Bachu

Try this $array_one = array('Champlain Valley TTC(1/31/2015)','Old Shelburne Town School5420 Shelburne RdSherburne, VT 05482','Friday - 6:30 - 10:00PM','Jack Devine802-877-3341','<-----NEXT_CLUB ----->','Green Mountain Table Tennis Club(GMTTC)(6/30/2014)','Knights of Columbus - Boys & Girls Club Gymnasium21 Merchants RowRutland, VT 05701','Wednesday - 6:00 - 10:00PM and 2 Fridays(6:00 - 10:00PM) & 2 Saturdays(7:00AM - 9:00PM) …

Did you dry just print the value directly like `<img id="imgs" src="../images/verificarimg/img.jpeg" width=700>` ? Then, If you don't get the image, please check the path of image/name of image/type of image(jpeg,jpg,png etc)

Try this [Click Here](http://www.9lessons.info/2009/09/pagination-with-jquery-mysql-and-php.html)

This link will help, Please check [Click Here](http://www.bageshsingh.com/bagesh-blog/2010/03/hdfc-bank-payment-gateway-code-hdfc-payment-gateway-php-code/)

Can you show me your code ? For reference [Click Here](http://www.9lessons.info/2009/09/php-login-page-example.html)

Did you give any height to this div in your CSS ? If yes, please remove; please check [Click Here](http://stackoverflow.com/questions/1709442/make-divs-height-expand-with-its-content)

Try this function setMyValue ($myValue) { $newValue = (int)$myValue; $myArray = array_reverse(str_split($newValue)); $myValue =""; foreach ($myArray as $key=>$myDigits) { $myValue.=($key>0 && $key%2==0) ? ":" : ""; $myValue.= $myDigits; } return strrev($myValue); } echo setMyValue(91819.0);

Check this [Click Here](http://www.91weblessons.com/php-ajax-country-state-city-drop-down/)

facebook will crawl your website URL for to get the title, description, and thumbnail. when you test with localhost, change the the URL http://localhost/ to http://127.0.0.1/.

You can set this in your CSS,like textarea { overflow-y: scroll; overflow-x: hidden; }

Try this <?php $yt_url="http://www.youtube.com/watch?v=ZvzwthHh-IQ"; function get_youtube_id_from_url($url) { preg_match("/^(?:http(?:s)?:\/\/)?(?:www\.)?(?:youtu\.be\/|youtube\.com\/(?:(?:watch)?\?(?:.*&)?v(?:i)?=|(?:embed|v|vi|user)\/))([^\?&\"'>]+)/", $url, $matches); return $matches[1]; } $variable= get_youtube_id_from_url($yt_url); echo "<div align='center'><iframe src=\"//www.youtube.com/embed/" . $variable. "\" style=\"background: #fff;\" frameborder=\"0\" height=\"300\" scrolling=\"auto\" width=\"480\"></iframe></div>"; ?>

Check the below link [Click Here](http://www.daniweb.com/web-development/php/threads/466125/upload-and-increment-filename)

Hai; Try this <form name="form1" method="post" action="#"> <label for="country">View outlets in which country? </label> <select name="country_chosen" onchange="document.form1.submit()" id="country"> <option value="">All Middle Eastern Countries</option> <option value="Iraq">Iraq</option> <option value="Kuwait">Kuwait</option> <option value="Bahrain">Bahrain</option> <option value="Saudi">Saudi Arabia</option> <option value="Qatar">Qatar</option> <option value="Oman">Oman</option> <option value="Yemen">Yemen</option> <option value="Jordan">Jordan</option> <option value="Israel">Israel</option> </select> </form> <br> <?php $country_selected = ''; $country_selected …

http://www.mamp.info/en/index.html

Change the above code to this <td><input type="text" name="c1" value=" <?php if($p_id!="")echo $a; else echo $p_name; ?>" ></td>

Check this link [Click Here](http://stackoverflow.com/questions/855577/br-tags-appearing-in-textarea-output)

Try this <?php $result = mysql_query("SELECT DATE_FORMAT(`joindate`,'%m/%Y') AS MonthYear FROM member GROUP BY YEAR(joindate), MONTH(joindate);"); while($row = mysql_fetch_array($result)) { echo '<h2>'.$row['MonthYear'].' </h2>'; $resultTwo = mysql_query("SELECT * FROM member WHERE DATE_FORMAT(`joindate`,'%m/%Y') = '{$row['MonthYear']}';"); while($rowTwo = mysql_fetch_array($resultTwo)) { echo '<p>Member: '.$rowTwo['name'].' </p>'; echo '<p>Email: '.$rowTwo['email'].' </p>'; echo '<p>Joindate: '.$rowTwo['joindate'].' </p><hr />'; } …

I tested your above code in my local pc. It works fine.Make sure that "$data['content']" having contents or check your database table structure.

