php表单无法识别来自sql db的用户名[关闭]

My problem being this:

 <?php
 function user_exists($username) {
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT_COUNT('user_id') FROM `user` WHERE `username` = '$username'"), 0) == 1) ? true : false;
 }
 ?>

As you can see I have included my database name which IS correct and the user DOES exist and this is where I am calling it to pull from sql:

<?php
include 'core/init.php';

if (user_exists('dan') === true) {
echo 'exists';
} 

 die();

if (empty($_POST) === false) {
$username = $_POST['username'];
$password = $_POST['password'];

if(empty($username) === true || empty($password) === true {

    $errors[] = 'You need to enter a username and password';
} else if (user_exists($username) === false) {
    $errors[] = 'Sorry this user does not exist';
}
}
?>

I am not sure why I am getting a blank page and it isn's showing the message to say the user exists?

Aside from that silly typo in your query, the way you are using mysql API is indeed a terrible one.
Some critical flaws to be noted

• whatever "sanitize" function used to build a query should add quotes around returned value. Otherwise it will do no good and lead you to injection.
• you aren't checking for the mysql errors
• you are writing your code in one line making it extremely hard to read.

What it should be at the very least

function user_exists($username) {
    $username = sanitize($username); // remember it should add the quotes
    $sql = "SELECT COUNT(1) FROM `user` WHERE `username` = $username";
    $res = mysql_query($sql) or trigger_error(mysql_error().$sql);
    $row = mysql_fetch_row($res);
    return !empty($row[0]);
 }

I am not sure why I am getting a blank page.

Most likely it is caused by some PHP error.
Either tell PHP to show them on-screen or peek into server error log to read the error message

Or there is no such user in the database.

So, make your code like this

<?php
ini_set('display_errors',1);
error_reporting(E_ALL);

echo "testing<br>";

include 'core/init.php';

if (user_exists('dan')) {
   echo 'exists';
} else {
   echo 'not found';
}

here might be your syntax error which could not be fire

(mysql_result(mysql_query("SELECT_COUNT('user_id') FROM `user` WHERE `username` = '$username'"), 0) == 1) ? true : false;

give space between SELECT COUNT remove underscore "_" because it will give you mysql error

The problem is here

 SELECT_COUNT('user_id')
       ^ 

should be

 SELECT COUNT('user_id')