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PHP - Selecting Mysql Database Row Based On Form Input
We recently upgraded from PHP4 to PHP5 and the below script that was working perfectly in 4 has completely stopped working and I can't figure out why for the life of me. I'm not an experienced PHP programmer--I've done some forms, but this is the first time I've used a database.
What needs to (and was) happen: A user enters their username in the form and gets a readout of their participation so far that month. The problem(s): I know that it's storing the variable 'user' because it echoes it back properly, but the database is no longer allowing me to select the row based on that variable. I know it's not that I can't connect to the database because if instead of '$user' I change the code to a username I know is in there, I get the proper readout. This all started as soon as I transferred over to PHP5--before that, no problems at all. The database information is all correct, I just took it out for privacy's sake. <form id="feedback" method="post" action="index.php"> <input name="user" type="text" value="Enter user name" size="20" maxlength="50" /><br /> <input name="send" id="send" type="submit" value="Submit" /> </form> <?php if (isset($_POST['user'])) { $_session['user'] = $_POST['user']; } ?> <p>You entered your username as: <strong><? echo $_session['user'];?></strong>. If this is not correct or you do not see your information below, please re-enter your username and click Submit again.</p> <?php $con = mysql_connect("database","username","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("database", $con); $result = mysql_query("SELECT * FROM March WHERE Username='$user'") or die ('Error: '.mysql_error ()); while($row = mysql_fetch_array($result)) { echo "<table border='0'>"; echo "<tr>"; echo "<td><strong>Username:</strong> </td><td>" . $row['Username'] . "</td></tr>"; echo "<tr><td><strong>Mar. 2 Discussion:</strong></td><td>" . $row['Mar2Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 2 Poll:</strong></td><td>" . $row['Mar2P'] . "</td></tr>"; echo "<tr><td><strong>Mar. 9 Discussion:</strong></td><td>" . $row['Mar9Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 9 Poll:</strong></td><td>" . $row['Mar9P'] . "</td></tr>"; echo "<tr><td><strong>Mar. 16 Discussion:</strong></td><td>" . $row['Mar16Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 16 Poll:</strong></td><td>" . $row['Mar16P'] . "</td></tr>"; echo "<tr><td><strong>March Participation To-Date:</strong></td><td>" . $row['Participation'] . "</td></tr>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> ANY help would be greatly appreciated! I've got a couple hundred people who use this on a regular basis and are starting to ask why it's not working. Similar TutorialsAlright, I've been assigned a project at work. I did not develop the application and the individual who did used CodeIgnited framework and mysql as the db.
Here's the problem, I'm not given much OT to do this and in our meeting the best way to proceed was to replicate the database for different parts of the organization. Basically we are a subsidiary and have been using an application that other groups within the organization want to use. Usually I would reconfigure the db schema and add org ids and in the user table add the appropriate organization to go to. However, they are not giving me enough time to do that.
So what I'm thinking is to just create a copy of the database we use (just the structure) and create a new database.
What I want to know is how to use mysql to check to see if a user exists in one database and if they don't then to go on to the next database. I understand this is a very sloppy way to do it, but it's the way we are moving forward.
I found the code to connect to the db in CodeIgnitor... how can I connect to a database, check to see if the user exists, then close that db connection and try the next database?
/**
* Select the database
*
* @access private called by the base class
* @return resource
*/
function db_select()
{
return @mysql_select_db($this->database, $this->conn_id);
}
Thanks in advance.