Try this <?php //table Name $tableName = "MyTable"; //database name $dbName = "MyDatabase"; $conn = mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db($dbName) or die(mysql_error()); //get the first row fields $fields = ""; $fieldsInsert = ""; if (($handle = fopen("test.csv", "r")) !== FALSE) { if(($data = fgetcsv($handle, 1000, ",")) !== FALSE) { …

Check this link [Click Here](http://www.php.net/manual/en/class.simplexmlelement.php)

Try this [Click Here](http://designwoop.com/2012/07/tutorial-coding-a-jquery-popup-modal-contact-form/)

Hai; Make sure that your actual database name is **my_db**

Hai; Try this <?php include("config.php"); function runQry ($qry){ $rows = array(); $result = mysql_query($qry); if($result) { while($row = mysql_fetch_object($result)) { $rows[] = $row; } } else { $rows = null; } return $rows; } function getAllMenus ($level=0) { $menuList=array(); $sql="SELECT * FROM `tbl_structure` WHERE `PARENTID` = '$level'"; $mainMenus=runQry($sql); foreach($mainMenus as …

Hai; Try this <script src="http://code.jquery.com/jquery-1.8.1.js"></script> <?php //require_once("connect.php"); // DELETE if(isset($_POST['delete'])) { echo '<h1>Posted Value ='.$_POST['delete'].'</h1>'; } ?> <form name="form1" id="formId" action="" method="post"> <input name="delete" type="hidden" value="Deleted" /> <table border=0; cellpadding="1" cellspacing="1" bgcolor="" align="center" > <div style=text-align:center;margin-top:20px;> <input type="button" name="delete" value="Delete" onclick="formConfirm()"> </div> </table> </form> <script type="text/javascript"> function formConfirm() { if(!confirm("Do …

Try the following example: Create a folder in your root, Here we named it as upload.In this folder we are uploading text file. In you form html <html> <body> <table align="center"> <form action="upload_page.php" method="post" enctype="multipart/form-data"> <label for="file">Filename:</label> <input type="file" name="file" id="file"><br> <input type="submit" name="submit" value="Submit"> </form> </table> </body> </html> And …

Check this [Click Here](http://stackoverflow.com/questions/19452363/mysql-inner-join-2-tables-to-match-user-number-to-user-name)

Hai; This code also works <body> <style type="text/css"> table, th, td{ margin: 0; font: 16px helvetica, sans-serif; border: 1px solid black; } tr { display: block; } </style> <form name="teacher" method="post" action="teacher.php"> <table> <tr id="name"> <td> <label for="first_name">First Name</label> </td> <td> <input type="text" name="first_name"> </td> <td> <label for="last_name">Last Name</label> </td> …

Do the following changes <?php global $con; require_once("connect.php"); $productValue=""; $productWeight=""; $productCode = ""; if(isset($_POST['button1'])) { // Get values from form $sno =$_POST['txtsno']; $record_check ="SELECT * FROM test WHERE sno = '$sno' "; $result=mysqli_query($con, $record_check); $row = mysqli_fetch_array($result); if(!$row) die ('No record Found'); else { $productValue = $row['packing']; $productWeight = $row['weight']; …

I think, You assign the result to an array, and you try to use the array as an object. Check this link [Click Here](http://www.daniweb.com/web-development/php/threads/462518/codeigniter-foreach-trying-to-get-property-of-non-object).

Try this $username = (isset($_POST['Username'])) ? $_POST['Username'] : "";

You Can do this with AJAX Check the below example : Your form page (formpage.php) <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>BROWSE DATA</title> <script src="http://code.jquery.com/jquery-1.10.1.min.js"></script> <style type="text/css"> html { overflow:auto; } body { background-color:#FFFFFF; margin:0 auto; } #mypopup2 { float: …

<?php function hello() { print "<h1>HELLO!</h1>"; print "<p>Welcome to my web site</p>"; } hello(); ?>

Hai; You are forgotten to echo the option value. Please change this <option value="<?php $x['id']?>" name="<?php $x['id']?>"><?php echo $x['display_name']?></option> into <option value="<?php echo $x['id']?>" name="<?php echo $x['id']?>"><?php echo $x['display_name']?></option> If you want to print this result on same page.. please chage this <form action="post.php" method="post"> into <form action="" method="post">

@tqmd1 You can do this using AJAX POST. (You don't change html to php). Try the below code for index.html <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Connection Checking</title> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"> </script> </head> <body> <center> <h1>Connection Checker</h1> <fieldset style=width:300px;background-color:#E0FFFF> <legend sytle=font-size:12;>Click …

