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PHP - Moved: [need To Help] Insert To Database Error.
This topic has been moved to MySQL Help.
http://www.phpfreaks.com/forums/index.php?topic=354933.0 Similar TutorialsThis topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=350408.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=352154.0 i am trying to insert data into a database with the following code <?php $first_name=$_POST['first_name']; $middle_name=$_POST['middle_name']; $last_name=$_POST['last_name']; $gender=$_POST['gender']; $file_number=$_POST['file_number']; $character=$_POST['character']; $diagnosis=$_POST['diagnosis']; $description=$_POST['description']; $day = $_POST['day']; $month = $_POST['month']; $year = $_POST['year']; $date = date("Y-m-d", mktime(0,0,0,$month, $day, $year)); $con = mysql_connect("localhost","fathersh_search","f33321rh"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("fathersh_childsearch", $con); $sql="INSERT INTO child_info (first_name,middle_name,last_name,gender,birthdate,character,diagnosis,description,file_number) VALUES ('$_POST[first_name]','$_POST[middle_name]','$_POST[last_name]','$_POST[file_number]','$_POST[gender]','$date','$_POST[character]','$_POST[diagnosis]','$_POST[description]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> the error i get is Error: 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 'character,diagnosis,description,file_number) VALUES ('James','Anthony','Peters',' at line 1 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=342913.0 hello I want query from one table and insert in another table on another domain . each database on one domain name. for example http://www.site.com $con1 and http://www.site1.com $con. can anyone help me? my code is : <?php $dbuser1 = "insert in this database"; $dbpass1 = "insert in this database"; $dbhost1 = "localhost"; $dbname1 = "insert in this database"; // Connecting, selecting database $con1 = mysql_connect($dbhost1, $dbuser1, $dbpass1) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname1) or die('Could not select database'); $dbuser = "query from this database"; $dbpass = "query from this database"; $dbhost = "localhost"; $dbname = "query from this database"; // Connecting, selecting database $con = mysql_connect($dbhost, $dbuser, $dbpass) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname) or die('Could not select database'); //query from database $query = mysql_query("SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1"); while($row=mysql_fetch_array($query)){ $result=$row[0]; $text=$row[1]."</br>Size:(".$row[4].")"; $alias=$row[2]; $link = '<a target="_blank" href='.$row[3].'>Download</a>'; echo $result; } //insert into database mysql_query("SET NAMES 'utf8'", $con1); $query3= " INSERT INTO `jos_content` (`id`, `title`, `alias`, `) VALUES (NULL, '".$result."', '".$alias."', '')"; if (!mysql_query($query3,$con1)) { die('Error: text add' . mysql_error()); } mysql_close($con); mysql_close($con1); ?> how do I make the $image piece so that it inserts "images/$image.php" into the database? VALUES('$name', '$id', '$image','$event')"; I've begun to play around with object oriented php and I was wondering how I would go about writing a method that inserts values into a table.
I have a class, called queries, and I made this very simple function:
class queries {
public function insert($table, $field, $value){
global $dbh;
$stmt = $dbh->prepare("INSERT INTO $table ($field) VALUES (:value)");
$stmt->bindParam(":value", $value);
$stmt->execute();
}
}
$queries = new queries();
Hi guys, Thanks
if(isset($_POST['send'])){
$category = $_POST['category'];
$item_type = $_POST['item_type'];
$item_size = $_POST['item_size'];
$product_name = $_POST['product_name'];
$item_qty = $_POST['item_qty'];
$price = $_POST['price'];
$total_price = $_POST['total_price'];
$item_price = $_POST['item_price'];
$sql = "
INSERT INTO shopping(
trans_ref,
category,
product_name,
item_type,
item_size,
item_qty,
item_price,
price,
date
)
VALUES(
:trans_ref,
:category,
:product_name,
:item_type,
:item_size,
:item_qty,
:item_price,
:price,
NOW()
)
";
$stmt = $pdo->prepare($sql);
$stmt->execute(array(
':trans_ref' => $_SESSION['trans_ref'],
':category' => $category,
':product_name' => $product_name,
