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PHP - Help With Insert Query
Hi,
can anyone help me I'm new to coding and have been having trouble getting my code to insert into the database. I have checked the connection and that works fine but it says you're registered but it dosen't input it into the database. This is the code i have in my script: <?php echo "<h1>Register</h1>"; $submit = $_POST ['submit']; //form data $fullname = strip_tags($_POST ['fullname']); $username = strip_tags($_POST ['username']); $password = strip_tags($_POST ['password']); $repeatpassword = strip_tags($_POST ['repeatpassword']); $date = date ("Y-m-d"); if ($submit) { //check for existance if ($fullname&&$username&&$password&&$repeatpassword) { //check passwords match if ($password==$repeatpassword) { //check char length of username and fullname if (strlen($username)>25||strlen($fullname)>25) { echo "Maximum limit for username and fullname is 25 characters!"; } else { //check password length if (strlen($password)>25||strlen($password)<6) { echo "Your password must be between 6 and 25 characters!"; } else { //register the user //encrypt password $password = md5 ($password); $repeatpassword = md5 ($repeatpassword); //open database $connect = mysql_connect ("localhost","","") or die ("couldn't connect!"); mysql_select_db ("phplogin") or die ("couldn't find db!"); mysql_query ("INSERT INTO users VALUES ('','$fullname','$username','$password','$date')"); die ("You've been registered! <a href='index.php'>Click here</a> to return to login page"); } } } else echo "Your passwords do not match!"; } else echo "Please fill in <b>all</b> fields."; } ?> <html> <form action='register.php' method='POST'> <table> <tr> <td>Your Full name:</td> <td><input type='text' name='fullname' value='<?php echo $fullname ?>'></td> </tr> <tr> <td>Your Username:</td> <td><input type='text' name='username' value='<?php echo $username ?>'></td> </tr> <tr> <td>Your Password:</td> <td><input type='password' name='password'></td> </tr> <tr> <td>Repeat your password:</td> <td><input type='password' name='repeatpassword'></td> </tr> </table> <input type='submit' name='submit' value='Register'> </form> </html> Similar TutorialsIf you also have any feedback on my code, please do tell me. I wish to improve my coding base. Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference. Code: Code: [Select] mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); Full code: Code: [Select] <?php include_once("includes/config.php"); ?> <!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"> <head> <title><? $title; ?></title> <meta http-equiv="Content-Language" content="English" /> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <link rel="stylesheet" type="text/css" href="style.css" media="screen" /> </head> <body> <div id="wrap"> <div id="header"> <h1><? $title; ?></h1> <h2><? $description; ?></h2> </div> <? include_once("includes/navigation.php"); ?> <div id="content"> <div id="right"> <h2>Create</h2> <div id="artlicles"> <?php if(!$_SESSION['user']) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $name = mysql_real_escape_string($_POST['name']); $server_type = mysql_real_escape_string($_POST['type']); $description = mysql_real_escape_string($_POST['description']); if(!$username || !$password || !$server_type || !$description || !$name) { echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>"; echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'> <option value='Any'>Any</option> <option value='PvP'>PvP</option> <option value='Creative'>Creative</option> <option value='Survival'>Survival</option> <option value='Roleplay'>RolePlay</option> </select></td></tr> <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>"; echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>"; } elseif(strlen($password) < 8) { echo "Password needs to be higher than 8 characters!"; } elseif(strlen($username) > 13) { echo "Username can't be greater than 13 characters!"; } else { $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1"); if(mysql_num_rows($check1) < 0) { echo "Sorry, there is already an account with this username and/or server name!"; } else { $ip = $_SERVER['REMOTE_ADDR']; mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); echo "Server has been succesfully created!"; } } } else { echo "You are currently logged in!"; } ?> </div> </div> <div style="clear: both;"> </div> </div> <div id="footer"> <a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop </div> </div> </body> </html> I need to insert a query but with the ID # from another table. Here is the query I am using to select which works fine. // get reviews $strQuery = sprintf("SELECT * FROM messages WHERE intItemID = %d", intval($intProductID)); $queryGetReviews = db_query($strQuery); However my insert doesn't seem to be picking up the %d. //query to insert data into table $sql = " INSERT INTO messages SET firstname = '$name', message = '$message' WHERE intItemID=%d" ; $result = mysql_query($sql); if(!$result) { echo "Failed to insert record"; } else { echo "Record inserted successfully"; } } Please help! Basically I need to input data into two tables. I am running 3 different query's but only 2 of them work. The other one doesn't. None working query: mysql_query("INSERT INTO users(username, password, email, pin, key) VALUES('$username', '$password', '$email', '$key', '$pin')"); Working querys: mysql_query("DELETE FROM beta_keys WHERE keys_new='$key'"); **and** mysql_query("INSERT INTO beta_keys(keys_used) VALUES('$key')"); So any ideas why the top one doesn't work but the bottom two do?
