Blank Page on mysqli_query();? [duplicate] - php

Question

This question already has answers here:
Parse errors are not displayed
(8 answers)
Closed 8 years ago.
I am making a basic song recommendation site, and I have a form set up that leads to a page with this exact code on it:
<?php
ini_set('display_errors',1);
ob_start();
session_start();
$host = "localhost";
$user = "root";
$pass = "MYPASS";
$db = "tts";
$conn = mysqli_connect($host, $user, $pass, $db);
$song = $_POST['song'];
$artist = $_POST['artist'];
$album = $_POST['album'];
$linkitunes = $_POST['linkitunes'];
$artwork = $_POST['artwork'];
$song = stripslashes($song);
$artist = stripslashes($artist);
$album = stripslashes($album);
$linkitunes = stripslashes($linkitunes);
$artwork = stripslashes($artwork);
$sql = "INSERT INTO recommendation (user_id, song, artist, album, linkitunes, artwork, rating)";
$sql = $sql . "VALUES ($_SESSION['id'], '$song', '$artist', '$album', '$linkitunes', '$artwork', '$rating');";
print "Hello.";
$result = mysqli_query($sql) or die("Fail");
ob_flush();
?>
It always shows the "Hello." string until I add in the $sql value. I am thinking there is something wrong with the code syntax, but not sure. Have tried very many variations. Just in case, I have added the form code as well:
<form action="recommend-action.php" method="POST">
<div id="noP" align="center">
<h2>Make a new Recommendation</h2>
<p>Please search for your song before you recommend it.</p>
</div>
<div align="center">
<input required name="song" type="text" placeholder="Song" maxlength="50"></input>
<input required name="artist" type="text" placeholder="Artist" maxlength="50"></input>
<input name="album" type="text" placeholder="Album" maxlength="50"></input>
<input name="artwork" type="url" placeholder="Artwork" maxlength="500"></input>
<input name="linkitunes" type="url" placeholder="Link in iTunes" maxlength="500"></input>
<input id="submit" type="submit" value="Recommend"></input>
</div>

Answer 1

you should execute this...
$result = mysqli_query($con,$sql); or die("Fail");

Answer 2

You DO NOT have any line in PHP that gets converted to HTML... How will it display anything on screen. Write this at the end of your code.
echo "<br> Inserted Successfully";
The syntax of mysqli_query is wrong.
Use the following one.
$result=mysqli_query($conn,$sql) or die ("Fail");
Other possible Checks:
First Check your username and password and check if the connection to the database is successful by using this piece of code.
if (!$conn) {
die('Could not connect to MySQL: ' . mysql_error());
}
Use session_start() as the first line of your code or else your sessions will not work. This is not needed but may cause you a problem in future.

Answer 3

I did some modifications to your code, try this:
<?php
ini_set('display_errors',1);
ob_start();
session_start();
$host = "localhost";
$user = "root";
$pass = "MYPASS";
$db = "tts";
$conn = mysqli_connect($host, $user, $pass, $db);
$song = $_POST['song'];
$artist = $_POST['artist'];
$album = $_POST['album'];
$linkitunes = $_POST['linkitunes'];
$artwork = $_POST['artwork'];
$song = stripslashes($song);
$artist = stripslashes($artist);
$album = stripslashes($album);
$linkitunes = stripslashes($linkitunes);
$artwork = stripslashes($artwork);
$sql = "INSERT INTO recommendation (user_id, song, artist, album, linkitunes, artwork, rating)";
$sql = $sql . "VALUES (".$_SESSION['id'].", '$song', '$artist', '$album', '$linkitunes', '$artwork', '$rating');";
print "Hello.";
$result = mysqli_query($conn,$sql) or die("Fail");
ob_flush();
?>