':item_type' => $item_type,
':item_size' => $item_size,
':item_qty' => $item_qty,
':item_price' => $item_price,
':price' => $price
));
if ( $stmt->rowCount()){
echo '<div class="alert bg-success text-center">SHOPPING COMPLETED</div>';
}else{
echo '<div class="alert bg-danger text-center">SHOPPING NOT COMPLETED. A PROBLE OCCURED</div>';
}
}
Everything on this page works the form loads and i can insert info into the form but when i click submit it just reloads the page(action="this page") and nothing happens. thanks in advance -Adam Code: [Select] <?php require("connection/connection.php"); include("connection/functions.php"); ?> <HTML> <head> <meta content="yes" name="apple-mobile-web-app-capable" /> <meta content="index,follow" name="robots" /> <meta content="text/html; charset=utf-8" http-equiv="Content-Type" /> <link href="" rel="apple-touch-icon" /> <meta content="minimum-scale=1.0, width=device-width, maximum-scale=0.6667, user-scalable=no" name="viewport" /> <meta name="format-detection" content="telephone=no"/> <link href="css/style.css" rel="stylesheet" media="screen" type="text/css" /> <script src="javascript/functions.js" type="text/javascript"></script> <title>.: Bookmarks :.</title> <link href="" rel="apple-touch-startup-image" /> <meta content="apple-mobile-web-app-status-bar-style" content="default" /> </head> <body class="musiclist"> <div id="topbar"> <div id="title">.: Bookmarks :.</div> <div id="leftbutton"> <a href="default.html"><img src="images/home.png"></a> </div> <div id="rightbutton"> <a href="">Programs</a> </div> </div> <div id="tributton"> <div class="links"> <a id="pressed" href="Bookmarks/Bookmarks.php">Bookmarks</a><a href="Music/Music.html">Music</a><a href="">Movies</a> </div> </div> <div id="content" style="top: 36px"> <form method="post" action="AddBookmark.php"> <span class="graytitle">Add Bookmark</span> <ul class="pageitem"> <li class="bigfield"><input type="text" name="BookmarkName" placeholder="Bookmark Name" /></li> <li class="bigfield"><input type="text" name="BookmarkLink" placeholder="Bookmark Link" /></li> <li class="select"><select name="d"> <?php $getbookmarksections = mysql_query("SELECT * FROM bookmarksections"); confirm_query($getbookmarksections); while($getbm = mysql_fetch_array($getbookmarksections)){ $id = $getbm["ID"]; $sn = $getbm["SectionName"]; echo "<option value=\"$id\">$sn</option>"; } ?> </select><span class="arrow"></span></li> <li class="button"> <input name="AddBookmark" type="submit" value="Add Bookmark" /> </li> </ul> </form> <?php if(isset($_POST["AddBookmark"])){ $SectionID = $id; $Bookmark = $_POST["BookmarkName"]; $BookmarkLink = $_POST["BookmarkLink"]; $add = "INSERT INTO `bookmarks` (`ID` ,`SectionID` ,`Bookmark` ,`BookmarkLink`) VALUES ('NULL', '$SectionID', '$Bookmark', '$BookmarkLink')"; if(confirm_query($add)){ echo "<span class=\"graytitle\">Bookmark '$BookmarkName' in '$sn' Added.</span>"; } } ?> </div> <div id="footer"> </div> </body> </HTML> <?php //Close Connection if(isset($connection)){ mysql_close($connection); } ?> this is the code i suspect is not working Code: [Select] <?php if(isset($_POST["AddBookmark"])){ $SectionID = $id; $Bookmark = $_POST["BookmarkName"]; $BookmarkLink = $_POST["BookmarkLink"]; $add = "INSERT INTO `bookmarks` (`ID` ,`SectionID` ,`Bookmark` ,`BookmarkLink`) VALUES ('NULL', '$SectionID', '$Bookmark', '$BookmarkLink')"; if(confirm_query($add)){ echo "<span class=\"graytitle\">Bookmark '$BookmarkName' in '$sn' Added.</span>"; } } ?> Hi all, I've made a page the variables are being recieved and echo'd ok, however they are not being inserted into the database. below is the file with the insert statement. any help really appreciated. MsKazza Code: [Select] <?php // Database connect $con = mysql_connect("mysql1.myhost.ie","admin_book","root123"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("book_test", $con); $orderid=$_GET['orderid']; //Parse Values from Coupon.php Form $orderid = mysql_real_escape_string(trim($_POST['orderid'])); $design = mysql_real_escape_string(trim($_POST['design'])); $childname = mysql_real_escape_string(trim($_POST['childname'])); $address = mysql_real_escape_string(trim($_POST['address'])); echo $orderid; echo $design; echo $childname; echo $address; $sql="INSERT INTO details (orderid, design, childname, address) VALUES ('$orderid','$design','$childname','$address')"; ?> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Digital Scribe Books</title> <link href="style.css" rel="stylesheet" type="text/css" /> <script type="text/javascript"> function MM_preloadImages() { //v3.0 var d=document; if(d.images){ if(!d.MM_p) d.MM_p=new Array(); var i,j=d.MM_p.length,a=MM_preloadImages.arguments; for(i=0; i<a.length; i++) if (a[i].indexOf("#")!=0){ d.MM_p[j]=new Image; d.MM_p[j++].src=a[i];}} } function MM_swapImgRestore() { //v3.0 var i,x,a=document.MM_sr; for(i=0;a&&i<a.length&&(x=a[i])&&x.oSrc;i++) x.src=x.oSrc; } function MM_findObj(n, d) { //v4.01 var p,i,x; if(!d) d=document; if((p=n.indexOf("?"))>0&&parent.frames.length) { d=parent.frames[n.substring(p+1)].document; n=n.substring(0,p);} if(!(x=d[n])&&d.all) x=d.all[n]; for (i=0;!x&&i<d.forms.length;i++) x=d.forms[i][n]; for(i=0;!x&&d.layers&&i<d.layers.length;i++) x=MM_findObj(n,d.layers[i].document); if(!x && d.getElementById) x=d.getElementById(n); return x; } function MM_swapImage() { //v3.0 var i,j=0,x,a=MM_swapImage.arguments; document.MM_sr=new Array; for(i=0;i<(a.length-2);i+=3) if ((x=MM_findObj(a[i]))!=null){document.MM_sr[j++]=x; if(!x.oSrc) x.oSrc=x.src; x.src=a[i+2];} } </script> </head> <body onload="MM_preloadImages('images/buttons/home_over.png','images/buttons/books_over.png','images/buttons/cards_over.png','images/buttons/letters_over.png')"> <div id="snow"> <div id="wrapper"> <div id="header"> <div id="logo"><img src="images/digital_scripe.png" width="218" height="91" /></div> <div id="menu"><a href="index.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Home','','images/buttons/home_over.png',1)"><img src="images/buttons/home_act.png" name="Home" width="131" height="132" border="0" id="Home" /></a><a href="books.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Books','','images/buttons/books_over.png',1)"><img src="images/buttons/books_act.png" name="Books" width="131" height="132" border="0" id="Books" /></a><a href="cards.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Cards','','images/buttons/cards_over.png',1)"><img src="images/buttons/cards_act.png" name="Cards" width="131" height="132" border="0" id="Cards" /></a><a href="letters.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Letters','','images/buttons/letters_over.png',1)"><img src="images/buttons/letters_act.png" name="Letters" width="131" height="132" border="0" id="Letters" /></a></div> </div> <div id="content"> <div id="info_bar"><br /><br /><br /> <p>Your personal order number is:</p><br /> <br /> <p>Please fill out the details of where you would like your order to be shipped to.</p> </div> <form action="card_paynow.php" method="POST" name="custdetails" id="custdetails"> <table width="60%" border="0" cellspacing="0" cellpadding="0"> <tr> <td width="19%">orderid</td> <td width="81%"><input name="orderid" type="hidden" value="<? echo $orderid; ?>"></td> </tr> <tr> <td>name</td> <td><input name="name" type="text" id="name"></td> </tr> <tr> <td>surname</td> <td><input name="surname" type="text" id="surname"></td> </tr> <tr> <td>address 1 </td> <td><input name="add1" type="text" id="add1"></td> </tr> <tr> <td>address 2 </td> <td><input name="add2" type="text" id="add2"></td> </tr> <tr> <td>town</td> <td><input name="town" type="text" id="town"></td> </tr> <tr> <td>county</td> <td><input name="county" type="text" id="county"></td> </tr> <tr> <td>postcode</td> <td><input name="postcode" type="text" id="postcode"></td> </tr> <tr> <td>number</td> <td><input name="phone" type="text" id="phone"></td> </tr> <tr> <td>email</td> <td><input name="email" type="text" id="email"></td> </tr> <tr> <td> </td> <td> </td> </tr> <tr> <td> </td> <td><input type="submit" name="Submit" value="Continue..."></td> </tr> <tr> <td> </td> <td> </td> </tr> </table> </form> </div> <div id="footer" class="clear"><div id="sign"><div id="sign_text">Personalised<br /> Books</div> </div></div> </div></div> </body> </html> Hi everyone. I have a registration form on my website that returns errors if the form isnt filled out correctly. An example of the code that returns the errors is: // if first name is blank return error if (!