Search Results

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

Page 11/33 | < Previous Page | 7 8 9 10 11 12 13 14 15 16 17 18  | Next Page >

  • AFNetworking PostPath php Parameters are null

    - by Alejandro Escobar
    I am trying to send a username and password from an iOS app using AFNetworking framework to a php script. The iOS app continues to receive status code 401 which I defined to be "not enough parameters". I have tried returning the "username" from the php script to the iOS app and receive . Based on what I've been investigating so far, it seems as though: 1) The php script is not decoding the POST parameters properly 2) The iOS app is not sending the POST parameters properly The following is the iOS function - (IBAction)startLoginProcess:(id)sender { NSString *usernameField = usernameTextField.text; NSString *passwordField = passwordTextField.text; NSDictionary *parameters = [NSDictionary dictionaryWithObjectsAndKeys:usernameField, @"username", passwordField, @"password", nil]; NSURL *url = [NSURL URLWithString:@"http://localhost/~alejandroe1790/edella_admin/"]; AFHTTPClient *httpClient = [[AFHTTPClient alloc] initWithBaseURL:url]; [httpClient defaultValueForHeader:@"Accept"]; [httpClient setParameterEncoding:AFJSONParameterEncoding]; [httpClient postPath:@"login.php" parameters:parameters success:^(AFHTTPRequestOperation *operation, id response) { NSLog(@"operation hasAcceptableStatusCode: %d", [operation.response statusCode]); } failure:^(AFHTTPRequestOperation *operation, NSError *error) { NSLog(@"Error with request"); NSLog(@"%@",[error localizedDescription]); }]; } The following is the php script function checkLogin() { // Check for required parameters if (isset($_POST["username"]) && isset($_POST["password"])) { //Put parameters into local variables $username = $_POST["username"]; $password = $_POST["password"]; $stmt = $this->db->prepare("SELECT Password FROM Admin WHERE Username=?"); $stmt->bind_param('s', $username); $stmt->execute(); $stmt->bind_result($resultpassword); while ($stmt->fetch()) { break; } $stmt->close(); // Username or password invalid if ($password == $resultpassword) { sendResponse(100, 'Login successful'); return true; } else { sendResponse(400, 'Invalid Username or Password'); return false; } } sendResponse(401, 'Not enough parameters'); return false; } I feel like I may be missing something. Any assistance would be great.

    Read the article

  • form_dropdown in codeigniter

    - by Patrick
    I'm getting a strange behaviour from form_dropdown - basically, when I reload the page after validation, the values are screwed up. this bit generates 3 drop downs with days, months and years: $days = array(0 => 'Day...'); for ($i = 1; $i <= 31; $i++) { $days[] = $i; } $months = array(0 => 'Month...', ); for ($i = 1; $i <= 12; $i++) { $months[] = $i; } $years = array(0 => 'Year...'); for ($i = 2010; $i <= 2012; $i++) { $years[$i] = $i; echo "<pre>"; print_r($years); echo "</pre>";//remove this } $selected_day = (isset($selected_day)) ? $selected_day : 0; $selected_month = (isset($selected_month)) ? $selected_month : 0; $selected_year = (isset($selected_year)) ? $selected_year : 0; echo "<p>"; echo form_label('Select date:', 'day', array('class' => 'left')); echo form_dropdown('day', $days, $selected_day, 'class="combosmall"'); echo form_dropdown('month', $months, $selected_month, 'class="combosmall"'); echo form_dropdown('year', $years, $selected_year, 'class="combosmall"'); echo "</p>"; ...and generates this: <p><label for="day" class="left">Select date:</label><select name="day" class="combosmall"> <option value="0" selected="selected">Day...</option> <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> <option value="6">6</option> <option value="7">7</option> <option value="8">8</option> <option value="9">9</option> <option value="10">10</option> <option value="11">11</option> <option value="12">12</option> <option value="13">13</option> <option value="14">14</option> <option value="15">15</option> <option value="16">16</option> <option value="17">17</option> <option value="18">18</option> <option value="19">19</option> <option value="20">20</option> <option value="21">21</option> <option value="22">22</option> <option value="23">23</option> <option value="24">24</option> <option value="25">25</option> <option value="26">26</option> <option value="27">27</option> <option value="28">28</option> <option value="29">29</option> <option value="30">30</option> <option value="31">31</option> </select><select name="month" class="combosmall"> <option value="0" selected="selected">Month...</option> <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> <option value="6">6</option> <option value="7">7</option> <option value="8">8</option> <option value="9">9</option> <option value="10">10</option> <option value="11">11</option> <option value="12">12</option> </select><select name="year" class="combosmall"> <option value="0" selected="selected">Year...</option> <option value="2010">2010</option> <option value="2011">2011</option> <option value="2012">2012</option> </select></p> however, when the form is reloaded after validation, the same code above generates this: <!-- days and months... --> <select name="year" class="combosmall"> <option value="0" selected="selected">Year...</option> <option value="1">2010</option> <option value="2">2011</option> <option value="3">2012</option> </select> So basically the value start from 1 instead of 2010. The same happens to days and months but obviously it doesn't make any difference in this particular case as the values would start from 1 anyway. How can I fix this - and why does it happen? edit: validation rules are: $this->load->library('form_validation'); //...rules for other fields.. $this->form_validation->set_rules('day', 'day', 'required|xss_clean'); $this->form_validation->set_rules('month', 'month', 'required|xss_clean'); $this->form_validation->set_rules('year', 'year', 'required|xss_clean'); $this->form_validation->set_error_delimiters('<p class="error">', '</p>'); //define other errors if($this->input->post('day') == 0 || $this->input->post('month') == 0 || $this->input->post('year') == 0) { $data['error'] = "Please check the date of your event."; }