<?php
<!DOCTYPE html>
I am having an error with this code You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ''name','email','password','profile') SET ('Sasural','kill@1234.com','kill','ANDK' at line 1 I am stuck with this for last 5hrs my SQL Query wont execute on on following lines: Code: [Select] $result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')") or die (mysql_error()); i get the following error: 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 ''gallery' ('image', 'memberid', 'caption') VALUES ('gallery/9074849_1.jpg', '1',' at line 1 here is my full code: Code: [Select] <?php require_once('connect.php'); $rand = mt_rand(1,9999999); $rand2 = mt_rand(1,9999999); $member_id = $_SESSION['SESS_MEMBER_ID']; $caption = $_POST["caption"]; if(isset($_FILES['uploaded']['name'])) { $allowed_filetypes = array('.jpg','.gif','.bmp','.png','.jpeg'); $max_filesize = 524288; // Maximum filesize in BYTES (currently 0.5MB) $fileName = basename($_FILES['uploaded']['name']); $errors = array(); $target = "gallery/"; $fileBaseName = substr($fileName, 0, strripos($fileName, '.')); // Get the extension from the filename. $ext = substr($fileName, strpos($fileName,'.'), strlen($fileName)-1); //$newFileName = md5($fileBaseName) . $ext; $newFileName = $target . $rand . "_" . $member_id.$ext; // Check if filename already exists if(file_exists("gallery/" . $newFileName)) { $errors[] = "The file you attempted to upload already exists, please try again."; } // Check if the filetype is allowed. if(!in_array($ext,$allowed_filetypes)) { $errors[] = "The file you attempted to upload is not allowed."; } // Now check the filesize. if(!filesize($_FILES['uploaded']['tmp_name']) > $max_filesize) { $errors[] = "The file you attempted to upload is too large."; } // Check if we can upload to the specified path. if(!is_writable($target)) { $errors[] = "You cannot upload to the specified directory, please CHMOD it to 777."; } //Here we check that no validation errors have occured. if(count($errors)==0) { //Try to upload it. if(!move_uploaded_file($_FILES['uploaded']['tmp_name'], $newFileName)) { $errors[] = "Sorry, there was a problem uploading your file."; } } //Lets INSERT database information here //Here we check that no validation errors have occured. if(count($errors)==0) { $result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')") or die (mysql_error()); { $errors[] = "SQL Error."; } } //If no errors show confirmation message if(count($errors)==0) { echo "<div class='notification success png_bg'> <a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a> <div> The file {$newFileName} has been uploaded<br>\n </div> </div>"; //echo "The file {$fileName} has been uploaded"; echo "<br>\n"; echo "<a href='gallery.php'>Go Back</a>\n"; } else { //show error message echo "<div class='notification attention png_bg'> <a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a> <div> Sorry your file was not uploaded due to the following errors:<br>\n </div> </div>"; //echo "Sorry your file was not uploaded due to the following errors:<br>\n"; echo "<ul>\n"; foreach($errors as $error) { echo "<li>{$error}</li>\n"; } echo "</ul>\n"; echo "<br>\n"; echo "<a href='gallery.php'>Go Back</a>\n"; } } else { //Show the form echo "Use the following form below to add a new image to your gallery;<br />\n"; echo "<form enctype='multipart/form-data' action='' method='POST'>\n"; echo "Please choose a file: <input name='uploaded' type='file' /><br />\n"; echo "Caption: <input name='caption' type='text' /><br />\n"; echo "<input type='submit' value='Upload' />\n"; echo "</form>\n"; //Echo Tests! echo "<br /><br />Random FileName: "; echo $rand; echo "<br />"; echo "member ID: #"; echo $member_id; } ?> any help appreciated. its prob something simple. my table has the following fields: "gallery" id (primary Key, AUTO_INC) memberid (fetched from session) image (will store image name including extension) caption (from "caption" text field in form) Hi, I am creating a new menu (food) in my system. This consists of a menu, menu_items and menu_connection table. I can insert the menu name just fine and return its id just fine. When inserting the menu items, i need to get each of the menu_item_ids to use in the query that inputs the menu_connection. This is what i have so far:
if ($_SERVER['REQUEST_METHOD']=="POST") {
/////////////////////
//menu name insert //
/////////////////////
$mname = mysqli_real_escape_string($conn, $_POST['newMenuName']);
$stmt=$conn->prepare('
INSERT IGNORE INTO ssm_menu (menu_name) VALUES (?);
');
$stmt->bind_param('s',$mname);
$stmt->execute();
$menuInsId = $stmt->insert_id;
echo $menuInsId;
$stmt->close();
/////////////////////
//menu item insert //
/////////////////////
$mitname = $_POST['newMenuItem'];
$stmt=$conn->prepare('
INSERT IGNORE INTO ssm_menu_items (menu_item_name) VALUES (?);
');
foreach ($_POST['newMenuItem'] as $k => $nmItem) {
$mitname = mysqli_real_escape_string($conn, $nmItem);
$stmt->bind_param('s',$mitname);
$stmt->execute();
$menuItmInsId = $stmt->insert_id;
echo $menuItmInsId;
}
$stmt->close();
///////////////////////////
//menu connection insert //
///////////////////////////
$stmt=$conn->prepare('
INSERT IGNORE INTO ssm_menu_connection (menu_id, menu_item_id) VALUES (?,?)
