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PHP - Database Object Problem
For convenience and efficiency purposes I'm changing my MySQL database to store PHP Objects as opposed to tons of individuals variables. In converting the user/login system I've come across a really weird problem. I can register a new user to the site, but only the first one in the database will work. I can register others and they get put in the database correctly, but for some reason the app doesn't see them or even recognize that they exist. Only the first entry in the database will work as it's supposed to. For instance, if I drop the first database entry, the second one (now first) will begin to work.
In most cases, I'm pulling all Object entries from the database and looping through them with foreach() to find the specific one I want. I've just come off of an 8 hour coding spree and my brain is just shot at the moment. I really think I just need a new pair of eyes. I'm sure that it's something simple, but I can't see it atm. I'm attaching the User class file which handles login, registration, and anything that I think would be affecting this. Thanks for the help. Similar TutorialsHey, I'm wondering what the issue is, I'm new to using PDO and objects so I'm sure this is a common error but setting me on the right track would be greatly appreciated. Code: [Select] Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 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 'AND password = d41d8cd98f00b204e9800998ecf8427e' at line 1' in /Users/JPFoster/Sites/Jaipai.Blog/engine.php:47 Stack trace: #0 /Users/JPFoster/Sites/Jaipai.Blog/engine.php(47): PDO->query('SELECT * FROM u...') #1 /Users/JPFoster/Sites/Jaipai.Blog/index.php(7): SiteEngine->login('', '') #2 {main} thrown in /Users/JPFoster/Sites/Jaipai.Blog/engine.php on line 47 Code: [Select] public function login($user, $pw){ $user = addslashes($this->user); $pw = md5($pw); $query = "SELECT * FROM users WHERE username = ".$user." AND password = ".$pw; //set up query $results = $this->pdo->query($query); $num_rows = $results->num_rows; if ($num_rows == 1){ $row = $results->fetch_assoc(); echo "Worked!"; //$this->set_session($row['id'], $row['username'], $row['email']); } else { echo "Your Username and Password did not match"; } //if user + passs match { redirect (member page) and set up sessions // Else ... Retry... } } I don't see an issue with my query, and the object is instantiated by having the submit button isset() and i use login($_post[user], $_post[pw]) I've been starting to play around the mysqli class and I've been having trouble using it due to various error it gives me from simple queries like this one. I'm not sure what the error is really, I've been following the php manual. Any help would be greatly appreciated. Code: [Select] <?php $mysqli = new MySQLi('localhost', 'root', 'root', 'jaipai'); if ($mysqli->connect_errno) { echo "There was a connection error: ". $mysqli->connecterrno; } class testClass { private $db; function __construct($mysqli) { $this->db = $mysqli; } public function pageInfo() { $query = "SELECT * FROM users WHERE username = jaipai"; $results = $this->db->query($query); $result = $this->db->fetch_assoc($results); return $result['username']; } } $testClass = new testClass($mysqli); echo $testClass->pageInfo(); ?> This gives me this error: Code: [Select] Fatal error: Call to undefined method mysqli::fetch_assoc() in /Users/JPFoster/Sites/Research & Development/Programs/Object Sandbox/DatabaseConnection.php on line 30 Just to be a little more informative I've also tried this method Code: [Select] $results = $this->db->query($query); $result = $results->fetch_assoc(); return $result['username']; This gives me an error: Code: [Select] Fatal error: Call to a member function fetch_assoc() on a non-object in Sites/Research & Development/Programs/Object Sandbox/DatabaseConnection.php on line 30 I'm not sure which is on the best path to go. Any help would be greatly appreciated. Hey all, I want to have an object that has a property which is an object containing instances of other objects. I try this: Code: [Select] class Blog extends Posts { public $has_posts; public function __construct($a,$b,$c){ $has_posts = (object) array_merge((array) $a, (array) $b, (array) $c); } } class Posts { public $b; public function __construct($b){ $this->b = $b; } } $post1 = new Posts(1); $post2 = new Posts(2); $post3 = new Posts(3); $blog = new Blog($post1,$post2,$post3); var_dump($blog->has_posts); //null foreach($blog->has_posts as $post){ //Invalid argument supplied for foreach() echo $post->b; } But as you see, has_posts is null, not an object containing other objects. Thanks for response. Hi all Good morning / night / afternoon I have a duplication problem when I need to print my records in PHP First of all let's see the database table As you can see the duplication happened duo to the approvers