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  • Get all link id from html source code using PREG_MATCH_ALL

    - by Jeremy Dicaire
    Hi there, I know i shouldn't do this that way but its just to retrieve all id of my links since i have a lot of them Here is the patern: <a href="mylink.php?get=123456">Click 1</a> <a href="mylink.php?get=222222">Click 2</a> <a href="mylink.php?get=81456">Click 3</a> <a href="mylink.php?get=1700">Click 4</a> I want to get all "get=" values (123456, 222222, etc.) And also the "Click 1", "Click 2", etc values using Preg_match_all() Any idea? Thanks a lot guys!!!

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  • JSON onFailure issue

    - by Mikey1980
    I’m having trouble getting a AJAX/JSON function to work correctly. I had this function grabbing value from a drop down box but now I want to use an anchor tag to set it's value. I thought it would be easy to just use the onClick event to pass string to the function I was using for the drop down box but I get the alert in JSON onFailure event even though data gets added to MySQL. I tried removing the alert from onFailure event but then it doesn't add the data. The drop down still continues work work fine, no alerts. (I should note that removing the alert also broke my drop down box) 1st I add an onClick event… <a href="<?php echo Settings::get('app.webroot'); ?>?view=schedule&action=questions" onmouseout="MM_swapImgRestore();" onmouseover="MM_swapImage('bre','','template/images/schedule/bre_f2.gif',1)" onclick="assignCallType('testing')";> 2nd I check main.js.php function assignCallType(type) { alert(type); new Request.JSON({ url: "ajax.php", onSuccess: function(rtndata,txt){ if (rtndata['STATUS'] != 'OK') alert('Status was not okay'); }, onFailure : function () { alert("onFailure") } }).get({ 'action': 'assignCallType', 'call_type': type }); } 3rd Ajax.php: the variable is back in PHP and values get added to MySQL but I get the Alert in the JSON onFailure event if ($_GET['action'] == "assignCallType") { if ($USER->isInsideSales()) { $call_type = $_GET['call_type']; $_SESSION['callinfo']->setCallType($call_type); $_SESSION['callinfo']->save($callid); echo json_encode(array('STATUS'=>'OK')); } else { echo json_encode(array('STATUS'=>'DENIED')); } } Any idea where I am going wrong. The only difference between this and the working drop down is how the function was called, I used onchange="assignCallType(this.value)".

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  • Selecte and retrieve file path to image using dialog box

    - by Steven
    I'm expanding Wordpress to function more like a CMS. This includes relating image to a post. Currently I have a text field where I paste an image path. Now I want to click a button which opens a dialog box, then I navigate to the right folder and select a file to "open" (from server side). The file path is returned. How can I go about accomplishing this?

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  • Path is present, Permissions are okay, but still getting error

    - by N e w B e e
    I recently installed pdftk using instruction provided at stack overflow I installed it, and run the commanded whereis pdftk the result was /usr/local/bin/pdftk /usr/bin/pdftk I have the powerpannel access and I saw it through it that pdftk actually exists at the location i run the command pdftk --version, it was okay but when in php i use <?php $command = "pdftk --help"; system("PATH=/usr/local/bin/ && $command",$response); if ($response===FALSE){ echo 'sorry error occured'; } else{ echo $response; } ?> the output is 127 the version i am using is 1.41 and the output '127' is something that i cant understand can somebody guide me?

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  • Adding to database. No repeat on refresh

    - by kevstarlive
    I have this code: Episode.php <?$feedback = new feedback; $articles = $feedback->fetch_all(); if (isset($_POST['name'], $_POST['post'])) { $cast = $_GET['id']; $name = $_POST['name']; $email = $_POST['email']; $post = nl2br ($_POST['post']); $ipaddress = $_SERVER['REMOTE_ADDR']; if (empty($name) or empty($post)) { $error = 'All Fields Are Required!'; }else{ $query = $pdo->prepare('INSERT INTO comments (cast, name, email, post, ipaddress) VALUES(?, ?, ?, ?, ?)'); $query->bindValue(1, $cast); $query->bindValue(2, $name); $query->bindValue(3, $email); $query->bindValue(4, $post); $query->bindValue(5, $ipaddress); $query->execute(); } }?> <div align="center"> <strong>Give us your feedback?</strong><br /><br /> <?php if (isset($error)) { ?> <small style="color:#aa0000;"><?php echo $error; ?></small><br /><br /> <?php } ?> <form action="episode.php?id=<?php echo $data['cast_id']; ?>" method="post" autocomplete="off" enctype="multipart/form-data"> <input type="text" name="name" placeholder="Name" /> / <input type="text" name="email" placeholder="Email" /><small style="color:#aa0000;">*</small><br /><br /> <textarea rows="10" cols="50" name="post" placeholder="Comment"></textarea><br /><br /> <input type="submit" onclick="myFunction()" value="Add Comment" /> <br /><br /> <small style="color:#aa0000;">* <b>Email will not be displayed publicly</b></small><br /> </form> </div> Include.php class feedback { public function fetch_all(){ global $pdo; $query = $pdo->prepare("SELECT * FROM comments"); $query->bindValue(1, $cast); $query->execute(); return $query->fetchAll(); } } This code updates to the database as it is suppose to. But after submission it reloads the current page as mentioned in the form action. But when I refresh the page to see the comment being added it asks to re submit. If I hit submit then the comment adds again. How can I stop this from happening? Maybe I could hide the comment box and display a thank you message but that would not stop a repeat entry. Please help. Thank you. Kev