$first_name) { echo "Please enter your first name!<br />"; } // if first name contains invalid charcters return error if (preg_match('/[^a-zA-Z\s]/', $first_name)) { echo "Your first name contains invalid charcters!<br />"; } // set first name maximum length if ($first_name && !preg_match('/[^a-zA-Z0-9\s]/', $first_name)) { if (strlen($first_name) >25) { echo "The maximum length for first name is 25 characters!<br />"; } } Iv wrote it like this so that i can get multiple error messages appearing at the same time. When it comes to writing the form data to the database im using the following: if ($first_name && !preg_match('/[^a-zA-Z\s]/', $first_name) && strlen($first_name) <26 && $last_name && !preg_match('/[^a-zA-Z\s]/', $last_name) && strlen($last_name) <26 && $dob_day != "Day" && $dob_month != "Month" && $dob_year != "Year" && $gender != "Select" && $email && filter_var($email, FILTER_VALIDATE_EMAIL) && $confirm_email && $email == $confirm_email && $email_count == 0 && $town && !preg_match('/[^a-zA-Z0-9\s]/', $town) && strlen($town) < 26 && $location != "Select" && $postcode && !preg_match('/[^a-zA-Z0-9\s]/', $postcode) && $username && !preg_match('/[^a-zA-Z\s]/', $username) && strlen($username) >2 && strlen($username) <25 && $username_count == 0 && $password && strlen($password) >5 && strlen($password) <26 && $confirm_password && $password == $confirm_password) { // code to write to database here } I was wondering if there was a better way to confirm that all the form data is ok, and if so write the data to the database? Im new to php and would love to learn a way to make this easier instead of typing out loads of lines of code. Thanks to anyone that can help me. Hello, I am trying to get a news script to work, but I can't get it to insert anything into the database. I cannot figure it out for the life of me. Code: (newsupdate.php) [Select] <form name="shout" action="post.php" method="post"> <p><b>Name:</b><br/><input type="text" name="name" size="15"/></p> <p><b>Message</b>:<br/> <textarea wrap="physical" name="message" rows="3" cols="25">News goes here! </textarea></p> <p><input type="submit" value="Shout now!"/> <input type="reset" value="Clear"/></p> </form> Code: (post.php) [Select] <? $name = $_POST["name"]; $message = $_POST["message"]; include("dbconnect.php"); $date = date("M j y"); $menu = MYSQL_QUERY("INSERT INTO adminnews (id,name,date,message)". "VALUES ('NULL', '$name', '$date', '$message')"); echo("Shout-out added! You will be redirected to the main page shortly. If you are not, click <a href='index.php'>here</a>."); ?> dbconnect.php is just the configuration for connecting to the database, and I have checked that all ready. A log-in script uses it just fine. Hi guys, Is it possible to insert into two tables in a database? Thanks Hello, ive got the right output from the code below, only my mysql query doesnt seem to be working as it should. Im not too great with mysql so please any help or suggestions would be great. I have tried the code but when I check my database nothing has been inserted. !?! Code: [Select] <?php include('db.php'); include('func.php'); ?><html><head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Chained Select Boxes using PHP, MySQL and jQuery</title> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function() { $('#wait_1').hide(); $('#drop_1').change(function(){ $('#wait_1').show(); $('#result_1').hide(); $.get("func.php", { func: "drop_1", drop_var: $('#drop_1').val() }, function(response){ $('#result_1').fadeOut(); setTimeout("finishAjax('result_1', '"+escape(response)+"')", 400); }); return false; }); }); function finishAjax(id, response) { $('#wait_1').hide(); $('#'+id).html(unescape(response)); $('#'+id).fadeIn(); } </script> </head> <body> <p> <form action="" method="post"> Name: <input type="text" name="Name" /><br /> Phone: <input type="text" name="Phone" /><br /> Email: <input type="text" name="Email" /><br /> Postcode: <input type="text" name="Postcode" /><br /> Web Address: <input type="text" name="Website" /><br /><br /> <select name="drop_1" id="drop_1"> <option value="" selected="selected" disabled="disabled">Select a Category</option> <?php getTierOne(); ?