that mean if there is only one approver there will be no duplication For example when I want to pint request number 1 it will print 1 - Cris - Hadi 1 - Cris - Jack but I want it to be like this 1 - Cris - Hadi,Jack and thank you for your time Hey, i have a script in my game that picks up the output rate of a unit which if there is more than one unit it devides, i have 3 teams and the script picks up team 1 and 2, but any units in team 3 the rate shows 0? IE HEAVY SOLDIER 100% but with the 3rd team its always showing 0% Code: [Select] <font face="Verdana" color="#000000" size="2"><b>Output Rate</b></font></td></tr>'; $sql = "SELECT id,name FROM units WHERE class={$row[0]} AND visible=1"; $r2 = mysql_query($sql); while ($row2 = mysql_fetch_row($r2)) { echo '<tr><td align="center">'.$row2[1].'</td><td align="center"><input text="text" size="5" style="'.$boxStyle.'" name="a'.$row2[0].'" value='.getRate($userid,$row2[0]).'>%</td></tr>'; } echo '<tr><td colspan="2" align="center"><input type="submit" value="Set Rates for '.$row[1].'"></td></tr>'; echo '</table></form>'; } echo '</center>'; DB's Code: [Select] FOR UNITS. id name toground toair hp class evolvesto style cost visible 6 Mulanix 10 10 200 3 0 1 20 1 5 Numani 5 5 100 3 6 1 10 1 any help would be great it's annoying me and i can't solve it. Hi, if someone could assist me, it would be much appreciated. I have a database website. I generated it using AppGini. Unfortunatly, they haven't answered my emails. What I'm trying to do is change my database to not dynamically load the selected record, but to instead load the direct URL to the record. (Here is the website address: http://2g2egkxd3.site.aplus.net/updates/bad_view.php) If you hover above the record, you will see the full URL for that record. Example: http://2g2egkxd3.site.aplus.net/updates/bad_view.php?SelectedID=Buck%20Allen But when you click it, it doesn't load that URL into the browser. I'm guessing it dynamically loads it into the page. The reason I'm trying to do this is because I want to have a social/comment system for each record/prospect and it won't work if the record is dynamically loaded. I have a Soup object in a Bowl object in a Microwave object. The Microwave object has the methods: receiveItem(Bowl $b) cookItem(???) I want the Microwave to be able to set the Soup->temperature but I'm not how to do that? Make sense? TomTees Hi, For school, I need to build a site and this site has to have a few applications. One of the applications, I have to make is to make a script that makes it possible to upload a file and it has to put this file in a online database. Now, I'm just trying to put it in a local database. My code doesn't give an error, but it doesn't work either. The file doesn't get into the database. I can't figure out why. I hope you can help me? Thanks in advance! Lize hi to everyone.. i need help with my php scripting..i just can't connect to my mysql database..the message was this.."Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\HR\checklogin.php on line 10 cannot connect" i got the code from http://www.phpeasystep.com/workshopview.php?id=6 <?php session_start(); $host="localhost"; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name="test"; // Database name $tbl_name="members"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // Define $myusername and $mypassword $myusername=$_POST['myusername']; $mypassword=$_POST['mypassword']; // To protect MySQL injection (more detail about MySQL injection) $myusername = stripslashes($myusername); $mypassword = stripslashes($mypassword); $myusername = mysql_real_escape_string($myusername); $mypassword = mysql_real_escape_string($mypassword); $sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'"; $result=mysql_query($sql); // Mysql_num_row is counting table row $count=mysql_num_rows($result); // If result matched $myusername and $mypassword, table row must be 1 row if($count==1){ $_SESSION['logged'] = true; header("location:login_success.php"); exit;} else { echo "Wrong Username or Password"; } ob_end_flush(); ?> Dear friends i,m a php beginner and i got a problem with connecting to my database i created a database called (koora) with one table called (admins) and when i tried to connect to it (database ) ; it did not connect here is the code i used for that <?php $connectdb = mysql_connect('localhost','','') or die("not connected"); $selectdb = mysql_select_db("koora", $connectdb); if(!$selectdb) { die("error connecting table" .mysql_error()); } then when refreshing my phpmyadmin page i got that message error connecting tableAccess denied for user ''@'localhost' to database 'koora' koora is the name of the database so i need your help with this problem and what is the reason not to connect to the data base Thank you Google Chrome Developer Tool causes an error when I run the following and there are not any records. I keep getting errors when I try to change it.