');
foreach ($_POST['newMenuItem'] as $k => $nmItem) {
$stmt->bind_param('ii',$menuInsId, $menuItmInsId);
$stmt->execute();
$connectionInserId = $stmt->insert_id;
echo $connectionInserId;
}
$stmt->close();
}
Currently it is inserting each of the items in the connection table with the same id - i understand why but i dont know how to collect up all of the ids to use later Hi I am having troubles inserting item into a table with a primary key in it.
My problem is that every time I added a row into a table without specifying the primary key, mysql will not add it to the next available id.
For example, I have 5 rows initially and I added 10 rows into my table by using the insert into query. However when I delete the 10 rows and add another row in, the id of the additional row will become 16 instead of 6.
FYI, this is a php page that inserts my data into the table.
Hello there, I am trying to build my site more efficient by sending me error messages that occur. I have decided that most of my errors are mysql errors. Thinking about what to do, I tried to put an insert query in the die message. Here is my line of code below. Code: [Select] $globalsql=mysql_query("SELECT * FROM global") or die(' $error=mysql_query("INSERT INTO errors (name, identity, user,) VALUES ('"'Error 1'"', '"'global_functions.php'"', '"'System'"')") '); However, I get this error: Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in global_functions.php on line 4 Which is that line. Any ideas? Thanks! Hello every body,, i want to create a new database (auto generate duty assigned using form) in PHP and mysql..
i have four input fields:-
Name, Subject, Class & weekly lectures
now i want to insert name,subject & class into database, when i insert number of weekly lecture in field four..
insert automaticlly in database multiple time which i write in field four (weekly lecture)
database: table structure i have already is:::
id-name-subject-class-period-monday-tuesday-wednesday-thursday-friday-saturday
anybody please help me to create this database or just insert query ....
I have this PHP file that I intend to provide the answers to only four questions in a little quiz in a "Questions" table: QuestionID, QuestionText Questions: 1 Which is NOT part of the FAT TOM acronym? 2 Of the following choices, which has a better chance of creating a foodborne illness? 3 How should food NEVER be thawed? 4 Where should pesticides be stored? The answers should go in the Answers table: "QuestionID, OptionText, CorrectAnswer" I have hundreds of questions and answers, but I need to get this small sample working first. The following code runs without triggering an error but it doesn't insert any data. <?php $link = mysql_connect('PathToMyData', 'myUsername', 'MyPassword'); // specifics removed for security if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db('quiz'); $sql = "INSERT INTO ANSWERS (QuestionID,OptionText,CorrectAnswer) // CorrectAnswer is type BOOL (Correct=1 Wrong Answer=0) (VALUES (1,'food',0), (1,'alkalinity',1), (1,'time',0), (1,'temperature',0), (1,'oxygen',0), (1,'moisture',0), (2,'celery sticks',0), (2,'beef jerky',0), (2,'cranberry juice',0), (2,'baked potato',1), (2,'saltine cracker',0), (3,'in the refrigerator',0), (3,'in a pot in the kitchen at room temperator',1), (3,'as a part of the cooking process',0), (3,'under cool running water',0), (4,'close to the food preparation area for easy access',0), (4,'in a locked storage area away from food',1), (4,'in the dry storage area',0), (4,'in a bin or box under the sink',0);)"; $result = mysql_query($sql); // executes the query // close database mysql_close($link); ?> Then I go back to the MySQL admin area at my host's site and query up "SELECT * FROM Answers;" and nothing! It's still empty! As I said, no error was triggered, but nothing was inserted. This is discouraging. Hello Friends, Now i m stuck with simple insert query . here's the code Code: [Select] <?php include_once("conf.php"); $firstName=$_POST['fname']; $lastName=$_POST['lname']; $email=$_POST['email']; $dob=$_POST['dob']; $password=$_POST['pass']; $fname= stripslashes($firstName); $lname=stripslashes($lastName); $mail= mysql_real_escape_string($email); $password= mysql_real_escape_string($password); mysql_select_db('site'); $statement="Insert into Accounts(Name,lastName,emailId,DOB,password) VALUES($fname,$lname,$mail,$dob,$password)"; $query=mysql_real_escape_string($statement); mysql_query($query) or die("Cannot save data:</br> ".mysql_error()); echo "Data Saved Successfully"; ?