simply <? echo "This is banner Info"; ?> or whats if php file has some php n some html code in it Thanks all

Hi, I want to develop a website and working on homepage. I wanted to divide home page into columns (may be using "div"). Now lets say I have more than 7 div but all are not visible at a same time. so what i wanted to do is display 4 div then use "next or > sign" to show next div, at that time present divs will move to the left (and < sign will display) and next divs will display. So its like side show of div, moving from right to left. I dont knw if Jquery will do this or javascript. hope so my statement will clear what i really want to do. Plz let me know if anyone has tutorial regarding to this or any kind of help. thanks in advance

sorry I am new to PHP

Is there any way to store variable in array, I have set for loop in javascript to display textbox on form (textbox may be display from 1 to 5 depending on users requirement) Now I want to add those textbox value to database. is there any way to store that values in array so that I can use loop to run 'insert statement' Thanks

I am using select to display combo box with values from 1 to 5... I wanted to use that value for "for loop" to display no of text box... No of text boxes are depends upon what user has selected in select I am not sure is this purely done by php so what i did is I use javascript to get value n then wanted to use that in php for loop. <div id="divSubMenu" style="display:none"> No of submenu: <select id="noofSubmenu" onchange="printMe()"> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> javascript code is <script type="text/javascript"> function printMe() { var noOfSubmenu =document.getElementById('noofSubmenu').value; //alert ("Selected option is " + noOfSubmenu); } </script> PHP code: <?php echo "No of submenus are:".noOfSubmenu ; ?>

Is there any way to use javascript variable in php code?

@Mahngiel: Thanks for ur quick reply I did change if(!con) to if(! $con) bt no use...

I have simple login form which takes input from user, check that with database n goes to the next page for valid user n password. This is my php code <?php $username =$_POST['txtUsername']; $pass=$_POST['pass']; $hostname="localhost"; $db_user="root"; $db_password="admin"; $database="myproject"; $db_table="login"; $rowno=1; global $usertype; $con=mysql_connect($localhost,$db_user,$db_password); if(!con) { die('Error in connection:'.mysql_error()); } mysql_select_db($database,$con); if($_POST['butSubmit']==true) { $sql="SELECT * FROM $db_table WHERE username='$username' AND password='$pass'"; $result=mysql_query("SELECT * FROM $db_table WHERE username='$username' AND password='$pass'"); $row=mysql_fetch_array($result); if($rowno = mysql_num_rows($result)) { session_start(); $_SESSION['name'] = $row[2]; $_SESSION['usertype']=$row[1]; header('Location:contactUs.php'); } else { echo "Try again"; } } ?> This code was working fine before, today when I tried run this page its doing all the validation n going to the next page but while loading the login page its showing Notice: Undefined index: txtUsername in C:\webserver\Apache\htdocs\myProject\login.php on line 36 Notice: Undefined index: pass in C:\webserver\Apache\htdocs\myProject\login.php on line 37 Notice: Undefined variable: localhost in C:\webserver\Apache\htdocs\myProject\login.php on line 47 Notice: Use of undefined constant con - assumed 'con' in C:\webserver\Apache\htdocs\myProject\login.php on line 48 Notice: Undefined index: butSubmit in C:\webserver\Apache\htdocs\myProject\login.php on line 55 Thanks in advance