<?php
$host = 'localhost';
$user = 'root';
$pass = '';
$database = 'ecommerce';
$options = array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
PDO::ATTR_EMULATE_PREPARES => false
);
$keyword = $_GET['keyword'];
$dbo = new PDO("mysql:host=$host;dbname=$database", $user, $pass, $options);
$q1 = "SELECT * FROM products INNER JOIN keywords on keywords.keywordID = products.KeywordID and keywords.KeyWord1 = \"$keyword\" ";
$counter = 10;
$counter1 = 0;
foreach ($dbo->query($q1) as $row) {
I'm trying to simply display the contents of an entire database for the purpose of building an administrative super-user into the back-end to remove login credentials without having to resort to having to manually interact with Mariadb or PHPmyadmin. However, the code is only returning the last row that was entered into the database. I can't seem to figure out why it's not working as intended. The code is intended to simply display the entire contents of the database into an html table. The code i'm using:
<?php
$link = mysqli_connect("127.0.0.1", "database user", "password", "login");
if (!$link) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
if ($result = mysqli_query($link, "SELECT * FROM accounts", MYSQLI_STORE_RESULT)) {
while ($row = $result->fetch_assoc()) {
$id = $row["id"];
$username = $row["username"];
$pass = $row["password"];
$email = $row["email"];
$su = $row["su"];
}
}
mysqli_close($link);
?>
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<table>
<tr> <th>id</th><th><th>user name</th><th>password</th><th>email</th><th>elevated privlege</th></tr>
<tr>
<td> <?php echo $id; ?> </td>
<td> <?php echo $username; ?> </td>
<td> <?php echo $pass; ?> </td>
<td> <?php echo $email; ?> </td>
<td> <?php echo $su; ?> </td>
</tr>
</table>
</body>
</html>
Edited August 31, 2020 by bakertaylor28 correction of the code used. Code: [Select] //enter into btree ---------------------------------------------- $sql_plevel='select nlevel from btree where uniqueid = '.$formjoinedunder; $abc = mysql_query($sql_plevel , $db) $result_plevel = mysql_fetch_assoc($abc) or die(mysql_error()); $plevel = $result_plevel['nlevel'] + 1; $insert_btree = 'INSERT INTO btree (uniqueid,nlevel,pside) VALUES ('.$reguniqueid.','.$plevel.','.$formposition.')'; mysql_query($insert_btree , $db) or die(mysql_error()); Can any one please tell me what is the problem in the above code it does not insert into the table btree. hi, i was having difficulty and i'm quite confused how to insert these values to the database before it will become as it is, based on this article: http://roshanbh.com.np/2008/01/populate-triple-drop-down-list-change-options-value-from-database-using-ajax-and-php.html what i did was putting/saving the values directly in the database (mysql). but now i want those values to be coming from the user's input:the country,state and city so the system will just get those values, store it to the database with the fields of each table having the same content like that from article above..i really need some help here.. Hello everybody, I have a script that allows a user to set themselves a target and a script that allows them to edit the target if they decide to. Now everything works as it should until the data from the form (on the edit target page) should be updated in the database. I require that the user must update at least one field. For example they may want to extend the completion date so I would only want to update the completion date field in the database and the fields the user left blank should not be updated in the database. The problem I am having is