PHP - Whats Wrong With This Code?
Hi all,
Yeah just a fresh pair of eyes needed. It just wont insert into the table and the alert returns a blank. Only thing I can think is thats its posting variables from two different queries...but that shouldnt make a difference... Cheers Code: [Select] FORM PAGE************************* //*****************************LOGGED CHECKER********************* $sql = "SELECT * FROM users WHERE firstName= '$logged'"; if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { if ($row = mysql_fetch_assoc($result)) { //**************QUESTION FORM***************************** $sql2 = "SELECT * FROM questions WHERE questionID= '$orderCounter'"; if ($result2 = mysql_query($sql2)) { if (mysql_num_rows($result2)) { if ($row2 = mysql_fetch_assoc($result2)) { echo "<p>" . $row2['questionText'] . "</p>"; echo '<form name="form1" method="post" action="questionProcess.php">'; echo '<input type="radio" name="q1" value="' . $row2['answer1'] . '" />' . $row2['answer1'] . '<br />'; echo '<input type="radio" name="q1" value="' . $row2['answer2'] . '" />' . $row2['answer2'] . '<br />'; echo '<input type="radio" name="q1" value="' . $row2['answer3'] . '" />' . $row2['answer3'] . '<br />'; echo '<input type="radio" name="q1" value="' . $row2['answer4'] . '" />' . $row2['answer4'] . '<br />'; echo '<input type="hidden" name="userid" value="' . $row['userid'] . '" />'; echo '<input type="hidden" name="orderCounter" value="' . $row['orderCounter'] . '" />' echo '<input type="submit" name="Submit" id="Submit" value="Next" />'; echo '</label>'; echo '</form>'; } } } } } } ?> PROCESSING PAGE**************************************** $q1=$_POST['q1']; $userid=$_POST['userid']; $orderCounter=$_POST['orderCounter']; // To protect MySQL injection $q1 = stripslashes($q1); $userid = stripslashes($userid); $orderCounter = stripslashes($orderCounter); $q1 = mysql_real_escape_string($q1); $userid = mysql_real_escape_string($userid); $orderCounter = mysql_real_escape_string($orderCounter); //ALERT CHECKER function confirm($msg) { echo "<script langauge=\"javascript\">alert(\"".$msg."\");</script>"; } $msg = $userid; confirm($msg); //ALERT CHECKER if ($q1 == "" ) { $wrong = '<div id="blackText"><p>Please Answer the Question<p></div>'; } else { $sql="INSERT INTO $tbl_name (userid, value, orderCounter) VALUES ('$userid', '$q1', '$orderCounter')"; $result=mysql_query($sql); } ob_end_flush(); ?> Similar TutorialsWhats wrong with this code ? Code: [Select] <?php $dbc = mysqli_connect('localhost', 'boyyo_boyyo', 'KiaNNa11', 'boyyo_aliendatabase') or die('Error connecting to MySQL server.'); $first_name = $_POST['firstname']; $last_name = $_POST['lastname']; $email = $_POST['email']; $query = "INSERT INTO email_list (first_name, last_name, email) " . "VALUES ('$first_name', '$last)name', '$email')"; mysqli_query($dbc, $query ) or die('Error querying database.'); echo 'Customer added. ' ; mysqli_close($dbc); ?> I have no idea whats wrong with this code? $search
= $_POST['search']; $replace = $_POST['replace']; $zero = 0; $gibs = mysql_query("SELECT * FROM anime"); while ($gib = mysql_fetch_assoc($gibs)) { $new_description[] = str_replace($search,$replace,$gib['description']); $new_history[] = str_replace($search,$replace,$gib['history']); $anime_id[] = $gib['anime_id']; $numberlinks = ++$zero; } for($i=0; $i<$numberlinks; $i++) { $thequery[] = mysql_query("UPDATE anime SET description = '".mysql_real_escape_string($new_description[$i])."', history='".mysql_real_escape_string($new_history[$i])."' WHERE anime_id='".$anime_id[$i]."'"); } foreach ($thequery as $query){ mysql_query($query) or die (mysql_error()); echo "<span style='color:green'>Edited sucessfully <b>Anime ".++$one."</b> </span><br/>"; } echo "<meta http-equiv='refresh' content='2;url=http://www.website.com/forums/downloads-manager.php?task=fix-descriptions'>"; output: Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1[/b] Inputs: $search: ’ $replace: " Hi, I'm trying to copy some values from a table called "member" to a table called "report" when new members submit a php application form. I can't see what's wrong with it. I have attached screenshots of the 2 tables and their structure. Any help would be much appreciated. global $conn,$dal; $customerID = mysql_insert_id(); $strSQLInsert = "insert into report (appID, memberID, select_agent, agent_details, first_name, last_name, street_address, suburb, postcode) values (".