> Now please explain me why i m getting the following error Cannot save data: 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 '@gmail.com,1988/12/20,password)' at line 1 Any help will be highly appreciated! I have tried many ways of storing an array of data that comes from a multiple selection form into a mysql table, including serialization. I am currently trying to do it by forming a string, and using the string in the insert query, but keep getting errors, or just nothing happening. I am not sure what I am doing wrong. Code: [Select] $categoryString = array(); if ($categoryArray){ foreach ($categoryArray as $category){ $categoryString[] = $category.'<br />';} } $categoryquery= "INSERT INTO categoryname VALUES('$categoryString')"; mysql_query($categoryquery); Thank you for your time as always. Code: [Select] <?php include_once("includes/config.php"); ?> <!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"> <head> <title><? $title; ?></title> <meta http-equiv="Content-Language" content="English" /> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <link rel="stylesheet" type="text/css" href="style.css" media="screen" /> </head> <body> <div id="wrap"> <div id="header"> <h1><? $title; ?></h1> <h2><? $description; ?></h2> </div> <? include_once("includes/navigation.php"); ?> <div id="content"> <div id="right"> <h2>Create</h2> <div id="artlicles"> <?php if(!$_SESSION['user']) { $username = $_POST['username']; $password = $_POST['password']; $name = $_POST['name']; $server_type = $_POST['type']; $description = $_POST['description']; if(!$username || !$password || !$server_type || !$description || !$name) { echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings! All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>"; echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'> <option value='Any'>Any</option> <option value='PvP'>PvP</option> <option value='Creative'>Creative</option> <option value='Survival'>Survival</option> <option value='Roleplay'>RolePlay</option> </select></td></tr> <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40'></textarea></td></tr>"; echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>"; } elseif(strlen($password) < 8) { echo "Password needs to be higher than 8 characters!"; } elseif(strlen($username) > 13) { echo "Username can't be greater than 13 characters!"; } else { $check1 = $db->prepare("SELECT username,name FROM servers WHERE username = ? OR name = ? LIMIT 1"); $check1->execute(array($username, $name)); if($check1->rowCount() > 0) { echo "Sorry, there is already an account with this username and/or server name!"; } else { $ip = $_SERVER['REMOTE_ADDR']; $insertQuery = $db->prepare("INSERT INTO servers (id, username, password, name, type, description, ip, votes, beta) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)"); $insertQuery->execute(array(null, $username, $password, $name, $server_type, $description, $ip, 0, 1)); if($insertQuery->errorInfo() > 0) { echo "Sorry, there was the following error:"; print_r($insertQuery->errorInfo()); } else { echo "Server has been succesfully created!"; } } } } else { echo "You are currently logged in!"; } ?> </div> </div> <div style="clear: both;"> </div> </div> <div id="footer"> <a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop </div> </div> </body> </html> Whats wrong with that page? :/ Hi, Here is an extract of my while loop while ($personquery = mysql_fetch_array($personfetch, MYSQL_ASSOC)) { echo "$personfetch[first_name]"; } This will list each person now I want to do an INSERT to the database on every single member that is being listed I can only get it to INSERT for the one. Thanks Hi All Again I am building a job site with a CV match system. Part of the match is education. I have a number of select boxes in a form for the number of qualifications for each individual applicant. What I would like to know is what is the correct way to put this into a MySQL database and I do not mean the insert code! i.e. English Grade 3. Maths Grade 2. I am getting them to put in total number of grades over 1, then 2 etc. When the query is made I would like the recruiter have the option of not just selecting the grades but subject with it. Should I put English then a , then the grade? It needs to be set up so I can run the query. Thanks for looking and as per last request thanks for your time and help Im sure this is simple, but I cannot see what my problem is! I am hitting an error on my insert query Code: [Select] 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 '1' at line 1 I know the output of $user_id is 1, so my error is on $mysavepath Code: [Select] $mysavepath = $folder.'