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Search found 815 results on 33 pages for 'isset'.

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  • Session in php are not enough clear to me

    - by Lulzim
    I find sessions in php kind of confusing, can anybody of you explain those to me. I have an example which is not working in my case: I register sessions this way, would you please tell me is this the right way of registering sessions //this is the page from where i register myusername in sessions if($count==1){ session_start(); $_SESSION['myusername'] = $_POST['myusername']; include("enterpincover.php"); } else { echo "Wrong Pin"; } here i check first whether the username is registered in sessions in oder to open his account , otherwise open again login. It works, if user is not loged in, it will show login page which is right, if user is loged it shows welcome message but not the Welcome the name of the user as I want. for ex: Welcome David <?php session_start(); if(isset($_SESSION['myusername'])) { echo 'Welcome '.$_SESSION['myusername']; } else { include("leftmodules.php"); include("rightmodules.php"); include("login.php"); } ?>

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  • Search for values in nested array

    - by dardub
    I have an array as follows array(2) { ["operator"] => array(2) { ["qty"] => int(2) ["id"] => int(251) } ["accessory209"] => array(2) { ["qty"] => int(1) ["id"] => int(209) } ["accessory211"] => array(2) { ["qty"] => int(1) ["id"] => int(211) } } I'm trying to find a way to verify an id value exists within the array and return bool. I'm trying to figure out a quick way that doesn't require creating a loop. Using the in_array function did not work, and I also read that it is quite slow. In the php manual someone recommended using flip_array() and then isset(), but I can't get it to work for a 2-d array. doing something like if($array['accessory']['id'] == 211) would also work for me, but I need to match all keys containing accessory -- not sure how to do that Anyways, I'm spinning in circles, and could use some help. This seems like it should be easy. Thanks.

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  • Automatically update php loop with data pulled from database

    - by John Svensson
    SQL STRUCTURE CREATE TABLE IF NOT EXISTS `map` ( `id` int(11) NOT NULL AUTO_INCREMENT, `x` int(11) NOT NULL, `y` int(11) NOT NULL, `type` varchar(50) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ; http://localhost/map.php?x=0&y=0 When I update the x and y via POST or GET, I would like to pull the new data from the database without refreshing the site, how would I manage that? Could someone give me some examples, because I am really stuck here. <?php mysql_connect('localhost', 'root', ''); mysql_select_db('hol'); $startX = $_GET['x']; $startY = $_GET['y']; $fieldHeight = 6; $fieldWidth = 6; $sql = "SELECT id, x, y, type FROM map WHERE x BETWEEN ".$startX." AND ".($startX+$fieldWidth). " AND y BETWEEN ".$startY." AND ".($startY+$fieldHeight); $result = mysql_query($sql); $positions = array(); while ($row = mysql_fetch_assoc($result)) { $positions[$row['x']][$row['y']] = $row; } echo "<table>"; for($y=$startY; $y<$startY+$fieldHeight; $y++) { echo "<tr>"; for($x=$startX; $x<$startX+$fieldWidth; $x++) { echo "<td>"; if(isset($positions[$x][$y])) { echo $positions[$x][$y]['type']; } else { echo "(".$x.",".$y.")"; } echo "</td>"; } echo "</tr>"; } echo "</table>"; ?>

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  • header location won't work in php

    - by Jayden Kelly
    I am making a login page for my website but the header location won't work. here is the code of login.php: <?php include ( './includes/header.php' ); if (isset($_POST['submit'])) { $username = $_POST['username']; $password = $_POST['password']; $check_username = mysql_query("SELECT username FROM users WHERE username='$username'"); $numrows = mysql_num_rows($check_username); if ($numrows != 1) { echo 'That User doesn\'t exist.'; } else { $check_password = mysql_query("SELECT password FROM users WHERE password='$password' && username='$username'"); while ($row = mysql_fetch_assoc($check_password)) { $password_db = $row['password']; if ($password_db == $password) { $_SESSION['username'] = $username; header("Location: members.php"); } } } } ?> <h2>Login to Your Account</h2> <form action='login.php' method='POST'> <input type='text' name='username' value='Username ...' onclick='value=""'/><p /> <input type='password' name='password' value='Password ...' onclick='value=""'/><p /> <input type='submit' name='submit' value='Login to my Account' /> </form> I would really appreciate it if someone could help me, thanks. P.S. If you need the php part of the header file it is here: <?php session_start(); include ( './includes/functions.php' ); include ( './includes/connect_to_mysql.php' ); ?>