$ID.", ".$ID.", '".$values["select_agent"]."', '".$values["agent_details"]."', '".$values["first_name"]."', '".$values["last_name"]."', '".$values["street_address"]."', '".$values["suburb"]."', '".$values["postcode"]."')"; db_exec($strSQLInsert,$conn); [attachment deleted by admin] I have an upload site whereby people get direct file links. I want to output links as bit.ly My code stands as following: Code: [Select] <?php $content = file_get_contents("http://api.bit.ly/v3/shorten?login=andre1990 &apiKey=R_56a95c84d089129012516e24806c3649 &longUrl='.urlencode($final).'&format=txt"); ?> //Input type because i want a text box with the link in it. <input type="text" size="28" onClick=select() value="<?php echo $content; ?>" READONLY><p /> where $final is the uploaded file URL. But it just outputs nothing, no bit.ly link. Any help? Cant Figure Out Whats Wrong With it.. Code: Code: [Select] <?php if(isset($_SESSION['user'])) { $bb = mysql_query("SELECT * FROM main WHERE id=". $_SESSION['id']) or die ("An error has occured: " . mysql_error()); while($n=mysql_fetch_array($bb)) { ?> <div class='cs_article'> <div class='left'> <h2>User Control Panel</h2> <p><a href="usercp.php">User CP</a><?php if($n['rights'] == 1) { echo ", <a href="modcp.php">Mod CP</a>"; }?> <?php if($n['rights'] == 2) { echo ", <a href="aa/">Admin CP</a>"; }?> <div class='button'><a href='usercp.php'>Read more</a></div> </div> <div class='right'> <h1>User CP</h1> </div> </div> <?php }} ?> What its doing: Mod-Justin Compared to Mod-Justin hi all me again i have this bit of code mysql_connect($db_hostname,$db_username,$db_password); @mysql_select_db($db_database) or die( "Unable to select database"); $whereClauses = array(); if(isset($_GET['bi'])) { $whereClauses[] = "bi=1"; } if(isset($_GET['print'])) { $whereClauses[] = "print=1"; } if(isset($_GET['online'])) { $whereClauses[] = "online=1"; } $query = "SELECT * FROM `project` ORDER BY `project`.`position` ASC "; if(count($whereClauses)) { $query .= " WHERE " . implode(" AND ", $whereClauses); } with this error msg There was a problem with the SQL query: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE online=1' at line 3 I get this kinda error a lot as im still learning is there any way to better find out whats going wrong i use error_reporting(E_ALL); ini_set('display_errors', '1'); at the start of my scripts but it dont tell me a lot. i can work out it some thing to do with the $whereClauses as the page displays ok on show all. Im bit stumped here not even sure what to call the prob is it sql query error or php array problem thanks for any help can you tell me whats wrong with this? Code: [Select] <?php $age= 60; if( isset($_SESSION['logedin']) ) { $q = mysql_query('SELECT id=`$id`, DATE_FORMAT(`last_activity`,"%a, %b %e %T") as `last_activity`,UNIX_TIMESTAMP(`last_activity`) as `last_activity_stamp`FROM `mymembers`WHERE `$logOptions_id` <> "'.($_SESSION['logedin']).'"'); $isonlinecheck = mysql_query($q); if ($isonlinecheck ="last_activity_stamp + $age "< time()){ $isonline = "is <font color='green'>online!</font>";} else { $isonline = "is<font color='red'> offline!</font>"; } } ?> Its a page 'Downloads'. Works great at localhost but doesnt show up @ web hosting. Code: [Select] <?php define('IN_PHPBB', true); $page = !isset($_GET["page"]) ? "None" : $_GET['page']; $phpbb_root_path = (defined('PHPBB_ROOT_PATH')) ? PHPBB_ROOT_PATH : './'; $phpEx = substr(strrchr(__FILE__, '.'), 1); include($phpbb_root_path . 