    Read the article

  • PHP & MySQL Undefined variable problem

    - by comma
    I keep getting the following error Undefined variable: id on line 91 can some one help me correct this problem? The error is on this line. $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; MySQL database tables. CREATE TABLE tags ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, skill VARCHAR(255) NOT NULL, experience VARCHAR(255) NOT NULL, years VARCHAR(255) NOT NULL, PRIMARY KEY (id) ); CREATE TABLE users_skills ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, skill_id INT UNSIGNED NOT NULL, user_id INT UNSIGNED NOT NULL, date_created DATETIME UNSIGNED NOT NULL, PRIMARY KEY (id) ); Here is the PHP & MySQL code. if (isset($_POST['info_submitted'])) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON learned_skills.id = users_skills.skill_id WHERE user_id='$user_id'"); if (!$dbc) { print mysqli_error($mysqli); return; } $user_id = '5'; $skill = $_POST['skill']; $experience = $_POST['experience']; $years = $_POST['years']; $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON users_skills.skill_id = learned_skills.id WHERE users_skills.user_id='$user_id'"); if (mysqli_num_rows($dbc) == 0) { if (isset($_POST['skill']) && trim($_POST['skill'])!=='') { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $query1 = mysqli_query($mysqli,"INSERT INTO learned_skills (skill, experience, years) VALUES ('" . $skill . "', '" . $experience . "', '" . $years . "')"); if (mysqli_query($mysqli, $query1)) { print mysqli_error($mysqli); return; } $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT id FROM learned_skills WHERE id='" . $skill . "' AND experience='" . $experience . "' AND years='" . $years . "'"); if (!$dbc) { print mysqli_error($mysqli); } else { while($row = mysqli_fetch_array($dbc)){ $id = $row["id"]; } } $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; } }

    Read the article

  • PHP Using session variables in array(s)