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  • Should I change $_REQUEST to $_POST

    - by Scarface
    Hey guys quick question, I have a checkbox system where a list of items can be checked and deleted on the click of a button. I currently use request and it does the job but I was wondering if $_REQUEST was some sort of security risk or improper. If anyone has any advice I would appreciate it. Should I change to $_POST? If so, what is the best way to go about it? foreach ($_REQUEST as $key=>$value) { if (substr($key,0,3)==="img") { $id = substr($key,3); if(isset($_REQUEST['Delete'])) { $sql = 'SELECT file_name,username FROM images WHERE id=?'; $stmt = $conn->prepare($sql); $result=$stmt->execute(array($id)); while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { $image=$row['file_name']; $user=$row['username']; $myFile = "$user/images/$image"; unlink($myFile); } <input id=\"img".$id."\" name=\"img".$id."\" type=\"checkbox\">

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  • A problem of a repeated parameter in the pagination links?

    - by myaccount
    The problem is that when I load page 2 for example the URL becomes: http://domain.com/index.php?restaurant-id=45&currentpage=2 And that's fine but when I get to page 3 it becomes: http://domain.com/index.php?restaurant-id=45&currentpage=2&currentpage=3 And so on, it adds one more currentpage parameter everytime a new page is loaded from the pagination links! I wonder how this problem can be fixed? Here's some of the pagination function's code /****** build the pagination links ******/ // Getting current page URL with its parameters $current_page_url = ($_SERVER["PHP_SELF"].(isset($_SERVER["QUERY_STRING"])?"?".htmlentities($_SERVER["QUERY_STRING"]):"")); // Determine which sign to use (? or &) before the (currentpage=xx) parameter $sign = preg_match('/\?/', $current_page_url) ? '&' : '?'; $pagination_links = ''; // if not on page 1, don't show back links if ($currentpage > 1) { // show << link to go back to page 1 $pagination_links .= " <a href='{$current_page_url}{$sign}currentpage=1'>First page</a> "; // get previous page num $prevpage = $currentpage - 1; // show < link to go back 1 page $pagination_links .= " <a href='{$current_page_url}{$sign}currentpage=$prevpage'>previous</a> "; } else { $pagination_links .= "? ?"; }// end if

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  • How to check Isavailablity in ajax?

    - by udaya
    Hi This is what i have in my view page <td width=""><input type="text" name="txtUserName" id="txtUserName" /></td> <td><input type="button" name="CheckUsername" id="CheckUsername" value="Check Availablity" onclick="Check_User_Name();"/></td> Onclick of the button the Check_User_Name function in my ajax.js loads This is the Check_User_Name function function Check_User_Name(source) { var UserName = document.getElementById('txtUserName').value; if(window.ActiveXObject) User_Name = new ActiveXObject("Microsoft.XMLHTTP"); else if(window.XMLHttpRequest) User_Name = new XMLHttpRequest(); var URL = newURL+"ssit/system/application/views/ssitAjax.php"; URL = URL +"?CheckUsername="+UserName; User_Name.onreadystatechange = User_Name_Fun; User_Name.open("GET",URL,true); User_Name.send(null); } function User_Name_Fun() { document.getElementById('User_div').innerHTML=User_Name.responseText; } Then I can have the value in echo username then the $result has all the user name Hows can i checkIsavailablity of username from here if(($_GET['CheckUsername']!="") || (isset($_GET['CheckUsername']))) { echo $UserName = $_GET['CheckUsername'];//echo username $_SESSION['state'] = $State; $queryres = "SELECT dUser_name FROM tbl_login WHERE dIsDelete='0'"; $result = mysql_query($queryres,$cn) or die("Selection Query Failed !!!");

