eMonk Posted April 23, 2011 Share Posted April 23, 2011 The php/mysql code below is called after a user fills out a html form and click on submit: insert-model.php <html> <head> <title>Untitled Document</title> </head> <body> <h1>Model Entry Results</h1> <?php // create short variable names $name=$_POST['name']; $age=$_POST['age']; $height=$_POST['height']; $hair=$_POST['hair']; $measurements=$_POST['measurements']; $weight=$_POST['weight']; $eyes=$_POST['eyes']; $service=$_POST['service']; $nationality=$_POST['nationality']; $location=$_POST['location']; $city_1=$_POST['city_1']; $city_2=$_POST['city_2']; $city_3=$_POST['city_3']; $city_4=$_POST['city_4']; $phone=$_POST['phone']; $email_1=$_POST['email_1']; $email_2=$_POST['email_2']; $website=$_POST['website']; $description=$_POST['description']; $availability=$_POST['availability']; $thumbnail=$_POST['thumbnail']; $url=$_POST['url']; $status=$_POST['status']; $views=$_POST['views']; $expiry_date=$_POST['expiry_date']; $notes=$_POST['notes']; if (!$name || !$thumbnail || !$url || !$views || !$expiry_date) { echo "You have not entered all the required details.<br />" ."Please go back and try again."; exit; } if (!get_magic_quotes_gpc()) { $name = addslashes($name); $height = addslashes($height); $hair = addslashes($hair); $measurements = addslashes($measurements); $eyes = addslashes($eyes); $nationality = addslashes($nationality); $location = addslashes($location); $phone = addslashes($phone); $email_1 = addslashes($email_1); $email_2 = addslashes($email_2); $website = addslashes($website); $description = addslashes($description); $availability = addslashes($availability); $thumbnail = addslashes($thumbnail); $url = addslashes($url); $expiry_date = addslashes($expiry_date); $notes = addslashes($notes); } @ $db = new mysqli('host', 'username', 'password', 'database'); // these values were removed if (mysqli_connect_error()) { echo "Error: Could not connect to database. Please try again later."; exit; } $query = "insert into model values ('".$name."', '".$age."', '".$height."', '".$hair."', '".$measurements."', '".$weight."', '".$eyes."', '".$service."', '".$nationality."', '".$location."', '".$city_1."', '".$city_2."', '".$city_3."', '".$city_4."', '".$phone."', '".$email_1."', '".$email_2."', '".$website."', '".$description."', '".$availability."', '".$thumbnail."', '".$url."', '".$status."', '".$views."', '".$expiry_date."', '".$notes."')"; $result = $db->query($query); if ($result) { echo $db->affected_rows." service provider inserted into the database."; } else { echo "An error has occurred. The model was not added."; } $db->close(); ?> </body> </html> I keep getting the following error: "An error has occurred. The model was not added." Any ideas? Quote Link to comment Share on other sites More sharing options...
Rommeo Posted April 23, 2011 Share Posted April 23, 2011 For to create short variables, you may write extract($_POST), you dont need to write that long. and for insert query : "insert into model (dbfieldage, dbfieldname, ) values ( '$postedage', ... ) " Quote Link to comment Share on other sites More sharing options...
eMonk Posted April 23, 2011 Author Share Posted April 23, 2011 I added in the following code but get the same error: $query = "insert into model (model_name, age, height, hair, measurements, weight, eyes, service, nationality, location, city_1, city_2, city_3, city_4, phone, email_1, email_2, website, description, schedule, thumbnail, url, status, views, expiry_date, notes) values ('$name', '$age', '$height', '$hair', '$measurements', '$weight', '$eyes', '$service', '$nationality', '$location', '$city_1', '$city_2', '$city_3', '$city_4', '$phone', '$email_1', '$email_2', '$website', '$description', '$availability', '$thumbnail', '$url', '$status', '$views', '$expiry_date', '$notes')"; $result = $db->query($query); Quote Link to comment Share on other sites More sharing options...