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  • mysql_fetch_array() expects parameter 1 to be resource problem

    - by user225269
    I don't get it, I see no mistakes in this code but there is this error, please help: mysql_fetch_array() expects parameter 1 to be resource problem <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM student WHERE IDNO=".$_GET['id']); ?> <?php while ($row = mysql_fetch_array($result)) { ?> <table class="a" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#D3D3D3"> <tr> <form name="formcheck" method="get" action="updateact.php" onsubmit="return formCheck(this);"> <td> <table border="0" cellpadding="3" cellspacing="1" bgcolor=""> <tr> <td colspan="16" height="25" style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Update Students</td> <tr> <td width="30" height="35"><font size="2">*I D Number:</td> <td width="30"><input name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $_GET['id']; ?>"></td> </tr> <tr> <td width="30" height="35"><font size="2">*Year:</td> <td width="30"><input name="yr" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["YEAR"]; ?>"></td> <?php } ?> I'm just trying to load the data in the forms but I don't know why that error appears. What could possibly be the mistake in here?

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  • Ajax post failing in asp

    - by Dave Kiss
    hey guys, this might be really stupid, but hopefully someone can help. I'm trying to post to an external script using ajax so i can mail the data, but for some reason my data is not making it to the script. $(document).ready(function() { $("#submitContactForm").click(function () { $('#loading').append('<img src="http://www.xxxxxxxx.com/demo/copyshop/images/loading.gif" alt="Currently Loading" id="loadingComment" />'); var name = $('#name').val(); var email = $('#email').val(); var comment = $('#comment').val(); var dataString = 'name='+ name + '&email=' + email + '&comment=' + comment; $.ajax({ url: 'http://www.xxxxx.com/demo/copyshop/php/sendmail.php', type: 'POST', data: '?name=Dave&[email protected]&comment=hiiii', success: function(result) { $('#loading').append('success'); } }); return false; }); }); the php script is simple (for now - just wanted to make sure it worked) <?php $name = $_POST['name']; $email = $_POST['email']; $comment = $_POST['comment']; $to = '[email protected]'; $subject = 'New Contact Inquiry'; $message = $comment; mail($to, $subject, $message); ?> the jquery is embedded in an .aspx page (a language i'm not familiar with) but is posting to a php script. i'm receiving emails properly but there is no data inside. am i missing something? i tried to bypass the variables in this example, but its still not working thanks

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  • Wordpress is_front_page if statment

    - by Anders Kitson
    I have the following code below. I am trying to get rid of the box articles div when I am on the home page the code works when the html is outside of the else portion of the IF statement, as soon as I put it inside the page goes blank, I can't seem to figure out where I have broken the code. Any help would be great. <?php if(is_front_page()) { if(function_exists('wp_content_slider')) { wp_content_slider(); } } else{ ?> <div class="box articles"> <div class="block" id="articles"> <h2><?php the_title(); ?></h2> <div class="article"> <div class="entry"> <?php the_content(); ?> </div> <?php edit_post_link('Modifca Contenuto.', '<p>', '</p>'); ?> </div> </div> </div> <?php endwhile; endif; ?> <? } ?>

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  • MySQL Database Query - Codeigniter