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  • Variable wont echo

    - by jonnnnnnnnnie
    I have the following code, where the var $username doesn't echo, when you type in a value. //TODO: SET AUTH TOKEN as random hash, save in session $auth_token = rand(); if (isset($_POST['action']) && $_POST['action'] == 'Login') { $errors = array(); //USED TO BUILD UP ARRAY OF ERRORS WHICH ARE THEN ECHOED $username = $_POST['username']; if ($username = '') { $errors['username'] = 'Username is required'; } echo $username; // var_dump($username) returns string 0 } require_once 'login_form.html.php'; ?> login_form is this: <form method="POST" action=""> <input type="hidden" name="auth_token" value="<?php echo $auth_token ?>"> Username: <input type="text" name="username"> Password: <input type="password" name="password1"> <input type="submit" name="action" value="Login"> </form> The auth token part isn't important, it just when I type in a value in username textbox and press the login button, the username wont echo, var_dump returns string (0) and print_r is just blank.

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  • PHP & MySQL username validation and storage problem.

    - by php
    For some reason when a user enters a brand new username the error message <p>Username unavailable</p> is displayed and the name is not stored. I was wondering if some can help find the flaw in my code so I can fix this error? Thanks Here is the PHP code. if($_POST['username'] && trim($_POST['username'])!=='') { $u = "SELECT * FROM users WHERE username = '$username' AND user_id <> '$user_id'"; $r = mysqli_query ($mysqli, $u) or trigger_error("Query: $u\n<br />MySQL Error: " . mysqli_error($mysqli)); if (mysqli_num_rows($r) == TRUE) { echo '<p>Username unavailable</p>'; $_POST['username'] = NULL; } else if(isset($_POST['username']) && mysqli_num_rows($r) == 0 && strlen($_POST['username']) <= 255) { $username = mysqli_real_escape_string($mysqli, $_POST['username']); } else if($_POST['username'] && strlen($_POST['username']) >= 256) { echo '<p>Username can not exceed 255 characters</p>'; } }

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  • PHP Session doesn't get read in next page after login validation, Why?

    - by NetStar
    I have a web site and when my users login it takes them to verify.php (where it connects to the DataBase and matches email and password to the user input and if OK puts client data into sessions and take the client to /memberarea/index.php ELSE back to login page with message "Invalid Email or password!") <?php ob_start(); session_start(); $email=$_POST['email']; $pass=md5($_POST['pass']); include("conn.php"); // connects to Database $sql="SELECT * FROM `user` WHERE email='$email' AND pass='$pass'"; $result=mysql_query($sql); $new=mysql_fetch_array($result); $_SESSION['fname']=$new['fname']; $_SESSION['lname']=$new['lname']; $_SESSION['email1']=$new['email1']; $_SESSION['passwrd']=$new['passwrd']; $no=mysql_num_rows($result); if ($no==1){ header('Location:memberarea/index.php'); }else { header("Location:login.php?m=$msg"); //msg="Invalid Login" } ?> then after email id and password is verified it takes them to ` /memberarea/index.php (This is where the problem happens.) where in index.php it checks if a session has been created in-order to block hackers to enter member area and sends them back to the login page. <? session_start(); isset($_SESSION['email'])` && `isset($_SESSION['passwrd'])` The problem is the client gets verified in verify.php (the code is above) In varify.php only after I put ob_start(); ontop of session_start(); It moves on to /memberarea/index.php , If I remove ob_start() It keeps the client on the verify.php page and displays error header is alredy SENT. after I put ob_start() it goes in to /memberarea/index.php but the session is blank, so it goes back to the login page and displays the error ($msg) "Invalid Login" which I programed to display. Can anyone tell me why the session cant pass values from verify.php to /memberarea/index.php

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  • Display error message at top of form

    - by moustafa
    Hello, I'm trying to get the following error to show when some once presses the submit button and has not filled in the required field/s. My PHP code is. <?php require_once("includes/database.php"); require_once("includes/functions.php"); if(isset($_POST['full_name'])) { $required = array('full_name','user_name','email','pwd','pwd2'); $missing = array(); $validation = array( 'full_name' => 'Please provide your full name', 'user_name' => 'Please provide your username', 'email' => 'Please provide your valid email address', 'pwd' => 'Please provide your password', 'pwd2' => 'Please confirm your password', 'userdup' => 'Username already registered', 'emaildup' => 'Email address already registered', 'mismatch' => 'Passwords do not match' ); //Sanitise and clean function $full_name = escape($_POST['full_name']); $user_name = escape($_POST['user_name']); $email = escape($_POST['email']); $pwd = escape($_POST['pwd']); $pwd2 = escape($_POST['pwd2']); foreach($_POST as $key => $value) { $value = trim($value); if(empty($value) && in_array($key,$required)) { array_push($missing,$key); } else { ${$key} = escape($value); } }