    - by Chris
    Hello, My question is how do i put these session variables into a array? I have tried countless ways but none of them work. Not really sure what to put in a array and what no and how to adress them. Currently when i fill in the form the data gets displayed in a table. Next when i press the hyperlink that takes me back to the same form, i wish to enter data again. This data should be added in a new row in the same display table. Best Regards. The code below (pardon me that it is not english). <?php session_start(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-Strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <title>ExpoFormulier</title> <body> <?php if (!empty($_POST)) { $standnaam = $_POST["standnaam"]; $oppervlakte = $_POST["oppervlakte"]; //value in the form van checkboxes op 1 zetten! $verdieping = isset($_POST["verdieping"]) ? $_POST["verdieping"] : 0; $telefoon = isset($_POST["telefoon"]) ? $_POST["telefoon"] : 0; $netwerk = isset($_POST["netwerk"]) ? $_POST["netwerk"] : 0; if (is_numeric($oppervlakte)) { $_SESSION["standnaam"]=$standnaam; $_SESSION["oppervlakte"]=$oppervlakte; $_SESSION["verdieping"]=$verdieping; $_SESSION["telefoon"]=$telefoon; $_SESSION["netwerk"]=$netwerk; header("Location:ExpoOverzicht.php"); } else { echo "<h1>Foute gegevens, Opnieuw invullen a.u.b</h1>"; } } ?> <form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post" id="form1"> <h1>Vul de gegevens in</h1> <table> <tr> <td>Standnaam:</td> <td><input type="text" name="standnaam" size="18"/></td> </tr> <tr> <td>Oppervlakte (in m^2):</td> <td><input type="text" name="oppervlakte" size="6"/></td> </tr> <tr> <td>Verdieping:</td> <td><input type="checkbox" name="verdieping" value="1"/></td> </tr> <tr> <td>Telefoon:</td> <td><input type="checkbox" name="telefoon" value="1"/></td> </tr> <tr> <td>Netwerk:</td> <td><input type="checkbox" name="netwerk" value="1"/></td> </tr> <tr> <td><input type="submit" name="verzenden" value="Verzenden"/></td> </tr> </table> </form> Second File: <?php session_start(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-Strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <title>ExpoOverzicht</title> <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <link href="StyleSheetExpo.css" rel="stylesheet" type="text/css" /> </head> <body> <h1>Overzicht van de ingegeven standen in deze sessie</h1> <?php $standnaam = $_SESSION["standnaam"]; $oppervlakte = $_SESSION["oppervlakte"]; $verdieping = $_SESSION["verdieping"]; $telefoon = $_SESSION["telefoon"]; $netwerk = $_SESSION["netwerk"]; $result1 = 0; $result2 = 0; $result3 = 0; $prijsCom = 0; $prijsVerdieping = 0; for ($i=1; $i <= $oppervlakte; $i++) { if($i <= 10) { $tarief1 = 1 * 100; $result1 += $tarief1; } if($i > 10 && $i <= 30) { $tarief2 = 1 * 90; $result2 += $tarief2; } if($i > 30) { $tarief3 = 1 * 80; $result3 += $tarief3; } } $prijsOpp = $result1 + $result2 + $result3; if($verdieping == 1) { $prijsVerdieping = $oppervlakte * 120; } if(($telefoon == 1) || ($netwerk == 1)) { $prijsCom = 20; } if(($telefoon == 1) && ($netwerk == 1)) { $prijsCom = 30; } $totalePrijs = $prijsOpp + $prijsVerdieping + $prijsCom; echo "<table class=\"tableExpo\">"; echo "<th>Standnaam</th>"; echo "<th>Oppervlakte</th>"; echo "<th>Verdieping</th>"; echo "<th>Telefoon</th>"; echo "<th>Netwerk</th>"; echo "<th>Totale prijs</th>"; echo "<tr>"; echo "<td>".$standnaam."</td>"; echo "<td>".$oppervlakte."</td>"; echo "<td>".$verdieping."</td>"; echo "<td>".$telefoon."</td>"; echo "<td>".$netwerk."</td>"; echo "<td>".$totalePrijs."</td>"; echo "</tr>"; echo "</table>"; ?> <a href="ExpoFormulier.php">Terug naar het formulier</a> </body> </html> </body> </html>

    Read the article

  • can I have multiline text with GD and PHP?

    - by alekone
    hello! I'm trying to output multiline text with GD+PHP but can't get it working. my php knowledge is really basic. here's the code, any idea on how to output 2 or 3 lines of text? $theText = (isset($_GET['caption']))? stripslashes($_GET['caption']) :''; imagettftext($baseImage, $textSize, $textAngle, $textXposition, $textYposition, $textColor, $fontName, $theText);

