chuddyuk Posted March 14, 2006 Share Posted March 14, 2006 I have a form which submits info into a database, it is very simple at the moment once i have solved this problem i will add the rest of my feilds. It submits, but on the variable name like it will show $title an $news in the database, i know its somethin v simple but cant figure it out please help, here is the page::[code]<? if(isset($_GET['commented'])){ // Tell the user it has been submitted (optional) echo('Your comment has been posted.');include "common.php";// Set global variables to easier names$titles = $_GET['title']; $new = $_GET['news']; // Connect to Mysql, select the correct database, and run teh query which adds the data gathered from the form into the databasemysql_connect($dbhost,$dbuser,$dbpasswd) or die(mysql_error());mysql_select_db($dbname) or die(mysql_error());$add_all = 'INSERT INTO `news` (`title`, `news`) VALUES (\'$titles\', \'$new\')'; mysql_query($add_all) or die(mysql_error());}else{// If the form has not been submitted, display it!?><form method="get" action="<?php echo $PHP_SELF?>">Title : <input type='text' name='title'><br><br>News : <input type='textarea' name='news'><br><br><input type='hidden' name='commented' value='set'><input name='submit' type='submit' value='Post your comment'></form> <?}?> [/code]Thanks Quote Link to comment Share on other sites More sharing options...
shocker-z Posted March 14, 2006 Share Posted March 14, 2006 Instead of[code]$add_all = 'INSERT INTO `news` (`title`, `news`) VALUES (\'$titles\', \'$new\')';[/code]use[code]$add_all = "INSERT INTO `news` (`title`, `news`) VALUES ('$titles', '$new')";[/code]And thats it :) Quote Link to comment Share on other sites More sharing options...
chuddyuk Posted March 14, 2006 Author Share Posted March 14, 2006 Works a Treat thanks !!!!Knew it was something really simple!!!! Quote Link to comment Share on other sites More sharing options...
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