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  • codeigniter avoiding html div's

    - by rabidmachine9
    Hello there!Is there a proper syntax to avoid div's in codeigniter? I don't really like opening and closing tags all the time... <div class="theForm"> <?php echo form_open('edit/links');//this form uploads echo "Enter the Name: ". form_input('name','name'); echo "Enter the Link: ". form_input('url','url'); echo " ".form_submit('submit', 'Submit'); echo form_close(); if (isset($linksQuery) && count($linksQuery)){ foreach($linksQuery as $link){ echo anchor($link['link'], $link['name'].".", array("class" => "links")); echo form_open('edit/links',array('class' => 'deleteForm')); echo form_hidden('name',$link['name']); echo " ".form_submit('delete','Delete'); echo form_close(); echo br(2); } } ?> </div>

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  • Php Syntax Error

    - by Jeff Cameron
    I'm trying to update a table in php and I keep getting syntax errors. Here's what I've got: if (isset($_POST['inspect'])) { // get gis_id from pole table to update fm_poles $sql = "select gis_id from poles where pole_number = '".$_GET['polenumber']."'"; $rs = pg_query($sql) or die('Query failed: ' . pg_last_error()); $gisid = $row['gis_id']; pg_free_result($rs); // update fm_poles $sql = "update fm_poles set inspect ='".$_POST['inspect']."',co_date = '".$_POST['co_date']."',size = '".$_POST['size']."',date = ".$_POST['date'].",brand ='".$_POST['brand']."',backspan = ".$_POST['backspan']." WHERE gis_id = ".$gisid.""; print $sql."<BR>\n"; $rs = pg_query($sql) or die('Query failed: ' . pg_last_error()); pg_free_result($rs); } This is the error it gives me: update fm_poles set inspect ='20120208',co_date = '20030710',size = '30-5',date = 0,brand ='test',backspan = 300 WHERE gis_id = The error message: Query failed: ERROR: syntax error at end of input at character 129

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  • Is fetching data from database a get-method thing?

    - by theva
    I have a small class that I call Viewer. This class is supposed to view the proper layout of each page or something like that... I have a method called getFirstPage, when called the user of this method will get a setting value for which page is currently set as the first page. I have some code here, I think it works but I am not really shure that I have done it the right way: class Viewer { private $db; private $user; private $firstPage; function __construct($db, $user) { $this->db = $db; if(isset($user)) { $this->user = $user; } else { $this->user = 'default'; } } function getFistPage() { $std = $db->prepare("SELECT firstPage FROM settings WHERE user = ':user'"); $std->execute(array(':user' => $user)); $result = $std->fetch(); $this->firstPage = $result['firstPage']; return $this->firstPage; } } My get method is fetching the setting from databse (so far so good?). The problem is that then I have to use this get method to set the private variable firstPage. It seems like I should have a set method to do this, but I cannot really have a set method that just fetch some setting from database, right? Because the user of this object should be able to assume that there already is a setting defined in the object... How should I do that?

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  • Show Color Picker Value Code Within Input Text Field in Wordpress