    Read the article

  • json null error help in php

    - by bobby
    I get 'json is null' as error My php file: <?php if (isset($_REQUEST['query'])) { $query = $_REQUEST['query']; $url='https://www.googleapis.com/urlshortener/v1/'; $key='ApiKey'; $result= $url.($query).$key; $ch = curl_init($result); curl_setopt($ch, CURLOPT_HEADER, 0); curl_setopt($ch, CURLOPT_POST, 1); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,1); $resp = curl_exec($ch); curl_close($ch); echo $resp; } ?> My html: <html> <head> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function(){ // when the user clicks the button $("button").click(function(){ $.getJSON("shortner.php?query="+$('#query').attr("value"),function(json){ $('#results').append('<p>Id : ' + json.id+ '</p>'); $('#results').append('<p>Longurl: ' + json.longurl+ '</p>'); }); }); }); </script> </head> <body> <input type="text" value="Enter a place" id="query" /><button>Get Coordinates</button> <div id="results"></div> Edited : <?php if (isset($_REQUEST['query'])) { $query = $_REQUEST['query']; $url='https://www.googleapis.com/urlshortener/v1/'; $key='Api'; $key2='?key='; $result= $url.$query.$key2.$key; $requestData= json_encode($result); echo var_dump($query); $ch = curl_init($requestData); curl_setopt($ch, CURLOPT_HEADER, 0); curl_setopt($ch, CURLOPT_POST, 1); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,1); $resp = curl_exec($ch); curl_close($ch); echo $resp; } ?>

    Read the article

  • PHP error handling : my code is not optimized

    - by Tristan
    Hello, I must warn you, this code will heart your eyes, so please don't judge me, i'm trying to improve the way I handle errors all my tests are like this : if ($something < 27) { $error_IP= '<div class="error_message">something bad</div> '; }else{ $erreur_IP=''; } and here's the ugliest thing : if( !isset($_POST) || ($erreur_captcha !='') || ($erreur_email !='') || ($erreur_hebergeurVide != '') || ($erreur_paysVide != '') || ($erreur_slotVide != '') || ($erreur_rconVide != '') || ($erreur_tick != '') + a lot more :d ) What do you suggest to me to optimize my errors handling ? Thank you

    Read the article

  • PHP & MySQL - saving and looping problems.

    - by R.I.P.coalMINERS
    I'm new to PHP and MySQL I want a user to be able to store multiple names and there meanings in a MySQL database tables named names using PHP I will dynamically create form fields with JQuery every time a user clicks on a link so a user can enter 1 to 1,000,000 different names and there meanings which will be stored in a table called names. Since I asked my last question I figured out how to store my values from my form using the for loop but every time I loop my values when I add one or more dynamic fields the second form field named meaning will not save the value entered also my dynamic form fields keep looping doubling, tripling and so on the entered values into the database it all depends on how many form fields are added dynamically. I was wondering how can I fix these problems? On a side note I replaced the query with echo's to see the values that are being entered. Here is the PHP code. <?php if(isset($_POST['submit'])) { $mysqli = mysqli_connect("localhost", "root", "", "site"); $dbc = mysqli_query($mysqli,"SELECT * FROM names WHERE userID='$userID'"); $name = $_POST['name']; $meaning = $_POST['meaning']; if(isset($name['0']) && mysqli_num_rows($dbc) == 0 && trim($name['0'])!=='' && trim($meaning['0'])!=='') { for($n = 0; $n < count($name); $n++) { for($m = 0; $m < count($meaning); $m++) { echo $name[$n] . '<br />'; echo $meaning[$m] . '<br /><br />'; break; } } } } ?> And here is the HTML code. <form method="post" action="index.php"> <ul> <li><label for="name">Name: </label><input type="text" name="name[]" id="name" /></li> <li><label for="meaning">Meaning: </label><input type="text" name="meaning[]" id="meaning" /></li> <li><input type="submit" name="submit" value="Save" /></li> </ul> </form> If needed I will place the JQuery code.

    Read the article

  • Variables variable and inheritance

    - by Xack
    I made code like this: class Object { function __get($name){ if(isset($this->{$name})){ return $this->{$name}; } else { return null; } } function __set($name, $value){ $this->{$name} = $value; } } If I extend this class (I don't want to repeat this code every time), it says "Undefined variable". Is there any way to do it?

    Read the article

  • Stop $_POST from having the same value when I refresh

    - by Cortopasta
    When a vote value is changed, the form POSTs the change, then refreshes the page. This is called on the top of the page upon load: if (isset($_POST['q'.$question_id])) { $user->updateQuestionVotes($question_id, $_POST['q'.$question_id]); } Why does it update every time I refresh after the first time? Do I need to unset it somehow?

    Read the article

  • How to avoid my this facebook app api login page?