Rommeo Posted April 23, 2011 Share Posted April 23, 2011 $query = "insert into model (model_name, age, height, hair, measurements, weight, eyes, service, nationality, location, city_1, city_2, city_3, city_4, phone, email_1, email_2, website, description, schedule, thumbnail, url, status, views, expiry_date, notes) values ('$name', '$age', '$height', '$hair', '$measurements', '$weight', '$eyes', '$service', '$nationality', '$location', '$city_1', '$city_2', '$city_3', '$city_4', '$phone', '$email_1', '$email_2', '$website', '$description', '$availability', '$thumbnail', '$url', '$status', '$views', '$expiry_date', '$notes')"; after these, write these : $result = mysql_result ( $query ) or die ( mysql_error() ) ; now it will tell you what the problem is, copy it here if you can not figure it out. Quote Link to comment Share on other sites More sharing options...
eMonk Posted April 23, 2011 Author Share Posted April 23, 2011 Warning: Wrong parameter count for mysql_result() in /usr/www/virtual/user/domain/v1/admin/insert-model.php on line 75 line 75 is: $result = mysql_result ( $query ) or die ( mysql_error() ) ; Quote Link to comment Share on other sites More sharing options...
Rommeo Posted April 23, 2011 Share Posted April 23, 2011 my mistake, it has to be mysql_query, not the result. $result = mysql_query( $query ) or die ( mysql_error() ) ; Quote Link to comment Share on other sites More sharing options...
eMonk Posted April 23, 2011 Author Share Posted April 23, 2011 Warning: mysql_query() [function.mysql-query]: Access denied for user 'root'@'localhost' (using password: NO) in /usr/www/virtual/user/domain/v1/admin/insert-model.php on line 75 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /usr/www/virtual/user/domain/v1/admin/insert-model.php on line 75 Access denied for user 'root'@'localhost' (using password: NO) Quote Link to comment Share on other sites More sharing options...
Rommeo Posted April 23, 2011 Share Posted April 23, 2011 You are not using password, or your password is wrong as it says. Quote Link to comment Share on other sites More sharing options...
eMonk Posted April 23, 2011 Author Share Posted April 23, 2011 @ $db = new mysqli('host', 'username', 'password', 'database'); // these values were removed I just logged into the mysql server via putty with this info and it worked. Quote Link to comment Share on other sites More sharing options...
Rommeo Posted April 23, 2011 Share Posted April 23, 2011 Did you create a DB and set the values ? added the user ? or is that a test server that you can use empty values ? Generally by default host is "localhost" and you set the rest. Quote Link to comment Share on other sites More sharing options...
eMonk Posted April 23, 2011 Author Share Posted April 23, 2011 I have it working now...forgot to set NULL as the first value as I'm using model_id int unsigned not null auto_increment primary key No idea why it was giving me those login/password errors as it was correct. Quote Link to comment Share on other sites More sharing options...
eMonk Posted April 23, 2011 Author Share Posted April 23, 2011 By the way, what did you mean by extract($_POST) Do I just use that instead of my short variable names? Quote Link to comment Share on other sites More sharing options...
Rommeo Posted April 23, 2011 Share Posted April 23, 2011 // create short variable names $name=$_POST['name']; $age=$_POST['age']; $height=$_POST['height']; $hair=$_POST['hair']; $measurements=$_POST['measurements']; $weight=$_POST['weight']; $eyes=$_POST['eyes']; $service=$_POST['service']; $nationality=$_POST['nationality']; $location=$_POST['location']; $city_1=$_POST['city_1']; $city_2=$_POST['city_2']; $city_3=$_POST['city_3']; $city_4=$_POST['city_4']; $phone=$_POST['phone']; $email_1=$_POST['email_1']; $email_2=$_POST['email_2']; $website=$_POST['website']; $description=$_POST['description']; $availability=$_POST['availability']; $thumbnail=$_POST['thumbnail']; $url=$_POST['url']; $status=$_POST['status']; $views=$_POST['views']; $expiry_date=$_POST['expiry_date']; $notes=$_POST['notes']; echo $notes; instead, you just write extract($_POST); echo $notes; Same thing. Quote Link to comment Share on other sites More sharing options...
eMonk Posted April 23, 2011 Author Share Posted April 23, 2011 Oh wow, that is awesome.. thanks bro! Quote Link to comment Share on other sites More sharing options...
Pikachu2000 Posted April 23, 2011 Share Posted April 23, 2011 Regarding security vulnerabilities and the use of extract: Do not use extract() on untrusted data' date=' like user input[/color'] (i.e. $_GET, $_FILES, etc.). If you do, for example if you want to run old code that relies on register_globals temporarily, make sure you use one of the non-overwriting extract_type values such as EXTR_SKIP and be aware that you should extract in the same order that's defined in variables_order within the php.ini. Quote Link to comment Share on other sites More sharing options...
eMonk Posted April 24, 2011 Author Share Posted April 24, 2011 Would it be safer to just use my old short variable names instead of extract($_POST)? I'm not too sure what the php manual meant on extract(). Quote Link to comment Share on other sites More sharing options...
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