Try this <?php include("connect.php"); $check_insert_query = "SELECT * FROM ghee WHERE sno='18'"; $check_insert_result = mysql_query($check_insert_query); $num_rows = mysql_num_rows($check_insert_result); if($num_rows>0) { die ("<p>Current SNO=18 already exists</p>"); } else { $insert_query = "INSERT INTO ghee (sno,pack,weight) VALUES (18,'kk',80)"; $insertion_result = mysql_query($insert_query); //check whether the data insertion was successful if(!$insertion_result) { die("Sorry! Something …

Check the below link [Click Here](http://phpsnips.com/4/Simple-User-Login#.UnnH4RDpyFw)

Try this <?php $dates = array("1/28/13","1/29/13","1/30/13","1/31/13"); foreach($dates as $myDate) { $dueDate = date('n/j/y', strtotime("+1 months", strtotime($myDate))); $myDateMonth = date ('n',strtotime($myDate)); $dueDateMonth = date('n', strtotime($dueDate)); $myDateYear = date ('Y',strtotime($myDate)); $dueDateYear = date ('Y',strtotime($dueDate)); if(($dueDateMonth - $myDateMonth)>1) { if($myDateYear==$dueDateYear) $nextMonth = $myDateMonth+1; else if($myDateYear<$dueDateYear) $nextMonth = 1; $dueDateOrg = date("n/t/y", strtotime($nextMonth."/1/".$dueDateYear)); } …

Check the below link.Very simple code for Export MySQL table to CSV using PHP. [Click Here](http://www.a2zwebhelp.com/export-data-to-csv)

You can do this with ajax form submition. I have done some changes and it works. Try the following example. In your form page <Html> <head> <title>Add two numbers on same form</title> <script src="http://code.jquery.com/jquery-1.8.1.js"></script> <body> <center> <h1><font color="blue"> Sum Two Numbers</font></h1> <form name="thisform" method="post" id="myForm" action="getsum2.php" > <table cellpadding=3 cellspacing=3> …

Try this <?php mysql_connect("localhost", "root", "") or die("Connection failed! " . mysql_error()); mysql_select_db("your_database") or die("Error in database selection" . mysql_error()); function qryTry ($parentId) { $find_query = "SELECT Id FROM Table_1 WHERE parent = '$parentId' "; $strSQL = mysql_query($find_query); while($row = mysql_fetch_array($strSQL)) { echo $row['id'] . "<br />"; } $strSQL2 = …

You need to over write your url using .htaccess file After that in on submit call a function like <input type="submit" name="search_pro_btn" id="search_pro_btn" onclick="rewrite_form();" value="Search" /> In rewrite_form function, function rewrite_form() { //Create custom link here ---------------------------- ---------------------------------- //create form submit action var url = '/search/' +'your custom link'; document.getElementById('FormId').action …

You did double exicution of query.. I have done the changes and it works for me. Please check ... <?php mysql_connect("localhost", "root", "") or die("Connection failed! " . mysql_error()); mysql_select_db("asia") or die("Error in database selection" . mysql_error()); echo "Connection successful!"; $find_query = "select *FROM ghee where sno=3"; $strSQL = mysql_query($find_query); …

Hai; Use the following codes for rename the file name, if any file with same name exist on upload path. $tempFile = $_FILES['txtFile']['tmp_name']; $targetPath = 'uploadfolder/'; $newFileName =$_FILES['txtFile']['name']; $targetFile = str_replace('//','/', $targetPath) . $newFileName; $cpy=0; do { if($cpy>0) $fileName="copy_".$cpy.str_replace(' ','_', ((isset($_FILES['file']['tmp_name']) && is_uploaded_file($_FILES['file']['tmp_name']))? $_FILES['file']['name'] : $_FILES['Filedata']['name'])); else $fileName=str_replace(' ','_', ((isset($_FILES['file']['tmp_name']) …

http://codex.wordpress.org/Moving_WordPress

PHP is server Side script and Javascript client-side.So You can't do this.But you can use AJAX to call a PHP function from Javascript.

http://php.net/manual/en/function.header.php

<?php session_start(); # We require the library require("facebook.php"); # Creating the facebook object $facebook = new Facebook(array( 'appId' => 'XXXXXXXXXXXXXXX', 'secret' => 'XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX', 'cookie' => true )); $session = $facebook->getSession(); if ($session) { $uid = $facebook->getUser(); $me = $facebook->api('/me'); $friends = $facebook->api('/me/friends'); } foreach ($friends as $key=>$value) { echo '<ul …

Hai; check the below link [Click Here](http://www.daniweb.com/web-development/php/threads/360450/php-video-upload-script)

Hai; You use a hidden field for store this total value in your form.Like <input type="hidden" name="hdnTotal" id="hdnTotal" value="" /> And also add the below code in your calculateSum(). $("#hdnTotal").val(sum5.toFixed(2)); In your action page you will get the total value as $_POST['hdnTotal'] //if you use post methode $_GET['hdnTotal'] // if …

Hai; The below link will help you `http://jenniferperrin.com/blog/jquery-tutorial-popup-modal-contact-form/`