    - by user1035140
    I got a problem regrading with my apps which is once I go to my apps, it sure will show me a login page instead of allow page? it always display the login page 1st then only display allow page, I had tried other apps, if I am 1st time user, It sure will appear the allow page only, it did not show me the login page. my question is how to I avoid my login page direct go to allow page? here is my login page picture here is my apps link https://apps.facebook.com/christmas_testing/ here is my facebook php jdk api coding <?php $fbconfig['appid' ] = "XXXXXXXXXXXXX"; $fbconfig['secret'] = "XXXXXXXXXXXXX"; $fbconfig['baseUrl'] = "myserverlink"; $fbconfig['appBaseUrl'] = "http://apps.facebook.com/christmas_testing/"; if (isset($_GET['code'])){ header("Location: " . $fbconfig['appBaseUrl']); exit; } if (isset($_GET['request_ids'])){ //user comes from invitation //track them if you need header("Location: " . $fbconfig['appBaseUrl']); } $user = null; //facebook user uid try{ include_once "facebook.php"; } catch(Exception $o){ echo '<pre>'; print_r($o); echo '</pre>'; } // Create our Application instance. $facebook = new Facebook(array( 'appId' => $fbconfig['appid'], 'secret' => $fbconfig['secret'], 'cookie' => true, )); //Facebook Authentication part $user = $facebook->getUser(); $loginUrl = $facebook->getLoginUrl( array( 'scope' => 'email,publish_stream,user_birthday,user_location,user_work_history,user_about_me,user_hometown' ) ); if ($user) { try { // Proceed knowing you have a logged in user who's authenticated. $user_profile = $facebook->api('/me'); } catch (FacebookApiException $e) { //you should use error_log($e); instead of printing the info on browser d($e); // d is a debug function defined at the end of this file $user = null; } } if (!$user) { echo "<script type='text/javascript'>top.location.href = '$loginUrl';</script>"; exit; } //get user basic description $userInfo = $facebook->api("/$user"); function d($d){ echo '<pre>'; print_r($d); echo '</pre>'; } ?

    Read the article

  • PHP problems when transfering code from Windows to OS X

    - by Makka95
    I have recently bought a new MacBook Pro. Before I had my MacBook Pro I was working on a website on my desktop computer. And now I want to transfer this code to my new MacBook Pro. The problem is that when I transfered the code (I put it on Dropbox and simply downloaded it on my MacBook Pro) I started to see lots of error messages in my PHP code. The error message I”m receiving is: Warning: Cannot modify header information - headers already sent by (output started at /some/file.php:1) in /some/file.php on line 23 I have done some research on this and it seems that this error is most frequently caused by a new line, simple whitespace or any output before the <?php sign. I have looked through all the places where I have cookies that are being sent in the HTTP request and also where I'm using the header() function. I haven’t detected any output or whitespace that possibly could interfere and cause this problem. Noteworthy is that the error always says that the output is started at line 1. Which got me thinking if there is some kind of coding differences in the way that the Mac OS X and Windows operating systems handle new lines or white spaces? Or could the Dropbox transfer messed something up? The code on one of the sites(login.php) which produces the error: <?php include "mysql_database.php"; login(); $id = $_SESSION['Loggedin']; setcookie("login", $id, (time()+60*60*24*30)); header('Location: ' . $_SERVER['HTTP_REFERER']); ?> login function: function login() { $connection = connecttodatabase(); $pass = ""; $user = ""; $query = ""; if (isset($_POST['user']) && $_POST['user'] != null) { $user = $_POST['user']; if (isset($_POST['pass']) && $_POST['pass'] != null) { $pass = md5($_POST['pass']); $query = "SELECT ID FROM Anvandare WHERE Nickname='$user' AND Password ='$pass'"; } } if ($query != "") { $id = $connection->query($query); $id = mysqli_fetch_assoc($id); $id = $id['ID']; $_SESSION['Loggedin'] = $id; } closeconnection($connection); } Complete error: Warning: Cannot modify header information - headers already sent by (output started at /Users/name/GitHub/website/login.php:1) in /Users/namn/GitHub/website/login.php on line 9