Hello. Many thanks for your help. I am writing a PHP/MySQL dating-site and have hit a programming impass. I have a database full of members and a search form consisting of checkboxes. So to search, a member ticks say...gender: female; age: 21,22,23,24,25,26; height: 5'4",5'5",5'6",5'7"; county: cornwall,devon,somerset How can a run a check on the database selecting all entries that fall into the selected criteria. For example a 23 year old female of 5'5" living in Cornwall and a 26 year old female of 5'4" living in Somerset? The key index of my database is 'id' and the fields a age,height,county The names of the form checkboxes a Gender: male, female; Age: 21,22,23,24 etc; Height: 5_4,5_5,5_6 etc; county: cornwall,devon etc So I'm trying to basically trying to make an "advanced search" function in PHP/mysql that will allow users to search by a number of different options like search by zipcode, username, gender, etc. Now my question is how do I vary my mysql query so that it searches for all these things based on whatever the user inputs? For example, the user might need to search zipcode and username but not gender, but in another case, might need to search username and gender, but not zipcode. Obviously I could just do some if/else statements, but that would be increasingly more difficult as there are more fields. What can I do? im wanting to change the sql in my query based on if a field is empty or not. if the field is empty then i dont want it to post anything, but if its not empty then it can post the content. heres what i have so far but it does post the content even if the field is empty. (!empty($password)) ? $pass_sql = "u_password = '$password'," : $pass_sql = null; $link->query("UPDATE ".TBL_PREFIX."users SET $pass_sql u_allow_user_pm = '".$_POST['allow_user_pm']."' WHERE u_username = '".$_POST['user_name']."'") or die(print_link_error()); how can i stop it from updating if the field is empty? Hi there, I found a Javascript with what I want, but I want it to be in PHP because if people don't have Javascript enabled, they won't see the login. Here is what I have, but I need it to be converted to PHP: Code: [Select] function loginArea() { val = document.loginForm.password.value; switch(val) { case "password1": document.location = 'http://www.google.com/password1-page/'; break; case "password2": document.location = 'http://www.google.com/password2-page/'; break; default: document.location ='http://www.google.com/sorry/'; break; } } Code: [Select] <form name="loginForm" id="loginForm" method="post" action=""> <input name="password" type="text" id="password" maxlength="5" /> <input name="login" type="button" id="login" value="Check" onclick="loginArea()" /> </form> Help? - Steph Not sure why this isnt working. Code: [Select] <?php session_start(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <script src="http://js.nicedit.com/nicEdit-latest.js" type="text/javascript"></script> <script type="text/javascript">bkLib.onDomLoaded(nicEditors.allTextAreas);</script> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <LINK REL=StyleSheet HREF="inc/dyn_style.css" TYPE="text/css" MEDIA=screen /> <?php ?> <?php include('logic.inc'); mysqlConnect(); ?> <script type="text/javascript" src="bbeditor/ed.js"></script> <link rel="stylesheet" type="text/css" href="dyn_style.css" /> <title> Social </title> <script type="text/javascript"> function changeTitle(title) { document.title = title; } </script> </head> <body> <?php $inc = 'new_story.php'; $view = 'Newest '; $by = 'added'; $where = " "; $where2 = " "; $order = "ASC"; $gen = "All"; $rat = 'All'; $blerg = ""; $sort = 'newest'; //---------------------------------------------------------------------- if(isset($_GET['sub'])) { $sort = $_GET['sort']; switch($sort) { case "Most Popular"; $by = 'views'; $view = 'Most Popular '; $order = 'DESC'; break; case "Most Reviewed"; $view= 'Most Reviewed '; $by = 'reviews'; $order = 'DESC'; break; case "Newest"; $by = 'added'; $view = 'Newest'; $order = 'ASC'; break; } $genre = mysql_real_escape_string($_GET['cat']); $rating = mysql_real_escape_string($_GET['rat']); if($gen == 'All') { $where = " "; $blerg = ""; } else { $where = "WHERE cat='$gen'"; } if ($rat == "All") { $where2 = ' '; $blerg = 'AND'; } else { $where2 = $blerg ." rating = '$rat' "; } } //---------------------------------------------------------------------- ?> <?php serch(); ?> <form action="story.php" method="get"> <label id='inline'> Order By: </label> <select name='sort'> <option selected='yes' label='Currently Selected' > <?php echo $view; ?> </option> <option> Newest </option> <option> Most Popular</option> <option> Most Reviewed </option> </select> <input type='hidden' value='spec_view' name='p' /> <label id='inline'> Genre/Catagory: </label> <select name='gen'> <option selected='yes' label = 'Selected Genre - <?php echo $gen; ?>'> <?php echo $gen; ?