regardless of if the user enter a value into the field it updates the database so if the user leaves a field blank it updates the database will a blank value. Here is my code: //If the user clicks the update button on a WEIGHT LOSS target run the following code to update the database if(isset($_POST['target_weight_loss_update'])){ $target_weight_loss_update = $_POST['target_weight_loss_update']; $weight_loss_complete_update = $_POST['weight_loss_complete_update']; $weight_loss_comment_update = $_POST['weight_loss_comment_update']; $weight_loss_user_id_update = $_POST['weight_loss_user_id_update']; //Check that at least one value has been entered into the form if(empty($target_weight_loss_update) && empty($weight_loss_complete_update) && empty($weight_loss_comment_update)){ $error_msg = 'Please enter at least one value to update your '.$target_type.' target'; } //Check to see what variables have values in them so we know what fields to update in datebase //If the user has entered a value for their TARGET WEIGHT LOSS update that value in the database if(isset($target_weight_loss_update) || !empty($target_weight_loss_update)){ $target_weight_loss_update_sql = mysql_query("UPDATE user_targets SET target_weight='$target_weight_loss_update' WHERE id='$target_id' LIMIT 1") or die (mysql_error()); //Check that the update was successful if($target_weight_loss_update_sql){ $error_msg = 'Target '.$target_type.' updated successfully'; } else { $error_msg = 'Could not update target. Please try again.'; } } //If the user has entered a value for their TARTGET WEIGHT LOSS DATE update that value in the database if(isset($weight_loss_complete_update) || !empty($weight_loss_complete_update)){ $weight_loss_completion_date_sql = mysql_query("UPDATE user_targets SET target_date='$weight_loss_complete_update' WHERE id='$target_id' LIMIT 1") or die (mysql_error()); //Check that the update was successful if($weight_loss_completion_date_sql){ $error_msg = 'Target '.$target_type.' updated successfully'; } else { $error_msg = 'Could not update target. Please try again.'; } } //If the user has entered a value for their TARGET WEIGHT LOSS COMMENT update that value in the database if(isset($weight_loss_comment_update) || !empty($weight_loss_comment_update)){ $weight_loss_comment_update_sql = mysql_query("UPDATE user_targets SET extra_info='$weight_loss_comment_update' WHERE id='$target_id' LIMIT 1") or die (mysql_error()); //Check that the update was successful if($weight_loss_comment_update_sql){ $error_msg = 'Target '.$target_type.' updated successfully'; } else { $error_msg = 'Could not update target. Please try again.'; } } } So the question I ask is how can I stop entering blank values into the database and only update the information the user enters into the form? Also I was thinking of having one mysql query at the end and using the IF statements to build the query. Thoughts? Thanks for any advice. I am trying to insert with a form the information below into the database, but I don't know how to link it to a specific user in the DB. Below is the SQL dump and my code for the insert page. Any help greatly appreciated. also, I get the error ,"duplicate entry for key 2" Code: [Select] - phpMyAdmin SQL Dump -- version 2.8.0.1 -- http://www.phpmyadmin.net -- -- Host: custsql-sl07.eigbox.net -- Generation Time: Aug 28, 2011 at 07:42 AM -- Server version: 5.0.91 -- PHP Version: 4.4.9 -- -- Database: `photo_artists` -- -- -------------------------------------------------------- -- -- Table structure for table `images` -- CREATE TABLE `images` ( `image_id` bigint(20) unsigned