    Read the article

  • php code is not fetching data from mysql database using wamp server

    - by john
    I want to display a table from database in phpMyAdmin by putting the following conditions that in every different options in drop down menu it displays different table from database by pressing the button of search. But it is not doing so. <p class="h2">Quick Search</p> <div class="sb2_opts"> <p></p> <form method="post" action="" > <p>Enter your source and destination.</p> <p>From:</p> <select name="from"> <option value="Islamabad">Islamabad</option> <option value="Lahore">Lahore</option> <option value="murree">Murree</option> <option value="Muzaffarabad">Muzaffarabad</option> </select> <p>To:</p> <select name="To"> <option value="Islamabad">Islamabad</option> <option value="Lahore">Lahore</option> <option value="murree">Murree</option> <option value="Muzaffarabad">Muzaffarabad</option> </select> <input type="submit" value="search" /> </form> </form> </table> <?php $con=mysqli_connect("localhost","root","","test"); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } if(isset($_POST['from']) and isset($_POST['To'])) { $from = $_POST['from'] ; $to = $_POST['To'] ; $table = array($from, $to); switch ($table) { case array ("Islamabad", "Lahore") : $result = mysqli_query($con,"SELECT * FROM flights"); echo "</flights>"; //table name is flights break; case array ("Islamabad", "Murree") : $result = mysqli_query($con,"SELECT * FROM isb to murree"); echo "</isb to murree>"; //table name isb to murree ; break; case array ("Islamabad", "Muzaffarabad") : $result = mysqli_query($con,"SELECT * FROM isb to muzz"); echo "</isb to muzz>"; break; //..... //...... default: echo "Your choice is nor valid !!"; } } mysqli_close($con); ?>

    Read the article

  • how to bind parameters correctly in example below in mysqli?

    - by user1421767
    In old mysql code, I had a query below which worked perfectly which is below: $questioncontent = (isset($_GET['questioncontent'])) ? $_GET['questioncontent'] : ''; $searchquestion = $questioncontent; $terms = explode(" ", $searchquestion); $questionquery = " SELECT q.QuestionId, q.QuestionContent, o.OptionType, an.Answer, r.ReplyType, FROM Answer an INNER JOIN Question q ON q.AnswerId = an.AnswerId JOIN Reply r ON q.ReplyId = r.ReplyId JOIN Option_Table o ON q.OptionId = o.OptionId WHERE "; foreach ($terms as $each) { $i++; if ($i == 1){ $questionquery .= "q.QuestionContent LIKE `%$each%` "; } else { $questionquery .= "OR q.QuestionContent LIKE `%$each%` "; } } $questionquery .= "GROUP BY q.QuestionId, q.SessionId ORDER BY "; $i = 0; foreach ($terms as $each) { $i++; if ($i != 1) $questionquery .= "+"; $questionquery .= "IF(q.QuestionContent LIKE `%$each%` ,1,0)"; } $questionquery .= " DESC "; But since that old mysql is fading away that people are saying to use PDO or mysqli (Can't use PDO because of version of php I have currently got), I tried changing my code to mysqli, but this is giving me problems. In the code below I have left out the bind_params command, my question is that how do I bind the parameters in the query below? It needs to be able to bind multiple $each because the user is able to type in multiple terms, and each $each is classed as a term. Below is current mysqli code on the same query: $questioncontent = (isset($_GET['questioncontent'])) ? $_GET['questioncontent'] : ''; $searchquestion = $questioncontent; $terms = explode(" ", $searchquestion); $questionquery = " SELECT q.QuestionId, q.QuestionContent, o.OptionType, an.Answer, r.ReplyType, FROM Answer an INNER JOIN Question q ON q.AnswerId = an.AnswerId JOIN Reply r ON q.ReplyId = r.ReplyId JOIN Option_Table o ON q.OptionId = o.OptionId WHERE "; foreach ($terms as $each) { $i++; if ($i == 1){ $questionquery .= "q.QuestionContent LIKE ? "; } else { $questionquery .= "OR q.QuestionContent LIKE ? "; } } $questionquery .= "GROUP BY q.QuestionId, q.SessionId ORDER BY "; $i = 0; foreach ($terms as $each) { $i++; if ($i != 1) $questionquery .= "+"; $questionquery .= "IF(q.QuestionContent LIKE ? ,1,0)"; } $questionquery .= " DESC "; $stmt=$mysqli->prepare($questionquery); $stmt->execute(); $stmt->bind_result($dbQuestionId,$dbQuestionContent,$dbOptionType,$dbAnswer,$dbReplyType); $questionnum = $stmt->num_rows();

    Read the article

  • MySQL only returning one result. Im Baffled.