    - by user2450349
    I am building an application with Codeigniter and need some help with a DB query. I have a table called users with the following fields: user_id, user_name, user_password, user_email, user_role, user_manager_id In my app, I pull all records from the user table using the following: function get_clients() { $this->db->select('*'); $this->db->where('user_role', 'client'); $this->db->order_by("user_name", "Asc"); $query = $this->db->get("users"); return $query->result_array(); } This works as expected, however when I display the results in the view, I also want to display a new column called Manager which will display the managers user_name field. The user_manager_id is the id of the user from the same table. Im guessing you can create an outer join on the same table but not sure. In the view, I am displaying the returned info as follows: <table class="table table-striped" id="zero-configuration"> <thead> <tr> <th>Name</th> <th>Email</th> <th>Manager</th> </tr> </thead> <tbody> <?php foreach($clients as $row) { ?> <tr> <td><?php echo $row['user_name']; ?> (<?php echo $row['user_username']; ?>)</td> <td><?php echo $row['user_email']; ?></td> <td><?php echo $row['???']; ?></td> </tr> <?php } ?> </tbody> </table> Any idea of how I can form the query and display the manager name is the view? Example: user_id user_name user_password user_email user_role user_manager_id 1 Ollie adjjk34jcd [email protected] client null 2 James djklsdfsdjk [email protected] client 1 When i query the database, i want to display results like this: Ollie [email protected] James [email protected] Ollie

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  • Form Submitting Incorrect Information to MySQL Database

    - by ThatMacLad
    I've created a form that submits data to a MySQL database but the Date, Time, Year and Month fields constantly revert to the exact same date (1st January 1970) despite the fact that when I submit the information to the database the form displays the current date, time etc to me. I've already set it so that the time and date fields automatically display the current time and date. Could someone please help me with this. Form: <html> <head> <title>Blog | New Post</title> <link rel="stylesheet" href="css/newposts.css" type="text/css" /> </head> <body> <div class="new-form"> <div class="header"> <a href="edit.php"><img src="images/edit-home-button.png"></a> </div> <div class="form-bg"> <?php if (isset($_POST['submit'])) { $month = htmlspecialchars(strip_tags($_POST['month'])); $date = htmlspecialchars(strip_tags($_POST['date'])); $year = htmlspecialchars(strip_tags($_POST['year'])); $time = htmlspecialchars(strip_tags($_POST['time'])); $title = htmlspecialchars(strip_tags($_POST['title'])); $entry = $_POST['entry']; $timestamp = strtotime($month . " " . $date . " " . $year . " " . $time); $entry = nl2br($entry); if (!get_magic_quotes_gpc()) { $title = addslashes($title); $entry = addslashes($entry); } mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); $sql = "INSERT INTO php_blog (timestamp,title,entry) VALUES ('$timestamp','$title','$entry')"; $result = mysql_query($sql) or print("Can't insert into table php_blog.<br />" . $sql . "<br />" . mysql_error()); if ($result != false) { print "<p class=\"success\">Your entry has successfully been entered into the blog. </p>"; } mysql_close(); } ?> <?php $current_month = date("F"); $current_date = date("d"); $current_year = date("Y"); $current_time = date("H:i"); ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <input class="field" type="text" name="date" id="date" size="2" value="<?php echo $current_month; ?>" /> <input class="field" type="text" name="date" id="date" size="2" value="<?php echo $current_date; ?>" /> <input class="field" type="text" name="date" id="date" size="2" value="<?php echo $current_year; ?>" /> <input type="text" name="time" id="time" size="5"value="<?php echo $current_time; ?>" /> <input class="field2" type="text" id="title" value="Title Goes Here." name="title" size="40" /> <textarea class="textarea" cols="80" rows="20" name="entry" id="entry" class="field2"></textarea> <input class="field" type="submit" name="submit" id="submit" value="Submit"> </form> </div> </div> </div> <div class="bottom"></div> </body> </html>

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  • highlight page selected in pagination?

    - by Mahmoud
    Hey There i have a php page, that shows all products that are stored on the database, so everything in fine, but my problem is that i am trying to highlight or give a color when a selected number is clicked, example, let say i all my product are showing and there are 2 pages, so when the user clicks on next the next fade and number 2 is colored yellow this well help the user to on which page he is below is my php code <?php echo"<a style:'color:#FFF;font-style:normal;'> "; include "im/config.php"; include('im/ps_pagination.php'); $result = ("Select * from product "); $pager = new PS_Pagination($conn, $result, 5, 6, "product=product"); $rs = $pager->paginate(); echo"<div id='image_container'>"; echo $pager->renderFullNav(); echo "<br /><br />\n"; while($row = mysql_fetch_array($rs)){ echo" <div class='virtualpage hidepiece'><a href='gal/".$row['pro_image']."' rel='lightbox[roadtrip]' title='".$row['pro_name']." : ".$row['pro_mini_des']."' style='color:#000'><img src='thumb/".$row['pro_thumb']."' /> </a></div>"; } echo"</div>"; echo "<br /><br />\n"; echo $pager->renderFullNav(); echo"</a>"; ?> Thank everyone

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  • Where can I find a jQuery article rotator / carousel / slider?