> </select> <span id="wait_1" style="display: none;"> <img alt="Please Wait" src="ajax-loader.gif"/> </span> <span id="result_1" style="display: none;"></span> <br /> </form> </p> <p> <?php if(isset($_POST['submit'])){ $drop = $_POST['drop_1']; $tier_two = $_POST['Subtype']; echo "You selected "; echo $drop." & ".$tier_two; } $Name = $_POST["Name"]; $Phone = $_POST["Phone"]; $Email = $_POST["Email"]; $Postcode = $_POST["Postcode"]; $Website = $_POST["Website"]; echo "<br>"; echo $Name; echo "<br>"; echo $Website; mysql_query ("INSERT INTO business (Name, Type, Subtype, Phone, Email, Postcode, Web Address) VALUES ('$Name', '$drop', '$tier_two' , '$Phone', '$Email', '$Postcode', '$Website')"); ?> Im not sure it makes a difference but, I am adding data into each column of my database with the exception of the 1st column named 'ID' which is set to auto_increment. hi, i was having difficulty and i'm quite confused how to insert these values to the database before it will become as it is, based on this article: http://roshanbh.com.np/2008/01/populate-triple-drop-down-list-change-options-value-from-database-using-ajax-and-php.html what i did was putting/saving the values directly in the database (mysql). but now i want those values to be coming from the user's input:the country,state and city so the system will just get those values, store it to the database with the fields of each table having the same content like that from article above..i really need some help here.. i have this code... and i want to make it more dynamic.... Code: [Select] <?php $content = $content[1].$content[2].$content[3].$content[4].$content[5].$content[6].$content[7].$content[8].$content[9].$content[10].$content[11].$content[12] .$content[13].$content[14].$content[15].$content[16] .$content[17].$content[18].$content[19].$content[20].$content[21].$content[22].$content[23].$content[24]; $sql = mysql_query ("insert into database (content) values ('$content')") or die(mysql_error()); ?> for example... instead content[1]. content[2]. content[3] ...etc... i want something like this....... $content = $content[from 1 to 24]; or $content = $content[from 1 to 777]; any solutions? how do i insert this into a database? i need the php code please. Code: [Select] <select class="select" name="sex" id="sex"> <option value="0">Select Gender</option> <option value="1">Female</option> <option value="2">Male</option></select>I need it to insert what ever one they select Thanks I created this database Code: [Select] <?php $mysqli = mysqli_connect('localhost', 'admin', 'jce123', 'php_class'); if(mysqli_connect_errno()) { printf("connection failed: %s\n", mysqli_connect_error()); exit(); }else{ $q = mysqli_query($mysqli, "DROP TABLE IF EXISTS airline_survey"); if($q){echo "deleted the table airline_survey....<br>";} else{echo "damm... ".mysqli_error($mysqli);} $sql = "CREATE TABLE airline_survey ( id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, staff CHAR(10) NOT NULL, luggage CHAR(10) NOT NULL, seating CHAR(10) NOT NULL, clean CHAR(10) NOT NULL, noise CHAR(10) NOT NULL )"; $res = mysqli_query($mysqli, $sql); if($res === TRUE) { echo "table created"; } else { printf("Could not create table: %s\n", mysqli_error($mysqli)); } mysqli_close($mysqli); } ?> When I look at it it looks fine. I have a form that sends data to this script: Code: [Select] <?php $con = mysql_connect('localhost', 'admin', 'abc123'); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("php_class", $con); foreach ($_POST as $key => $value) { $staff = ""; $luggage = ""; $seating = ""; $clean = ""; $noise = ""; switch($key){ case "staff": $staff = $value; break; case "luggage": $luggage = $value; break; case "seating": $seating = $value; break; case "clean": $clean = $value; break; case "noise": $noise = $value; break; default: echo "we must be in the twilight zone"; } echo $staff."<br>"; [color=red] mysql_query("INSERT INTO airline_survey (staff, luggage, seating, clean, noise) VALUES ($staff, $luggage, $seating, $clean, $noise)");[/color] } ?