> </option> <option> All </option> <option> Fantasy </option> <option> Adventure </option> <option> Science Fiction</option> <option> Drama</option> <option> Fable </option> <option> Horror</option> <option> Humor</option> <option> Realistic Fiction </option> <option> Tall Tale</option> <option> Mystery </option> <option> Mythology </option> <option> Poetry </option> <option> Shorty Story </option> <option> Romance </option> </select> <label id='inline'> Rating: </label> <select name='rat'> <option selected='yes' label = "Selected Genre - <?php echo $rat; ?>"> <?php echo $gen; ?> </option> <option> All </option> <option> C </option> <option> C13 </option> <option> YA </option> <option> A </option> </select> <input type='submit' value='Go!' name = 'go' /> </form> <?php $query = " SELECT * FROM story_info ORDER BY $by $order $where $where2 "; echo $query; $select = mysql_query($query) or die(mysql_error()); while($rows = mysql_fetch_assoc($select)) { $viewsdb = $rows['views']; $titledb = $rows['title']; $userdb = $rows['user']; $catdb = $rows['cat']; $ratdb = $rows['rating']; $id_db = $rows['story_id']; $sumdb = shorten($rows['sum']); echo "<h3><a href='?p=page&id=$id_db'> $titledb </a> </h3>"; echo "<div id='fun_info'>"; echo "$sumdb <br />"; echo "By <a href='?p=profile&user=$userdb'> $userdb </a> <br /> "; echo "$viewsdb Views | Rated: $ratdb | Catagory: <a href='?p=cat_view&gen=$catdb'> $catdb </a> </div>"; } ?> </div> </body> </html> What I'm trying to do here is generate a table based on a form in which the user selects two options. The first option tells the script which database entries to put in the table, the second option tells it how to arrange them. The first works perfectly, the second not at all - it doesn't produce an error, it just doesn't do anything. I've found a number of tutorials that seem to suggest that the problem is somewhere in my punctuation around ORDER BY '$POST[sort]' but I've been unable to find a solution that actually works. Any help would be much appreciated, my code is below. Thank you! mysql_select_db($database, $con); $result = mysql_query("SELECT * FROM main WHERE state='$_POST[state]' ORDER BY '$POST[sort]'"); if(mysql_num_rows($result)==0){ echo "<p align='left'>View Pantries by State</p><div id='stateform'> <form action='../admin/viewstate.php' method='post'> <select name='state' /> <option value='AL'>Alabama</option> <option value='AK'>Alaska</option> <option value='AZ'>Arizona</option> </select> <select name='sort' /> <option value='name'>Name</option> <option value='id'>Id</option> <option value='city'>City</option> <option value='zip'>Zip</option> <option value='timestamp'>Timestamp</option> </select> <input type='submit' value='Go'></input></form> </div>"; } else{ echo "<p align='left'>View Pantries by State</p> <div id='stateform'> <form action='../admin/viewstate.php' method='post'> <select name='state' /> <option value='AL'>Alabama</option> <option value='AK'>Alaska</option> <option value='AZ'>Arizona</option> </select> <select name='sort' /> <option value='name'>Name</option> <option value='id'>Id</option> <option value='city'>City</option> <option value='zip'>Zip</option> <option value='timestamp'>Timestamp</option> </select> <input type='submit' value='Go'></input></form> </div>"; echo "<div id='fptext'><span class='h1'>Food Pantries for " . $_POST['state'] . "</span><br><br></div>"; } echo "<table border='1' align='center' cellpadding='3' width='900px'> <tr> <th>ID</th> <th>Name</th> <th>Type</th> <th>Address</th> <th>State</th> <th>Phone</th> <th>E-mail</th> <th>Website</th> <th>Hours</th> <th>Requirements</th> <th>Additional Information</th> <th>Lat</th> <th>Lng</th> <th>Update</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['id'] . "</td>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['type'] . "</td>"; echo "<td>" . $row['address'] . "</td>"; echo "<td>" . $row['state'] . "</td>"; echo "<td>" . $row['phone'] . "</td>"; echo "<td>"; echo "<a href=mailto:".$row['email'].">".$row['email']."</a>"; echo "</td>"; echo "<td>"; echo "<a href=".$row['website']."\>".$row['website']."</a>"; echo "</td>"; echo "<td>" . $row['hours'] . "</td>"; echo "<td>" . $row['requirements'] . "</td>"; echo "<td>" . $row['additional'] . "</td>"; echo "<td>" . $row['lat'] . "</td>"; echo "<td>" . $row['lng'] . "</td>"; echo "<td><a href='../public/updatepage.php?id=".$row['id']."'>Update Pantry</a></td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Hello sir, My name is Soji. I'm working on a php project on result transcript processing. I got to a stage where i need to rank student based on their score (Subject Position) and also Overall Position. I have tried all i could but i still don't get it I have a table called Subject position where with field (id, studentregNo, subjectid, levelid, armsid, yearid, total). what i want is if there is 3 student in a class and the first student score 50 in english and the second student score 45 and the third score 40. i want the system to tell me that the first student position in English is 1st and the second student position in English is 2nd and third student position in English is 3rd. So all this will be applicable to all subjects that the students in a particular level are offering.