/'.$worldname.'_'.date("dMjY"); echo $mysavepath; $savepath = mysql_query("INSERT INTO saves (user_id,savepath) VALUES ('$user_id','$mysavepath')"); echo '<br>'.$savepath; if(!mysql_query($savepath)) { die('<br>Error: ' . mysql_error()); } however it all echos out ok? Code: [Select] 188ea678f0dcdc8252aeb15e3c910408/world_15Jan152012 1 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 '1' at line 1 Can anyone see the problem? Cheers Dave i want the name of a picture stored in my db after i upload it the data is not stored in the db after i run this script, but i dont get errors either i print the two vars before sending them, and they get printed fine any help on this would be greatly appreciated thanks ! <?php error_reporting(E_ALL); ini_set("display_errors", 1); // INCLUDE THE CLASS FILE include('ImageLib.Class.php'); include("./includes/egl_inc.php"); $displayMessage = ''; if($_POST){ if(isset($_FILES['image_file'])){ // SEE THE MAGIC HAPPEN $destination_path = 'uploads/'; $post_file_name = 'image_file'; $width = 600; $height = 400; $scale = false; $trim = true; $uniqueName = true; $img = ImageLib::getInstance()->upload($post_file_name, $destination_path, $uniqueName)->resize($width, $height, $scale, $trim)->save(); $imgstr = mysql_real_escape_string ($img); $fileName = $_FILES['image_file']['name']; $displayMessage = '<div class="image"><img src="'.$destination_path.$fileName.'" /><br />Uploaded And Resized...With new file name : "'.$img.'"</div><br /><br />'; $playerid=$_SESSION['tid']; $matchdetails = mysql_fetch_array(mysql_query("SELECT id FROM ffa_matches WHERE status=2 and admin=$playerid")); $id = $matchdetails[id]; print $img;print $imgstr; print $id; mysql_query(" INSERT INTO ffa_screens (imgname,match) VALUES( '" . mysql_real_escape_string($imgstr) . "', '" . mysql_real_escape_string($id) . "' )"); }} ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>ImageLib Samples By Rahul Kate</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <style> body{font-family: arial; font-size:12px; color:#444444; padding:20px;} li{margin-top:10px;} .image{color:green;} .image img{margin-bottom:5px;} </style> </head> <body> <h1>ImageLib | Upload Image, move it to Uploads folder and Resize it and Trim</h1> <?php echo $displayMessage; ?> <form method="post" enctype="multipart/form-data"> Select Image<br /> <input type="file" name="image_file" id="image_file" /> <br /> <br /> <input type="submit" name="submit" value="Submit" /> <br /> <br /> <a href="index.html">Back TO Home</a> </form> </body> </html> I've been poking around the Internet for the answer to this without any success. This query APPEARS to run, but NOTHING GETS INSERTED. The QuestionID in the Questions table is set to auto-increment, so I only need the QuestionText data for this table. the QuestionID in the Answers table is not because there might be four or five options per question. The snippets of code under consideration is pasted below snipped from index.php: Code: [Select] <form action="./newquestion.php" method="post"> Question Text:<br><input type="text" size=100 name="QuestionText" /><br><br> <input type="submit" value="Enter next question"/> </form> Then snipped from newquestion.php mysql_select_db(quiz); $query = mysql_query("INSERT INTO Questions VALUES ('$_POST[QuestionText]'"); $result = mysql_query($query); Thank you all. I'm sure I'll learn this pretty fast, but I still need a little help. hi ..
hit little snag Ive got html <form name="formx" method="POST" action="connect.php" class='ajaxform'> and about 4 varibles in that form .. now how to properly
query table users so i can find out which prefix correspond with that user
Insert into table "upis" in this case that prefix with 3 variables..
everything in single,.. in this case connect.php file
I know how to do QUERY and INSERT INTO but i don't know how to connect with 2 different tbl or db and do the query and input of data.. every time I put 2 of them together .. never works out..
this is example.. i dont know is this right way to do it .. |
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