'common.' . $phpEx); // Start session management $user->session_begin(); $auth->acl($user->data); $user->setup(); page_header('Downloads'); if(strcmp($page,"org") == 0) { $template->set_filenames(array('body' => 'download/org.html',)); } else if(strcmp($page,"audi") == 0) { $template->set_filenames(array('body' => 'download/audi.html',)); } else if(strcmp($page,"bmw") == 0) { $template->set_filenames(array('body' => 'download/bmw.html',)); } else if(strcmp($page,"vw") == 0) { $template->set_filenames(array('body' => 'download/vw.html',)); } else if(strcmp($page,"nissan") == 0) { $template->set_filenames(array('body' => 'download/nissan.html',)); } else if(strcmp($page,"opel") == 0) { $template->set_filenames(array('body' => 'download/opel.html',)); } else if(strcmp($page,"ford") == 0) { $template->set_filenames(array('body' => 'download/ford.html',)); } else if(strcmp($page,"chevrolet") == 0) { $template->set_filenames(array('body' => 'download/chevrolet.html',)); } else if(strcmp($page,"other") == 0) { $template->set_filenames(array('body' => 'download/other.html',)); } else { $template->set_filenames(array('body' => 'download/org.html',)); } make_jumpbox(append_sid("{$phpbb_root_path}viewforum.$phpEx")); page_footer(); ?> and the html: Code: [Select] <table class="tablebg" cellspacing="1" width="100%"> <tr> <th>BMW</th> </tr> <tr class="row1"> <td align="center" style="padding:5px 5px 5px 5px;"> <center> <hr width="60%" size="3" /> <table width="70%"> <tr> <td width="10%"><a href="http://www.upload.ee/image/1626514/325i_2.jpg" target = "_blank"> <img src="http://www.upload.ee/thumb/1626514/325i_2.jpg" border="0"/></a> <a href="http://www.upload.ee/image/1626518/325i_3.jpg"> <img src="http://www.upload.ee/thumb/1626518/325i_3.jpg" border="0"/></a> <td width="0%"><a href="http://www.upload.ee/image/1626519/325i_1.jpg" target = "_blank"> <img src="http://www.upload.ee/thumb/1626519/325i_1.jpg" border="0"/></a> <a href="http://www.upload.ee/image/1626521/325i_4.jpg" target = "_blank"> <img src="http://www.upload.ee/thumb/1626521/325i_4.jpg" border="0"/></a></td></td> <td><b><font color="white">Name:</font></b> 1996 BMW 325i e36 Convertible</br> <b><font color="white">Original Author:</font></b> ikey07</br> <b><font color="white">Size:</font></b> 3.89MB</br></br> <b><font color="white">Description:</font></b></br> Tunable in WAA</br> 2 Body kits included</br> 1. BMW M e36 series + Hood Vents</br> 2. BMW M e46 series + Masked Lights</br> New 2 M series wheels e36/e46</br> Realistic heavy damaged model</br> Model accuracy as real car 96%</br> Openable petrolcap</br> If you replace this car as BF Injection, the front fan is working ( spining )</br></br> <a href="http://ikey07.c-rp.net/viewtopic.php?f=6&t=2"><b><font color="white">Comments</font></b></a></br> </br> <ul id="nav"> <a class="downl" href="http://ikey07.c-rp.net/download/1996_BMW_325i_e36.rar" title="Download"><span>{L_DOWN}</span></a> </div> </br> </td> </tr> </table> <hr width="60%" size="3" /> </center> </td> </tr> </table> <br /> Code: [Select] <?php function dbConnect($usertype, $connectionType = 'mysqli') { $host = 'localhost'; $db = 'gallery'; if ($usertype == 'read') { $user = 'psread'; $pwd = '####'; } elseif ($usertype == 'write') { $user = 'pswrite'; $pwd = '####'; } else { exit('Unrecognized connection type'); } if ($connectionType == 'mysqli') { return new mysqli($host, $user, $pwd, $db) or die ('Cannot open database'); } } ?> when i call upon the function i get this message Fatal error: Call to a member function query() on a non-object in C:\webs\phpsolutions\comments.php on line 8 from this code Code: [Select] <?php require_once('includes/connection.php'); // connect to MySQL $conn = dbConnect('write'); // prepare the SQL query $sql = 'SELECT * FROM images'; // submit the query and capture the result $result = $conn->query($sql) or die(mysqli_error()); // find out how many records were retrieved $numRows = $result->num_rows; ?> Not sure what is wrong with line 17
<?php $hostname="localhost"; //local server name default localhost $username="root"; //mysql username default is root. $password=""; //blank if no password is set for mysql. $database="login"; //database name which you created $con=mysql_connect($hostname,$username,$password); if(! $con) { die('Connection Failed'.mysql_error()); } mysql_select_db($database,$con); //include connect.php page for database connection Include('connect.php') //if submit is not blanked i.e. it is clicked. If(isset($_REQUEST['submit'])!='') { If($_REQUEST['name']=='' || $_REQUEST['email']=='' || $_REQUEST['password']==''|| $_REQUEST['repassword']=='') { Echo "please fill the empty field."; } Else { $sql="insert into student(name,email,password,repassword) values('".$_REQUEST['name']."', '".$_REQUEST['email']."', '".$_REQUEST['password']."', '".