    - by Tapha
    Here is te code: <?php //Starting session session_start(); //Includes mass includes containing all the files needed to execute the full script //Also shows homepage elements without customs require_once ('includes/mass.php'); $username = $_SESSION['username']; if (isset($username)) { //Query database for the users networths $sq_l = "SELECT * FROM user"; $sql_query_worth = mysql_query($sq_l); while ($row = mysql_fetch_assoc($sql_query_worth)) { $dbusername = $row['username']; } echo $dbusername; } ?>

    Read the article

  • limit PHP script to one domain per license

    - by Mac Os
    what is the best way to make my php code working on one domain and sure i will encode the whole code by ioncube i want function like function domain(){ } if($this_domain <> domain()){ exit('no'); } or $allowed_hosts = array('foo.example.com', 'bar.example.com'); if (!isset($_SERVER['HTTP_HOST']) || !in_array($_SERVER['HTTP_HOST'], $allowed_hosts)) { header($_SERVER['SERVER_PROTOCOL'].' 400 Bad Request'); exit; } now i want know the best way to do that may be will user strpos

    Read the article

  • showing content in profile box from another file

    - by fusion
    i'm stuck at this facebook application, not knowing how do i go ahead. for the canvas, the app works quite perfectly. i've two pages, quote.html and quote.php. in quote.html, quote.php [which gets the quotes from the database] is called through ajax which displays a quote randomly for 10 secs. however, for the wall tab/profile box, i can't seem to know where i'm going wrong. i created a page called 'profile_box.php', call quote.php for the quotes, setFBML. while this works on canvas [sans the timing], it doesn't display anything on the wall tab. <?php include 'quote.php'; $row = mysql_fetch_array($result, MYSQL_ASSOC); if ( !$row ) { echo "Empty"; } else{ $fb_box = "<p>" . h($row['cArabic']) . "</p>"; $fb_box .= "<p>" . h($row['cQuotes']) . "</p>"; $fb_box .= "<p>" . h($row['vAuthor']) . "</p>"; $fb_box .= "<p>" . h($row['vReference']) . "</p>"; } try{ $url="http://website/name/quote.html"; $facebook->api_client->fbml_refreshRefUrl($url); $fbml = "<fb:ref url='$url'/>"; if(isset($_REQUEST['fb_sig_page_id'])){ $page_id = $_REQUEST['fb_sig_page_id']; $profile_type = $_REQUEST['fb_sig_type']; $facebook->api_client->profile_setFBML($appapikey, $page_id, "$fbml", NULL, NULL, "$fb_box"); } else { $is_tab = isset($_REQUEST['fb_sig_in_profile_tab']); if( !$is_tab ) $user_id = $facebook->require_login(); $result = $facebook->api_client->profile_setFBML($appapikey, $user_id, "$fbml", NULL, NULL, "$fb_box"); } } catch(Exception $ex){ echo 'Caught exception: ', $ex->getMessage(), "\n"; } echo "<fb:add-section-button section=\"profile\" /><br />"; ?> can anyone please give any pointers?

    Read the article

  • Someone please can see why the following prepared statment returns nothing?

    - by jartaud
    $stmt = mysqli_prepare($link,"SELECT *FROM ads INNER JOIN dept ON dept.id_dept = ads.in_dpt INNER JOIN members ON members.idMem = ads.from_Mem INNER JOIN sub_cat_ad ON id_sub_cat = ads.ads_in_Cat INNER JOIN cat_ad ON idCat_ad = sub_cat_ad.from_cat_ad WHERE ads_in_Cat = ? "); if(isset($_GET['fromSCat'])){ $fromSCat = mysqli_real_escape_string($link,$_GET['fromSCat']);} mysqli_stmt_bind_param($stmt,'i',$fromSCat); mysqli_stmt_execute($stmt); mysqli_stmt_fetch($stmt); $tot=mysqli_stmt_num_rows($stmt); //Ouput: 0

    Read the article

  • Not allow a href tags in form textarea

    - by saquib
    Hello friends, How can i prevent user to enter any url or link in contact form text area, i have tried it with this but its not working - if (!isset($_POST['submit']) && preg_match_all('/<a.*>.*<\/a>/', $_POST['query'])) { echo "<h1 style='color:red;'>HTML Tag Not allowed </h1>"; } else { //sendmail } Please help me

    Read the article

< Previous Page | 7 8 9 10 11 12 13 14 15 16 17 18  | Next Page >