NOT NULL auto_increment, `gallery_user` int(11) NOT NULL, `filename` varchar(255) NOT NULL, `mime_type` varchar(255) NOT NULL, `file_size` int(11) NOT NULL, `file_data` longblob NOT NULL, PRIMARY KEY (`image_id`), UNIQUE KEY `image_id` (`image_id`), KEY `filename` (`filename`), KEY `gallery_user` (`gallery_user`) ) ENGINE=MyISAM AUTO_INCREMENT=38 DEFAULT CHARSET=latin1 AUTO_INCREMENT=38 ; -- Table structure for table `users` -- CREATE TABLE `users` ( `id` int(11) NOT NULL auto_increment, `status` varchar(20) NOT NULL, `lastname` varchar(50) NOT NULL, `dob` date NOT NULL, `gender` varchar(10) NOT NULL, `username` varchar(20) NOT NULL, `password` varchar(60) NOT NULL, `email` varchar(20) NOT NULL, `activationkey` varchar(100) NOT NULL, `sex` varchar(10) NOT NULL, `race` varchar(40) NOT NULL, `job` varchar(40) NOT NULL, `country` varchar(30) NOT NULL, `state` varchar(30) NOT NULL, `town` varchar(30) NOT NULL, `hobbies` mediumtext NOT NULL, `business` varchar(50) NOT NULL, `religion` varchar(40) NOT NULL, `social_groups` mediumtext NOT NULL, `political_groups` varchar(200) NOT NULL, `other_affiliations` varchar(200) NOT NULL, `buying` longtext NOT NULL, `selling` longtext NOT NULL, `links` varchar(100) NOT NULL, `likes` mediumtext NOT NULL, `dislikes` longtext NOT NULL, `firstName` varchar(30) NOT NULL, `zip` varchar(15) NOT NULL, PRIMARY KEY (`id`), UNIQUE KEY `username` (`username`), UNIQUE KEY `email` (`email`), UNIQUE KEY `activationkey` (`activationkey`) ) ENGINE=MyISAM AUTO_INCREMENT=54 DEFAULT CHARSET=latin1 AUTO_INCREMENT=54 ; -- -- Dumping data for table `users` -- INSERT INTO `users` VALUES (49, 'verify', '', '0000-00-00', '', 'user2', '32b3491336522e073489725b5daf298cd749007a', '1@aol.com', '34705256317118179901016541118525481318594628324', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', ''); INSERT INTO `users` VALUES (53, '', '', '0000-00-00', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', ''); INSERT INTO `users` VALUES (52, 'activated', '', '0000-00-00', '', 'user', 'e8ee80e34d93d48813b08949c07fc826a1c82b63', '1@hotmai', '153967352213317143421219669520489072241284552824', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', ''); and here is the code. note I don't know how to link the insert to the current user. Code: [Select] <?php require_once('globals.php'); include('bouncer.php'); $first = $_POST['first']; $last = $_POST['last']; $first = $_POST['dob']; $last = $_POST['gender']; $first = $_POST['ethnic']; $last = $_POST['country']; $first = $_POST['state']; $last = $_POST['town']; $first = $_POST['zip']; $last = $_POST['mail']; $first = $_POST['job']; $last = $_POST['business']; $first = $_POST['religion']; $last = $_POST['social']; $first = $_POST['political']; $last = $_POST['affiliations']; $first = $_POST['political']; $last = $_POST['affiliations']; $first = $_POST['buying']; $last = $_POST['selling']; $first = $_POST['likes']; $last = $_POST['dislikes']; $first = $_POST['links']; $currentUser = $_SESSION['myusername']; mysql_query ( "INSERT INTO users (lastname, firstName, dob, gender, email, race, job, country, state, town, zip, hobbies, business, religion, social_groups, political_groups, other_affiliations, buying, selling, links, likes, dislikes) VALUES ('$last','$first','$dob','$gender','$email','$ethnic','$job','$country','$state','$town','$zip','$hobbies','$business', '$religion','$social','$political','$affiliations','$buying','$selling','$links','$likes','$dislikes')"); echo mysql_error(); mysql_close(); ?