    - by Steven
    I want to make an article rotator similar to http://www.vmagazine.com/ And I want to use jQuery. I know there are carousels like jCarousel and jQuery Carousel. There is also the Coda-Slider, but I don't fint it very elegant. Can anyone suggest a good jQuery method / script for having an article carousel? I don't want to use MooTools, Prototype etc. since I'm already using jQuery. Any suggestions anyone?

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  • Problems using "at" with Apache

    - by Alex Padgett
    I'm trying to use a PHP script to create at jobs, but when it comes time to execute the jobs, nothing seems to be happening. I've tried to output any errors to log files, but have had no luck. It seems obvious that it's a permissions issue, because when I set apache to run as my personal user, everything works fine. However, when I exec wget directly from PHP, everything works fine so it seems that apache has the correct permissions to use it. The problem appears to be when using at in conjunction with apache. So I need to find a way to make this work with apache running as its own user. Here is the command I'm using: echo "wget -qO- http://example.com/" | at now + 1 minute 2>&1 Any ideas? EDIT: Apache can create the at jobs, it just seems that when they execute nothing is happening.

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  • convert mysql code to codeigniter

    - by Jethro Tamares Doble
    How can i convert this code into an acceptable codeigniter code: mysql_select_db($database_connection_ched, $connection_ched); $query_Institutions = "SELECT * FROM tb_institutional_profile ORDER BY tb_institutional_profile.institution_name ASC"; $Institutions = mysql_query($query_Institutions, $connection_ched) or die(mysql_error()); $row_Institutions = mysql_fetch_assoc($Institutions); $totalRows_Institutions = mysql_num_rows($Institutions); <td width="192"><select name="institution_id"> <?php do { <option value="<?php echo $row_Institutions['institution_id']?>" ><?php echo $row_Institutions['institution_name']?></option> <?php } while ($row_Institutions = mysql_fetch_assoc($Institutions)); ?> </select></td>

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  • variable $base_path is not working

    - by Nidhi Prasad
    I am trying to get the value of base_path variable in PHP (on lamp server) . I have kept the code insider beta_test directory inside www directly. i.e, base path function should return " /beta_test/ " . But it is returning just single slash ( "/" ) . The code that I tried is <script type="text/javascript" src="<?php print base_path(); ?>sites/all/themes/people10/slider/call.js"></script> Expected output is <script type="text/javascript" src="/beta_test/sites/all/themes/people10/slider/call.js"></script> But its giving <script type="text/javascript" src="/sites/all/themes/people10/slider/call.js"></script> I am using php version 5.3.3.Can anyone please help me in getting this issue solved? I am newbie to php and drupal .

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  • Redirection Still not working (updated on earlier question)

    - by NoviceCoding
    So earlier I asked this question: JQuery Login Redirect. Code Included The php file is sending the following: $return['error'] = false; $return['msg'] = 'You have successfully logged in!!'; I've tried all the suggestions, quoting the error on php and ajax end, 2 equals instead of 3, I've also tried DNE true which should be the same as an else statement: $(document).ready(function(){ $('#submit').click(function() { $('#waiting').show(500); $('#empty').show(500); $('#reg').hide(0); $('#message').hide(0); $.ajax({ type : 'POST', url : 'logina.php', dataType : 'json', data: { type : $('#typeof').val(), login : $('#login').val(), pass : $('#pass').val(), }, success : function(data){ $('#waiting').hide(500); $('#empty').show(500); $('#message').removeClass().addClass((data.error === true) ? 'error' : 'success') .text(data.msg).show(500) if(data.error != true) window.location.replace("http://blahblah.com/usercp.php"); if (data.error === true) $('#reg').show(500); $('#empty').hide() }, error : function(XMLHttpRequest, textStatus, errorThrown) { $('#waiting').hide(500); $('#message').removeClass().addClass('error') .text("There was an Error. Please try again.").show(500); $('#reg').show(500); $('#empty').hide(); Recaptcha.reload(); } }); return false; }); And it still wont work. Any ideas on how to make a redirection work if login is successful and error returns false? Also while I am asking, can I put a .delay(3000) 3s at the end of window.location.replace("http://blahblah.com/usercp.php")?

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