    - by Shwan Namiq
    Hi i have options page for my theme i used color picker jquery plugin in my options page as show in image below. i want when i change the color from color picker automatically show the color value code within the text field.how can do this? this is the code within my options page related to appearing the color picker and text field function to register the options setting function YPE_register_settings_sections_fields() { register_setting ( 'YPE_header_option_group', 'YPE_header_option_name', 'YPE_sanitize_validate_callback' ); add_settings_section ( 'YPE_header_section', 'Header Section', 'YPE_header_section_callback', 'YPE_menu_page_options' ); add_settings_field ( 'YPE_header_bg', 'Header Background', 'YPE_header_bg_callback', 'YPE_menu_page_options', 'YPE_header_section' ); } add_action('admin_init', 'YPE_register_settings_sections_fields'); function to appear the text field and color picker function YPE_header_bg_callback() { $YPE_options = get_option('YPE_header_option_name'); $YPE_header_bg = isset($YPE_options['YPE_header_bg']) ? $YPE_options['YPE_header_bg'] : ''; ?> <div class="input-group color-picker"> <input class="form-control" style="width:80px;" name="YPE_header_option_name[YPE_header_bg]" id="<?php echo 'YPE_header_bg'; ?>" type="text" value="<?php echo $YPE_header_bg; ?>" /> <span class="input-group-btn"> <div id="colorSelector"> <div nam style="background-color: #0000ff"> </div> </div> </span> </div> <script> $("#colorSelector").ColorPicker({ color: '#0000ff', onShow: function (colpkr) { $(colpkr).fadeIn(500); return false; }, onHide: function (colpkr) { $(colpkr).fadeOut(500); return false; }, onChange: function (hsb, hex, rgb) { $('#colorSelector div').css('backgroundColor', '#' + hex); }); </script> <?php }

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  • Sort the $_POST variables

    - by Jerry
    Hello guys Might be an easy for you guys. I am trying to sort the $_POST variables that were sent by a form and update the sorted result in mysql. I am not sure how to do it and appreciate it anyone can help me about it. My main.php //I have a loop here. (omitted) //$k will be increased by 1 every time the loop starts, so I will know the total times of the loops //the form will be submitted to update.php echo "<input type='hidden' name='pickTotal' value='".$k."' />"; echo "<input type='hidden' id='point' name='earnedPoint".$k."' value='".$point."' />"; echo "<input type='hidden' id='users' name='userName".$k."' value='".$userPick['user']."' />"; //loop ends My update.php if(isset($_POST['submit'])){ $pickTotal=$_POST['pickTotal']; //get the total loop for ($p=0;$p<=$pickTotal;$p++){ $userToBeUpdated=$_POST['userName'.$p]; $userPoint=$_POST['earnedPoint'.$p]; //sort the $userPoint here. //I need to find out who got the most points //and list user's place. 1st, 2nd, 3rd...etc. //update my mysql } Thanks for any helps.

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  • PHP Array to CSV

    - by JohnnyFaldo
    I'm trying to convert an array of products into a CSV file, but it doesn't seem to be going to plan. The CSV file is one long line, here is my code: for($i=0;$i<count($prods);$i++) { $sql = "SELECT * FROM products WHERE id = '".$prods[$i]."'"; $result = $mysqli->query($sql); $info = $result->fetch_array(); } $header = ''; for($i=0;$i<count($info);$i++) { $row = $info[$i]; $line = ''; for($b=0;$b<count($row);$b++) { $value = $row[$b]; if ( ( !isset( $value ) ) || ( $value == "" ) ) { $value = "\t"; } else { $value = str_replace( '"' , '""' , $value ); $value = '"' . $value . '"' . "\t"; } $line .= $value; } $data .= trim( $line ) . "\n"; } $data = str_replace( "\r" , "" , $data ); if ( $data == "" ) { $data = "\n(0) Records Found!\n"; } header("Content-type: application/octet-stream"); header("Content-Disposition: attachment; filename=your_desired_name.xls"); header("Pragma: no-cache"); header("Expires: 0"); print "$data"; Also, the header doesn't force a download. I've been copy and pasting the output and saving as .csv

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  • PHP SDK Requires two logins...doesn't recognize first

    - by Jay Konieczny
    I am using the following code to authenticate whether users are logged in or not. While users can log in, they have to click the login button twice. Additionally, sometimes even after they click the log-in button twice, my "user info" part of the page (earlier in the page than the content) shows them as logged out while the actual page shows them as logged in. Here is the code. Could someone suggest a better way of handling log-ins? function isLoggedIn($facebook) { if (isset($facebook) and $facebook->getUser() != 0) { // UserID exists, but user may still not be logged in. Let's check: try { $facebook->api('/me', 'GET'); // If this succeeds, then they are logged in. return true; } catch(FacebookApiException $e) { // Some kind of error, so not logged in. if(session_id() === '') session_destroy(); return false; } } else { if(session_id() === '') session_destroy(); return false; } } Thanks!