> <html> <body> <p>Login Successful</p> <p> </p> <form action="login_success.php" method="post"> <p> <input type="text" name="first" size="20" /> First name<br /> <input type="text" name="last" size="20" /> Last name<br /> <input name="dob" type="text" size="20" id="dob" /> Date of Birth<br /> <input type="text" name="gender" size="20" id="gender" /> Gender <br /> <input type="text" name="ethnic" size="20" id="ethnic" /> Ethnicity <br /> <input type="text" name="country" size="20" id="country" /> Country<br /> <input type="text" name="state" size="20" id="state" /> State<br /> <input type="text" name="town" size="20" id="town" /> Town<br /> <input type="text" name="zip" size="20" id="zip" /> Zip Code<br /> <br /> <input type="text" name="email" size="40" id="email" /> Email<br /> <textarea name="job" cols="40" id="job"></textarea> Job<br /> <textarea name="business" cols="40" id="business"></textarea> Business<br /> <input type="text" name="religion" size="60" id="religion" /> Religion</p> <p><br /> <textarea name="social" cols="100" id="social"></textarea> Social Groups<br /> <textarea name="political" cols="100" id="political"></textarea> Political groups<br /> <textarea name="affiliations" cols="100" id="affiliations"></textarea> Other Affiliations<br /> <textarea name="buying" cols="100" id="buying"></textarea> Items I am buying<br /> <textarea name="selling" cols="100" id="selling"></textarea> Items I am selling<br /> <textarea name="likes" cols="100" id="likes"></textarea> My likes <br /> <textarea name="dislikes" cols="100" id="dislikes"></textarea> My dislikes <br /> <textarea name="links" cols="100" id="links"></textarea> My links <br /> <input type="submit" value="Store in database and search" /> <input type="reset" value="Reset fields" /> </p> </p> </form> <p><a href="index.php">HOME</a></p> <p>Welcome to members page. to log out click <a href="Logout.php">HERE. </a></p> </body> </html> I got some help from this forum previously and am having some more issues. I created a database class that works and returns a PDO object. I am having trouble figuring out how to use the object in another class I want to use to access some CRUD functions. I get to the point where I start using the pdo object I create in the database class and my code fails. I am obviously calling the pdo object property incorrectly. I guess I don't understand the proper syntax. I've included the code for the database class and the CRUD class. The problem starts at the point where I try to run a prepared statement. I have include a comment "Problem stars here" to indicate that point. There is a lot of debug stuff still in the code. Thanks, --Kenoli
<?php
class Db {
public $pdo = '';
public $message = 'A message from db!<br><br>';
function __construct() {
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "tio-local";
$db_options = array(
PDO::ATTR_EMULATE_PREPARES => false,
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC
);
try {
$this->pdo = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password, $db_options);
// set the PDO error mode to exception
$this->pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} // End Try
catch(PDOException $e) {
echo "Error: " . $e->getMessage();
}
} // End __construct
} // End class definition DB.php
$db = new Db;
$pdo = $db->pdo;
print_r ($pdo);
?>
<?php // Db_functions.php
include_once ('Db.php'); // $db instantiated in class file
$pdo = $db->pdo;
class Db_functions {
public $pdo = '';
function __construct($pdo) {
$this->pdo = $pdo;
} // Close __construct
public function insert($table_name, $white_list, $insert_array) {
if ($white_list == '' && $table_name == 'Sites') {
