joshstevens19 Posted April 25, 2013 Share Posted April 25, 2013 so im connecting my netbeans php to phpmyadmin (database).. now in this i am trying to create a php code statement which will take the inserted data into the html fields and insert them into my database successful.. anybody know what im doing wrong here is the error message: Warning: mysql_query() expects parameter 1 to be string, resource given in C:\xampp\htdocs\PhpProject1\New Review.php on line 27 error inserting new record <?php if (isset($_REQUEST['submitted'])) { include('connect database.php'); $showtitle = $_POST['SHOWTITLE']; $showreview =$_POST['SHOWREVIEW']; $sqlinsert = "INSERT INTO reviews (SHOWTITLE,SHOWREVIEW) VALUES('$showtitle', '$showreview')"; if (!mysql_query($dbConn,$sqlinsert)) { die('error inserting new record'); } here is my html coding with it <form method="post" action="New Review.php"> <input type="hidden" name="submitted" value="true" /> <fieldset> <legend> New Reviews</legend> <label> Enter show title:<input type ="text" name="SHOWTITLE" /></label> <label> Enter Review: <input type ="text" name="SHOWREVIEW" /></label> </fieldset> <br /> <input type ="submit" value ="add new review" /> </form> Quote Link to comment Share on other sites More sharing options...
lemmin Posted April 25, 2013 Share Posted April 25, 2013 The problem is that mysql_query() expects parameter 1 to be string and you gave it a resource . mysqli_query() accepts the connection resource as the first parameter, not mysql_query(). You can fix it by taking the first variable out of the function call: if (!mysql_query($sqlinsert)) { Quote Link to comment Share on other sites More sharing options...
joshstevens19 Posted April 25, 2013 Author Share Posted April 25, 2013 so all i do is replace if (!mysql_query($dbConn,$sqlinsert)) { to your code? am i correct in saying that and thats the only issue involved?? Quote Link to comment Share on other sites More sharing options...
joshstevens19 Posted April 25, 2013 Author Share Posted April 25, 2013 i have tried that part of changing and now it comes up with this Notice: Undefined variable: INSERT INTO reviews (SHOWTITLE,SHOWREVIEW) VALUES('Match Of The Day', 'amazing') in C:\xampp\htdocs\PhpProject1\New Review.php on line 27 error inserting new record <?php if (isset($_POST['submitted'])) { include('connect database.php'); $showtitle = $_POST['SHOWTITLE']; $showreview =$_POST['SHOWREVIEW']; $sqlinsert = "INSERT INTO reviews (SHOWTITLE,SHOWREVIEW) VALUES('$showtitle', '$showreview')"; if (!mysql_query($$sqlinsert)) { die('error inserting new record'); } Quote Link to comment Share on other sites More sharing options...
lemmin Posted April 25, 2013 Share Posted April 25, 2013 Too many dollar signs. You should only have one dollar sign on $sqlinsert. Quote Link to comment Share on other sites More sharing options...
joshstevens19 Posted April 25, 2013 Author Share Posted April 25, 2013 Too many dollar signs. You should only have one dollar sign on $sqlinsert. what the dollar signs are making it not work :S and in the if statement do i need a result factor as at the moment it only says die if you look at my code! Quote Link to comment Share on other sites More sharing options...
lemmin Posted April 25, 2013 Share Posted April 25, 2013 Just paste the line I posted in place of your "if" line and it should work. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.