> as you can see right before the insert query I test one of the variables to see if it has the string I'm expecting and it does. The problem is the script runs without giving me an error message but the data never gets inserted into the table. Hi I have started my error checking for my form. I am understand that the date must be inserted in the format yyyy/mm/dd but on my form i need it to be inserted in the format dd/mm/yyyy. I have wrote some code but have only been doing php about a week properly so I can't see what my error is. Whenever I try to insert the date, the form its self does not error but if i check the database the date is simply 0000/00/00. Can anyone help? if(isset($_POST['submit'])) { $first_name = mysql_real_escape_string($_POST['first_name']); $last_name = mysql_real_escape_string($_POST['last_name']); $DOB = mysql_real_escape_string($_POST['DOB']); $sex = mysql_real_escape_string($_POST['sex']); $email = mysql_real_escape_string($_POST['email']); $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $agree = mysql_real_escape_string($_POST['agreed']); $creation_date = mysql_real_escape_string($_POST['creation_date']); $user_type = mysql_real_escape_string($_POST['member_type']); $access_level = mysql_real_escape_string($_POST['access_level']); $validation = mysql_real_escape_string($_POST['validation_id']); $club_user = mysql_real_escape_string($_POST['user_type']); $date_check = "/^([0-9]{2})/([0-9]{2})/([0-9]{4})$/"; if($first_name == "") { $message = "Please enter a first name"; $success = 0; } else if($last_name == "") { $message = "Please enter a surname"; $success = 0; } else if($DOB =="") { $message = "Please enter your date of birth."; $success = 0; } else if(!(preg_match("/^([0-9]{2})\/([0-9]{2})\/([0-9]{4})$/", $DOB))) { $message = "Please enter your birthday in the format dd/mm/yyyy"; $success = 0; } else if($email =="") { $message = "Please enter a correct email."; $success = 0; } else if($username =="") { $message = "Please enter a username."; $success = 0; } else if($password =="") { $message = "Please enter a password greater than 6 characters long."; $success = 0; } else { $DOB = $explode[2]."-".$explode[1]."-".$explode[0]; $password_md5 = md5($password); $insert_member= "INSERT INTO Members (`first_name`,`last_name`,`DOB`,`sex`,`email`,`username`,`password`,`agree`,`creation_date`,`usertype`,`access_level`,`validationID`) VALUES ('".$first_name."','".$last_name."','".$DOB."','".$sex."','".$email."','".$username."','".$password_md5."','".$agree."','".$creation_date."','".$user_type."','".$access_level."', '".$validation."')"; $insert_member_now= mysql_query($insert_member) or die(mysql_error()); $url = "thankyou.php?name=".$_POST['username']; header('Location: '.$url); Hi... Got a slight problem with my code, it keeps telling me that there is an error in my syntax, yet I have used the same code before perfectly, and for the life of me cannot see the problem? ....any ideas? <?php if (!isset($_SESSION)) { session_start(); } // Connects to your Database mysql_connect("localhost", "justair1_mick", "guru1969") or die(mysql_error()) ; mysql_select_db("justair1_justair") or die(mysql_error()) ; $filename = ($_FILES['image']['name']); $group = $_POST['group']; $make_model = $_POST['make_model']; $seats = $_POST['seats']; $transmission = $_POST['transmission']; $doors = $_POST['doors']; $lg_bags = $_POST['lg_bags']; $sm_bags = $_POST['sm_bags']; $aircon = $_POST['aircon']; $day1 = $_POST['1_day']; $days2 = $_POST['2_days']; $days3_6 = $_POST['3_6days']; $week = $_POST['week']; $days8_13 = $_POST['8_13days']; $days14 = $_POST['14_days']; if(isset($filename)){ //This is the directory where images will be saved $target = "images/"; $target2 = $target . basename( $_FILES['image']['name']); //This gets all the other information from the form $fileA=($_FILES['image']['name']); } $updateSQL = "INSERT INTO car_groups (group, make_model, transmission, seats, doors, lg_bags, sm_bags, aircon, image) VALUES ('$group', '$make_model', '$transmission', '$seats', '$doors', '$lg_bags', '$sm_bags', '$aircon', '$filename')"; mysql_query($updateSQL) or die(mysql_error()); //Writes the photo to the server move_uploaded_file($_FILES['image']['tmp_name'], $target2); header ("Location: carhireInsert.php"); ?> It keeps telling me... "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 'group, make_model, transmission, seats, doors, lg_bags, sm_bags, aircon, image) ' at line 1" ..? Really stuck on this one... Any help would be apreciated.. thanks.. Mick. |
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