$white_list = array('gone'=>'','site_name' =>'', 'site_address' =>'', 'grommets' =>'', 'tape' =>'', 'site_image' =>'', 'description' =>'', 'surface' =>'', 'tio_contact' =>'', 'site_contact' =>'','owner' =>'', 'lessee' =>'', 'contact_phone' =>'', 'contact_email' =>'', 'contact_date' =>'', 'comments' =>'');
} elseif ($white_list == '' && $table_name == 'Persons') {
$white_list = array('gone'=>'', 'fname'=>'', 'lname'=>'', 'tio'=>'', 'volunteer'=>'', 'general'=>'', 'artist_pic'=>'', 'email'=>'', 'website'=>'', 'telephone'=>'', 'address'=>'', 'city'=>'', 'state'=>'', 'zip'=>'', 'statement'=>'');
}
echo '<strong>***The following is the PDO object: </strong>';
print_r ($this->pdo);
echo '<p>The following is the $white_list:<br>';
echo '<pre>';
print_r ($white_list);
echo '</pre>';
echo '<strong>***The following is the PDO object: </strong>';
print_r ($this->pdo);
/** Test and remove any unpermitted columns **/
$insert_array = array_intersect_key($insert_array, $white_list);
echo '<strong>***The following is the PDO object: </strong>';
print_r ($this->pdo);
/** Generate variables to create prepared statements **/
foreach($insert_array as $key => $value) {
$col .= $key . ', ';
$val .= ':' .$key . ', ';
}
echo '$col = ' . $col . '<p>';
echo '$val = ' . $val . '<p>';
echo '<strong>***The following is the PDO object: </strong>';
print_r ($this->pdo);
/** Remove ', ' at end of arrays and create prepared statement */
$col = substr_replace($col ,"",-2);
$val = substr_replace($val ,"",-2);
$sql = "INSERT INTO Sites ($col) VALUES ($val)";
echo "<p>SQL = $sql<br><br>"; /** Debug **/
echo '<h3>More</h3<br>';
/** Register prepared statement */
/****** PROBLEM STARTS HERE *****/
$stmt = $this->pdo->prepare($sql);
echo '<h3>More2</h3>';
/** Create [:field, $value] pairs. */
foreach($insert_array as $key => $value) {
$param = ':' . $key;
$stmt->bindParam($param, $$value);
//}
/** Create [field => value] array */
foreach($insert_array as $key => $value) {
$insert[$key] = $value;
}
/** Execute statement using $insert array. **/
$stmt->execute($insert);
} // End insert function
} // Close class definition
$db_functions = new Db_functions($pdo);
$insert_array = array('fname' => 'John', 'lname' => 'Hancock');
$db_functions->insert('Persons', '', $insert_array);
echo '<pre>';
print_r ($db_functions);
echo '</pre>';
?>
Edited February 6 by kenoli Hi, how to make this form + php to make it work this way from the drop-down box choose the database title, or ID, and then the form in which I write some content will be added to the database that matches the title or ID that is sometimes so as not to invade a mistake and content that I type is not added to another title or ID I created something like this: <?php $db = new mysqli('localhost','xxxxx','xxxx','xxxx'); mysqli_query($db,'SET NAMES `utf8`'); $sqlnowe = mysqli_query($db,'SELECT * FROM `cc` ORDER BY `id` DESC '); ?> <form action="formu.php" method="post"> <select name="tytul"> <?php while ($rownowe = mysqli_fetch_array($sqlnowe)) { ?> <option><? echo $rownowe['tytul']; ?></option><?php };?> </select> opismax:<br /> <input type="text" name="opismax" /><br /> <textarea name="opismax" cols="50" rows="10">Proszę, wpisz tutaj jakiś komentarz...</textarea> <input type="submit" value="dodaj" /> </form> <?php // odbieramy dane z formularza $tytul = $_POST['tytul']; $opismax = $_POST['opismax']; if($tytul and $opismax) { // dodajemy rekord do bazy $ins = mysqli_query("INSERT tytul='$tytul' INTO publications SET opismax='$opismax'"); if($ins) echo "Good"; else echo "Bad"; } ?> |
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