mysqli_num_rows not returning accurate results - php

Question

I have a script that is supposed to take a form email value and query the database for duplicates. However, regardless of if I use an existing email (in the database) or a new one, I get "success".
My script looks like this:
if(!empty($_POST['email'])) {
$email = $_POST['email'];
// query the database for duplicates
$query = 'SELECT email FROM user_info WHERE email = "$email"';
$result = mysqli_query($db, $query);
if(mysqli_num_rows($result)){
echo "Email already taken";
} else {
echo "Success";
}
Obviously mysqli_num_rows is returning some kind of data - or none at all.
Does anyone understand why I would not get a returned row if I'm selecting an email that already exists in the database?
EDIT
$('#email').blur( function() {
if(!this.value){
$('#jquery_msg').html("Email can't be empty");
} else if (!filter.test(this.value)){
$('#jquery_msg').html("Not a valid email");
} else {
var emailVal = $('#email').val();
// make an ajax call
$.ajax({
url: 'validation.php',
type: 'post',
data: {'email': 'emailVal'},
success: function(data, status) {
$('#jquery_msg').html(data);
// console.log(data);
},
error: function(xhr, desc, err) {
console.log(xhr);
console.log("Details: " + desc + "\nError:" + err);
}
}); // end ajax call
}
});
The problem is in the data that's being passed to the PHP script via Ajax. What I thought was the value of $('#email').val(); was literally emailVal. Huh..
Now I need to solve how to pass the value into the Ajax call properly.
Thank you all for you help.

Answer 1

This is wrong:
$query = 'SELECT email FROM user_info WHERE email = "$email"';
Apart from the potential sql injection problem, variables do not get parsed when you use single quotes.
You should use a prepared statement instead:
$query = 'SELECT email FROM user_info WHERE email = ?';
Then you can prepare the statement, bind the placeholder and get the results. Check the manual for more information.
A quick solution would also be:
$query = 'SELECT email FROM user_info WHERE email = "'
. mysqli_real_escape_string($db, $email) . '"';
Edit: To trouble-shoot further problems, you should add error handling to your database calls. You can have mysqli throw exceptions telling you exactly what is wrong - if anything - by adding this to the top of your script:
ini_set('display_errors',1);
error_reporting(E_ALL | E_STRICT);
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);