Moreover, I need Rank student in a subject in that class, And this will work with each subject for each student in that class. Hi people. I have a form which inputs into a database. Here is the code that inserts a yes no option... Code: [Select] <select name = "consent"> <option value = "Yes" <?php if ($_POST['consent'] == 'Yes') { echo 'selected="selected"'; } ?>>Yes</option> <option value = "No" <?php if ($_POST['consent'] == 'No') { echo 'selected="selected"'; } ?>>No</option> </select> However, I have been asked if I can make it a yes or no checkbox instead. Please can you tell me how I need to code it so that the "yes" or "no" is recorded in the DB. At the moment I just have this Code: [Select] <input name="consent" type="checkbox" value="Yes" />Yes<br /> <input name="consent" type="checkbox" value="No" />No<br /> Thanks in advance VinceG Hi im new to php and I need help making webpage that queries a mysql database based on a 3 check boxes and displays results on the same page or on another page. The table being queried has 4 columns, name, gps, wifi, bluetooth. So for example a row in the table would be like, samsung galaxy s2, yes, yes, yes. The idea is for it to be a website that will display phones according to their features. So the idea is depending on if the boxes were ticked the samsung galaxy would be displayed as a result. So i need some help understanding how to make this. Some1 gave me the code below in attempt to help me (im not sure it works or not) but im not sure how fully use it, ie what pages i need to make and how i create the connection to the mysql database, and how to use the query that they wrote to display the results thanks code: Code: [Select] <form action="?do=filter" method="post"> <table cellspacing="0" cellpadding="3" border="1"> <tr> <td>GPS<input type="checkbox" name="gps" value="checked"></td> <td>Wifi<input type="checkbox" name="wifi" value="checked"></td> <td>Bluetooth<input type="checkbox" name="bluetooth" value="checked"></td> </tr> <tr><td><input type='submit' name='filter' value='Filter'></td></tr> </table> </form> </html> <?php function filterMe($filter){ if(isset($_POST[$filter])){ return "Yes"; }else{ return "No"; } } if(isset($_POST['filter'])){ echo "Gps - " . filterMe('gps'); echo " Wifi - " . filterMe('wifi'); echo " Bluetooth - " . filterMe('bluetooth'); } ?> All you need to do is use a query something like SELECT name,gps,wifi,bluetooth FROM `product` WHERE `gps`='".filterMe('gps')."' AND `wifi`='".filterMe('wifi')."' AND `bluetooth`='".filterMe('bluetooth')."' I wish to return the object in an array with the highest index where its index falls between between integers, and return null should one not exist. For instance, with the following and a min and max of 10,000 and 15,000, it should return OBJ4, and with a min and max of 15,000 and 20,000 it should return NULL. Any thoughts? Thanks
function getObj(int $min, int $max):?OBJ {
$list=[
12314=>'OBJ1',
321=>'OBJ2',
42142=>'OBJ3',
14314=>'OBJ4',
123=>'OBJ5',
13314=>'OBJ6'
];
return getIt($list, $min, $max);
}
As a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> Hello, all! I am trying to learn PHP and MySQL on my own, and need some debugging help. What exactly is going wrong here? I am following a tutorial and trying to write the code as it says, but am still having trouble with syntax. Running a WAMP, PHP5.3, and MySQL5.5. This is my code: Code: [Select] <html> <body> <form name = "newVenue" method = "post"> Establishment name: <input type = "text" name = "name"> <br> Street Address: <input type = "text" name = "streetAddress"> <br> City: <input type = "text" name = "city"> <br> State: <select name="state"> <option value="AL">AL</option> <option value="AK">AK</option> <option value="AZ">AZ</option> <option value="AR">AR</option> <option value="CA">CA</option> <option value="CO">CO</option> <option