$_REQUEST['repassword']."')"; $res=mysql_query($sql); If($res) { Echo "Record successfully inserted"; } Else { Echo "There is some problem in inserting record"; } } } ?> Edited by Metoriium, 25 October 2014 - 05:04 PM. hello. i have a column called "pageName" and on the each page i have put $pName = "the name of that page". i want to pull from the database all the information where the $pName on the page is the same as the pageName in the database its getting the $pName = "adminHome" but it thinks its the name of the column not the row. this is the error im getting: Quote DATABASE.PHP - confirm_query = MySQL Datatbase Query Failed: Unknown column 'adminHome' in 'where clause' Last SQL query: SELECT * FROM pages WHERE pageName=adminHome this is the function code Code: [Select] public static function find_by_pageName($pName=""){ global $database; $sql = "SELECT * FROM ".self::$table_name." WHERE pageName=".$database->escape_value($pName).""; $result_array = self::find_by_sql($sql); return !empty($result_array) ? array_shift($result_array) : false; } this is the page code Code: [Select] <?PHP require_once("../includes/initialize.php"); $pName = "adminHome"; $page = Pages::find_by_pageName($pName); ?> thanks rick I can't honestly see whats wrong with this code, it's print'ing correct with the print_r. Code: [Select] Array ( [value1] => 2 [option] => - [value2] => 1 ) however i can't actually echo out the value1,value2 or option? It says the vars are empty yet it can still print_r the values so i know there not empty Code: [Select] <?php if(isset($_POST["submit"])){ $value1=$_POST["value1"]; $value2=$_POST["value2"]; $option=$_POST["option"]; } print_r($_POST); echo"$value1"; ?> <form action="index.php" name="submit" method="post"> <input type="text" name="value1"> <select name="option"> <option>+</option> <option>-</option> <option>*</option> <option>/</option> </select> <input type="text" name="value2"> <input type="submit" value="submit"> </form> ive looked at this for a while now and im not sure whats wrong, the error occured when i entered $cat and catagory into it, without that the query works perfectly fine, any help appreciated. <?php INSERT INTO newnotes (uid, name, catagory, location) VALUES ('1', 'test', '1001 Laws - Polceing', '/home/mikeh/public_html/uzEr Upl0ds/Alyssa O'Leary.doc') the <php tag is only to give color and make it less dull the query is from a die($query) statment Hello,
I am trying to make 'onclick img' that will provide a Linux command when clicked.
I've been trying this code:
<body> <?php include("connect.php"); include('Net/SSH2.php'); *** CONNECTION TO DATABASE MYSQL CODE *** if(mysql_num_rows($result) > 0) { while($row = mysql_fetch_array($result)) { $_SESSION['logged'] = true; $ssh = new Net_SSH2('xx.xx.xx.xx'); if (!$ssh->login($info['username'], $info['password'])) { exit('Connection to Linux has failed, please try again.'); } echo '<form action="" method="GET">'; echo '<input type="image" src="start.png" width=100 height=100 name="start" />'; echo '</form>'; if(isset($_GET['start'])) { echo $ssh->setTimeout(5); echo $ssh->exec('(cd '.$info['username'].' ; nohup ./samp03svr &)'); } } } else { $_SESSION['logged'] = null; header("Location: loginfail.php"); } ?> </body>But when I clicked the image, it sends me to "loginfail.php". Please help >< hi all, Is this line right? Im trying to say if the cookie has not been created create it and set as default value = 0 however i dont think this is right: Code: [Select] $points = $_COOKIE["points"] ? $_COOKIE['points']:0; thanks, hi all, I have possibly up to 9 images that I want to find and show on a webpage. I am trying to put a $file in a loop and get all the .jpg images that are found in to an array and show them all. I think from the script you can see what I am attempting to do! Any pointers would be grateful for <?php while($i=0 $i < 9 $i++) { $file = "aircraft/".$_GET['reg'].$i.".jpg"; } foreach($file) { echo "<li class=\"active\"><img src=\"./aircraft/".$reg.$i.".jpg\" alt=\"Flowing Rock\" style=\"width: auto; height: 70px; margin-left: -28.5px; display: block;\" class=\"thumb\" ><br />"; } ?> This topic has been moved to CSS Help. http://www.phpfreaks.com/forums/index.php?topic=358008.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=307477.0 |