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  • How to pass values from array into mysql with php

    - by moustafa
    my original code is this <tr> <th> <label for="user_level"> User Level: * <?php echo isset($valid_user_level) ? $valid_user_level : NULL; ?> </label> </th> </tr> <td> <select name="user_level" id="user_level" class="sel"> <option value="">Select one…</option> <option value="1">User</option> <option value="5">Admin</option> </select> </td> this give me the option to select one of choice from the drop down menu i.e. user and when user is selected and the submit button is pressed this will insert the value 1 into the database which will when the user logs in tell the system that they are are normal user. I want to change the code to the following <tr> <td> <select name="user_level" id="user_level" class="sel"> <option value="">Select one…</option> <?php if(!empty($level)) { foreach($level as $value) { echo "<option value='{$value}'"; echo getSticky(2,'user_level',$value); echo ">{$value}</option>"; } } ?> </select> </td> </tr> With this being my array query $level = array('User','Admin'); How can I pass the values of 1 for user level and 5 for admin in this code so when the user is selected it inouts 1 into the database?

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  • excluding previously randomized integer, and randomize again without it

    - by Rob
    <?php if (isset($_POST['Roll!'])) { $sides = $_POST['sides']; $rolled = rand(1,$sides); echo "$rolled was rolled by the dice, it is now out!"; } ?> This is the code I currently have. After rolling that number, however, I want it to roll again, but without the previously rolled number, until it has rolled all number except one, which would be the winning number. I have no idea how to go about doing that. Any ideas? EDIT: I'm sorry, I should have been more clear, thank you all for the help so far, but I also need to echo each number rolled, such as echo "$rolledArray[0] was rolled, it lost.\n"; echo "$rolledArray[1] was rolled, it lost.\n"; echo "$rolledArray[2] was rolled, it lost.\n"; echo "$rolledArray[3] was rolled, it lost.\n"; echo "$rolledArray[x] was rolled, it lost.\n"; echo "$rolledArray[x] was rolled, it lost.\n"; echo "$rolledArray[50?] was rolled, it lost."; EDIT AGAIN: Also I only want them to have to click Roll! once, not multiple times until they've rolled all the numbers, meaning no need for session, I think, though I could be wrong, most of you are clearly more experienced than me. Sorry, I should have mentioned that before as well.

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  • detect click on submit button in PHP

    - by Remus Rigo
    hi all I have a php file that contains a form (witch contains 2 input boxes and a submit button) for updating a contact. I managed to fill the fields with the contact's data, but I can't detect if the submit button is clicked form looks like this echo "<form action=Contact.php><table>". "<tr><td>First Name</td><td><input type=text size=75% name=FirstName value='".$row['FirstName']."'></td></tr>". "<tr><td>Last Name</td><td><input type=text size=75% name=LastName value='".$row['LastName']."'></td></tr>". "<tr><td colspan=2><input type=submit name=UpdateContact value=Update></td></tr>". "</table></form>"; this code should output a "clicked" message if the button is clicked if (isset($_POST['UpdateContact'])) { echo "<p>clicked"; } else { echo "<p>not clicked"; } can anyone help me out or tell me what i've done wrong (I want from the same php file to fill the contact's data in a from and to update the database)

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  • Zend Framework How can I print a logged in user name from a Zend_Session_Namespace

