Post will not insert into MySQL Database - php

Question

I've been searching around for like 15 minutes and could not find anything that would fix this. Sorry if I just used wrong keywords or something it has been answered. Also to state this is not something that needs to be extremely secure, as anybody can view this.
So my PHP Post will not insert into MySQL Database.
Form:
<form method="post" action="./thankyou.php">
<h2>Please sign in</h2>
<input type="user"name="user" placeholder="Username">
<input type="textarea" class="" name="feedback" placeholder="Feedback for us."><br />
<button class="" type="submit" name="submitted">Submit Feedback</button>
</form>
Thank you: (Yes I replaced xxx with info)
<?php
$conn = mysql_connect('xxx', 'xxxx', 'xxxx');
if (!$conn) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("xxx") or die(mysql_error());
$user = $_POST['user'];
$fdb = $_POST['feedback'];
$insert = "INSERT INTO contact WHERE (user, feedback) VALUES ('".$user."', '".$fdb."')";
mysql_query($insert);
if(!$insert)
{
die('Could not enter data: ' . mysql_error());
}
echo $insert;
?>
Echo outputs correctly:
INSERT INTO contact WHERE (user, feedback) VALUES ('thisisauser', 'thisisfeedback')
I'm not sure what to do.

Answer 1

An insert does not, by definition, have a where clause.
Change your query as follows:
INSERT INTO contact (user, feedback) VALUES ('thisisauser', 'thisisfeedback')
OR, you can use this structure:
INSERT INTO contact SET user = 'thisisauser' , feedback = 'thisisfeedback'
Finally, this is bad for security. Use a different database API

Answer 2

You dont need to provide WHERE in insert its not correct
it should be
$insert = "INSERT INTO contact (user, feedback) VALUES ('".$user."', '".$fdb."')";
You had as
$insert = "INSERT INTO contact WHERE (user, feedback) VALUES ('".$user."', '".$fdb."')";
^.........here is the issue

Answer 3

The main issues are the SQL query and the way of you check the correct execution of the query
<?php
$conn = mysql_connect('xxx', 'xxxx', 'xxxx');
if (!$conn) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("xxx") or die(mysql_error());
$user = $_POST['user'];
$fdb = $_POST['feedback'];
$insert = "INSERT INTO contact (user, feedback) VALUES ('".$user."', '".$fdb."')";
$retval = mysql_query($insert, $conn);
if(!$retval) { //<---- You must check the result of the execution
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";