value="CT">CT</option> <option value="DE">DE</option> <option value="DC">DC</option> <option value="FL">FL</option> <option value="GA">GA</option> <option value="HI">HI</option> <option value="ID">ID</option> <option value="IL">IL</option> <option value="IN">IN</option> <option value="IA">IA</option> <option value="KS">KS</option> <option value="KY">KY</option> <option value="LA">LA</option> <option value="ME">ME</option> <option value="MD">MD</option> <option value="MA">MA</option> <option value="MI">MI</option> <option value="MN">MN</option> <option value="MS">MS</option> <option value="MO">MO</option> <option value="MT">MT</option> <option value="NE">NE</option> <option value="NV">NV</option> <option value="NH">NH</option> <option value="NJ">NJ</option> <option value="NM">NM</option> <option value="NY">NY</option> <option value="NC">NC</option> <option value="ND">ND</option> <option value="OH">OH</option> <option value="OK">OK</option> <option value="OR">OR</option> <option value="PA">PA</option> <option value="RI">RI</option> <option value="SC">SC</option> <option value="SD">SD</option> <option value="TN">TN</option> <option value="TX">TX</option> <option value="UT">UT</option> <option value="VT">VT</option> <option value="VA">VA</option> <option value="WA">WA</option> <option value="WV">WV</option> <option value="WI">WI</option> <option value="WY">WY</option> </select> <br> Zip: <input type = "text" name = "zip"> <br> email: <input type = "text" name = "email"> <br> password: <input type = "text" name = "password"> <br> <input type="submit" name="Submit" value="Submit"> </form> <?php //If the form isn't empty, assign the value to a variable if (!empty($_POST['name'])) { $name = $_POST['name']; $address = $_POST['streetAddress']; $city = $_POST['city']; $state = $_POST['state']; $zip = $_POST['zip']; $email = $_POST['email']; $password = $_POST['password']; //Connect to the 'Users' database and store the new bar into the 'Venue' table... mysql_connect ("localhost", "newbar", "Jpr5HJ2K5fWvPLXq") or die ('Oh, fuck: '.mysql_error()); mysql_select_db ("users"); $query = "INSTERT INTO venues VALUES ('NULL', 'testPic.jpg', '".$name."', '".$address."', '".$city."', '".$state."', '".$zip."', '".$email."', '".$password."', 0)"; mysql_query($query) or die ('Oh, fuck: '.mysql_error()); echo "Damn, Nathan. This shit actually worked..."; } ?> </body> </html>This is the error I receive: Code: [Select] Oh, fuck: 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 'INSTERT INTO venues VALUES ('NULL', 'testPic.jpg', 'Nathan's house', '666', 'DAY' at line 1This is my table: Code: [Select] id INT PRIMARY KEY pic_location VARCHAR name VARCHAR address VARCHAR city VARCHAR state VARCHAR zip VARCHAR email VARCHAR password VARCHAR event_name INT Any thoughts as to what is causing this error? Thanks in advance... Hi, I want to be able to generate visitor statistics for a blog I'm creating. I'm going to be collecting numerous pieces of data when a post is viewed, including a time stamp of the visit. I need to be able to select timestamps that were within the current day, the previous day, the day before that ect.. So that I can generate the statistics. Show it for the current week (current day and 6 previous days). So it would be the entries where the timestamp was made on the days: 11/1, 10/1, 9/1, 8/1, 7/1, 6/1, 5/1. For example. Not quite sure how I could do this. Thanks. Hey guys im making a blog and when i make a post it automatically takes the date and created an archive based on what ever i have in the database (date wise) anyways it works almost fine but if i make a date in the same month it duplicates like so: October 2010 October 2010 November 2010 Is there away where i can limit October by 1 but when i come to make a post next year it will still display like so: October 2010 November 2010 December 