    - by IrishStudent76
    Hi all I have created the following login controller for my site and it works fine in relation to logging users in a logging them out. The thing I want to do is echo the logged in users name into the FlashMessenger for the success page how ever as my code stands I only get the following message when redirected to the success page, "you have been successfully logged in as Array". Can I also ask the following does the line $session-user =$adaptergetResultArray('Password'); create an array of user information less the password value from the database. Many Thanks in advance, IrishStudent76 <?php class LoginController extends Zend_Controller_Action { public function init(){ $this->view->doctype('XHTML1_STRICT'); } // login action public function loginAction() { $form = new PetManager_Form_Login; $this->view->form = $form; /* check for valid input from the form and authenticate using adapter Add user record to session and redirect to the original request URL if present */ if ($this->getRequest()->isPost()) { if ($form->isValid($this->getRequest()->getPost())) { $values = $form->getValues(); $adapter = new PetManager_Auth_Adapter_Doctrine( $values['username'], $values['password'] ); $auth = Zend_Auth::getInstance(); $result = $auth->authenticate($adapter); if ($result->isValid()) { $session = new Zend_Session_Namespace('petmanager.auth'); $session->user = $adapter->getResultArray('Password'); if (isset($session->requestURL)) { $url = $session->requestURL; unset($session->requestURL); $this->_redirect($url); } else { $this->_helper->getHelper('FlashMessenger') ->addMessage('You have been successfully logged in as '.$session- >user); $this->_redirect('/login/success'); } } else { $this->view->message = 'You could not be logged in. Please try again.'; } } } } public function successAction() { if ($this->_helper->getHelper('FlashMessenger')->getMessages()) { $this->view->messages = $this->_helper ->getHelper('FlashMessenger') ->getMessages(); } else { $this->_redirect('/login'); } } public function logoutAction() { Zend_Auth::getInstance()->clearIdentity(); Zend_Session::destroy(); $this->_redirect('/'); } }

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  • Textbox auto generate by php and html

    - by user2892997
    i have few field of data such as product , amount and barcode.The system just show 1 row of data insert form that contain the 3 field.when i completed insert the first row, then second row of textbox will auto generate, i can do it by microsoft access, can i do so for php ? <?php $i=0; ?> <form method="post" action=""> <input type="text" name="<?php echo $i; ?>" /> </form> <?php if(isset($_POST[$i])){ $i++; ?> <form method="post" action=""> <input type="text" name="<?php echo $i; ?>" /> </form> <?php }?> it work for the first and second textbox, but how can i continue to create more textbox accordingly?

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  • How to limit a user to entering 10 keywords or less using PHP & MySQL?

    - by G4TV
    I'm trying to limit my users to entering at least 10 keywords and was wondering how would I be able to do this using PHP & MySQL with my current Keyword script? Here is the add keywords PHP MySQL code. if (isset($_POST['tag']) && trim($_POST['tag'])!=='') { $tags = explode(",", $_POST['tag']); for ($x = 0; $x < count($tags); $x++){ $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $query1 = "INSERT INTO tags (tag) VALUES ('" . mysqli_real_escape_string($mysqli, strtolower(htmlentities(trim(strip_tags($tags[$x]))))) . "')"; if (!mysqli_query($mysqli, $query1)) { print mysqli_error($mysqli); return; } $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT id FROM tags WHERE tag='" . mysqli_real_escape_string($mysqli, strtolower(htmlentities(trim(strip_tags($tags[$x]))))) . "'"); if (!$dbc) { print mysqli_error($mysqli); } else { while($row = mysqli_fetch_array($dbc)){ $id = $row["id"]; } } $query2 = "INSERT INTO question_tags (tag_id, question_id, user_id, date_created) VALUES ('$id', '$question', '$user', NOW())"; if (!mysqli_query($mysqli, $query2)) { print mysqli_error($mysqli); return; } } }

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  • Why can I not echo out the value from an input text?

    - by user3684783
    I am using Wordpress to do an auction website. Here is what the code looks like <form method="post" action="<?php echo ProjectTheme_post_new_with_pid_stuff_thg($pid, '1');?>"> <?php do_action('ProjectTheme_step1_before_title'); ?> <!--////////// Project Title /////////////--> <li> <h2><?php echo __('Your Project Title', 'ProjectTheme'); ?>: <img src="../../images/help-icon.png" width="16" height="16" id="showhelp1"/></h2> <div id="help1">Your Project Title should be informative and brief.</div> <p><input type="text" style="width:90%;" class="do_input" name="project_title" value="Enter an informative & brief title..." onfocus="this.value = this.value=='Enter an informative & brief title...'?'':this.value;" onblur="this.value = this.value==''?'Enter an informative & brief title...':this.value;" /></p> <input type="submit" class="post-button" name="project_submit1" value="<?php _e("Next Step", 'ProjectTheme'); ?> &raquo;" /> </form> I am using a step by step form for users to fill in and I wanted to do a preview page, however I tried to use: if(isset($_POST['project_submit1'])){ $Name = $_POST['project_title']; echo "$Name"; } It shows up nothing.

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