divinedesigns1 Posted July 29, 2012 Share Posted July 29, 2012 ok i been trying to fix this error, and i am not able to insert the information into the database, the information in the db is correct, and the names for the form fields are correct also, but im still having problems inserting the information into the database. so i want to know what can cause the insert to not put the information into the database. because i havent figure it out as yet, can someone please give me a hand Link to comment Share on other sites More sharing options...
birbal Posted July 29, 2012 Share Posted July 29, 2012 what does your error says? they have all the answers always. Link to comment Share on other sites More sharing options...
divinedesigns1 Posted July 29, 2012 Author Share Posted July 29, 2012 what does your error says? they have all the answers always.no errors at all, var_dump(); displayed the strings, print_r(); displayed the strings also, i even echo it out, no errors on and i have error all on also still nothing Link to comment Share on other sites More sharing options...
birbal Posted July 29, 2012 Share Posted July 29, 2012 have you check the mysql error? Link to comment Share on other sites More sharing options...
divinedesigns1 Posted July 29, 2012 Author Share Posted July 29, 2012 have you check the mysql error?how you do that? Link to comment Share on other sites More sharing options...
birbal Posted July 29, 2012 Share Posted July 29, 2012 (edited) mysql_error() or mysqli_error() if you are using one of those api http://php.net/mysql_errorhttp://php.net/mysqli_error Edited July 29, 2012 by birbal Link to comment Share on other sites More sharing options...
divinedesigns1 Posted July 29, 2012 Author Share Posted July 29, 2012 mysql_error() or mysqli_error() if you are using those api http://php.net/mysql_errorhttp://php.net/mysqli_error yeah, im using mysqli_error(); im not getting any errors from that either Link to comment Share on other sites More sharing options...
birbal Posted July 29, 2012 Share Posted July 29, 2012 can you post the codes? Link to comment Share on other sites More sharing options...
divinedesigns1 Posted July 29, 2012 Author Share Posted July 29, 2012 Note: The session_start() and connection to the db is at the top of the pageill post a structure of the table below this reply in a few <?phpif(isset($_SESSION['uid'])){// if log in display the below menu?><table align="center" width="800"><tr><td width="663"><?phpif(isset($_GET['action'])){$action = $_GET['action'];if($action == "news"){ // creating a new Blog echo '<h3>New Blog</h3>'; if(isset($_POST['submit'])){ $title = $_POST['title']; $article = $_POST['article']; if(empty($title) && empty($article)){ echo 'Please Fill In All Fields'; } if(!empty($title) && empty($article)){ echo 'Article: ' . 'Some text is required'; } if(empty($title) && !empty($article)){ echo 'Title: ' . 'A Title Is Required'; } if(!empty($title) && !empty($article)){ // filter the title and article $title = mysqli_real_escape_string($con, $title); $article = mysqli_real_escape_string($con, $article); $title = stripslashes($title); $title = strip_tags($title); $article = stripslashes($article); $article = strip_tags($article); $query = "INSERT INTO article(title, body, post_user) VALUES('$title', '$article', '" . $_SESSION['uid'] . "')"; if(!$query){ die('Error: ' . mysqli_error()); } echo 'Your Blog Have Been Post'; } }else{ ?> Link to comment Share on other sites More sharing options...
divinedesigns1 Posted July 29, 2012 Author Share Posted July 29, 2012 the table structure id int(11) ac primarytitle varchar(255)body textnid int(11) index Link to comment Share on other sites More sharing options...
birbal Posted July 29, 2012 Share Posted July 29, 2012 (edited) you are not using mysqli_query(). it is just assigning the query not executing it. you will check the return value of mysqli_query() and you will pass the $query as parameter in the mysqli_query() http://php.net/mysqli_query Edited July 29, 2012 by birbal Link to comment Share on other sites More sharing options...
divinedesigns1 Posted July 29, 2012 Author Share Posted July 29, 2012 you are not using mysqli_query(). it is just assigning the query not executing it. you will check the return value of mysqli_query() and you will pass the $query as parameter in the mysqli_query() http://php.net/mysqli_query thank you very much, i though it wouldnt matter if i had the mysqli_query() or not, thanks birbal Link to comment Share on other sites More sharing options...
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