2010 January 2011 ... October 2011 If i show you the code might make more sense haha:: public function create_archive() { // Loop through the database grabbing all the dates:: $sql = "SELECT blog_date FROM subarc_blog"; $stmt = $this->conn->prepare($sql); $stmt->execute(); $stmt->bind_result($date); $rows = array(); while($row = $stmt->fetch()) { $date = explode("-",$date); $month = date("F",mktime(0,0,0,$date['1'])); $year = $date[0]; $item = array( 'blog_month' => $month, 'blog_year' => $year ); $rows[] = $item; } $stmt->close(); return $rows; } and the sidebar looks like so:: <ul class="sidebar"> <?php $result = $Database->create_archive(); foreach($result as $row) : ?> <li><a href=""><?php echo $row['blog_month'] . " " . $row['blog_year']; ?></a></li> <?php endforeach; ?> </ul> Hope someone can help! Hey freaks! I have a problem, i can't figure out how to recieve one single string from the database, i tried alot of things. Code: [Select] public function getpass($name){ $q = sprintf("SELECT password FROM database WHERE name='%s'", mysql_real_escape_string($name)); $result = mysql_query($q) or die(mysql_error()); // And here i tried every single way to fetch the data. Wich one should i use when its only one slot in a row i need? } I hope you can help me! I can't figure out how to select something from the database that is under today's date. This is what I have: if($row['date'] == ".date('Y-m-d').' 00:00:00'."' AND date < '".date('Y-m-d').' 23:59:59'.") Any help would be greatly appreciated. Hello everyone,
I have a card database and everything works perfectly besides one thing..
I can't store
'in my database via my form, i however do can store them into the database via PHPMyAdmin. I dont know what i've been doing wrong and it really bothers me. If any of you guys could help me out. Here's all the code you would need to find the issue. This is the form file
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Edit Page</title>
</head>
<body>
<h1 align="center"> Add Cards</h1>
<form action="insert.php" method="POST">
<input type="text" name="name" placeholder="Name" />
<input type="text" name="color" placeholder="Color" />
<input type="text" name="type" placeholder="Type" />
<input type="text" name="subtype" placeholder="Sub Type" />
<input type="text" name="power" placeholder="Power" />
<input type="text" name="toughness" placeholder="Toughness" />
<br>
<input type="text" name="manacost" placeholder="Converted Mana Cost" />
<input type="text" name="rarity" placeholder="Rarity" />
<input type="text" name="expansion" placeholder="Expansion" />
<input type="text" name="foil" placeholder="Foil" />
<input type="text" name="stock" placeholder="Stock" />
<input type="submit" value="Save" />
</form>
</body>
</html>
This inserts it into my database.
<?php
($GLOBALS["___mysqli_ston"] = mysqli_connect("", "", "", , ))or die("cannot connect");
((bool)mysqli_query($GLOBALS["___mysqli_ston"], "USE e_industries"))or die("cannot select DB");
$name = $_POST['name'];
$color = $_POST['color'];
$type = $_POST['type'];
$subtype = $_POST['subtype'];
$power = $_POST['power'];
$toughness = $_POST['toughness'];
$manacost = $_POST['manacost'];
$rarity = $_POST['rarity'];
$expansion = $_POST['expansion'];
$foil = $_POST['foil'];
$stock = $_POST['stock'];
$sql="INSERT INTO Osiris (Name, Color, Type, Subtype, Power, Toughness, Manacost, Rarity, Expansion, Foil, Stock)
VALUES ('$name', '$color', '$type', '$subtype', '$power', '$toughness', '$manacost', '$rarity', '$expansion', '$foil', '$stock')";
$result=mysqli_query($GLOBALS["___mysqli_ston"], $sql);
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='add.html'>Back to main page</a>";
}
else {
echo "ERROR";
}
?>
If anyone could help me out that would be great!
Edited by OsirisElKeleni, 05 October 2014 - 12:29 PM. |
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