PHP通过变量从MySQL获取单个值

Question

我似乎无法通过UserEmail从Users表中找到UserID

我有一个简单的函数假设返回UserID显然也

function get_userID($UEml) 
{
// Check database connection
if( ($DB instanceof MySQLi) == false) {
return array(status => false, message => 'MySQL connection is invalid');
}
$qSQL = "SELECT UsID FROM Users WHERE UsEml=? LIMIT 1";
$qSQL = $DB->prepare($qSQL);
$UEml = $DB->real_escape_string($UEml);
$qSQL->bind_param("s", $UEml);
$qSQL->execute();
$result = $qSQL->get_result();
while ($row = $result->fetch_row()) {
return $row[0];
}
//  return $row[0];
if($qSQL) {
return array(status => true);
}
else {
return array(status => false, message => 'Not Found');
}
}

从php脚本调用check-User.php

<?php
require_once("db-config.php");
include 'functions.php';
...
$UsID = get_userID("[e-mail]");
echo 'UserID: <span style="color: blue">'. $UsID ."</span>";
...
?>

db-config.php

<?php
// Two options for connecting to the database:
define('HOST_DIRECT', 'example.com'); // Standard connection
define('HOST_LOCAL', '127.0.0.1');    // Secure connection, slower performance
define('DB_HOST', HOST_DIRECT);         // Choose HOST_DIRECT or HOST_STUNNEL, depending on your application's requirements
define('DB_USER', 'dbUser');    // MySQL account username
define('DB_PASS', 'SecretPas');    // MySQL account password
define('DB_NAME', 'DBName');     // Name of database

// Connect to the database
$DB = new MySQLi(DB_HOST, DB_USER, DB_PASS, DB_NAME);
if ($DB->connect_error) {
die("Connection failed: " . $DB->connect_error);
}
//echo "Connected successfully";
?>

我尝试了很多变化，但没有运气，也检查了许多类似的帖子在这里，但就是不能让它工作。由于

Answer 1

这里有几个错误:

• $DB在函数
中不可用
• echo语句错误

这是没有这两个错误的代码:

function get_userID($DB, $UEml) 
{
// Check database connection
if ( ($DB instanceof MySQLi) == false) {
return array(status => false, message => 'MySQL connection is invalid');
}
$qSQL = "SELECT UsID FROM Users WHERE UsEml=? LIMIT 1";
$qSQL = $DB->prepare($qSQL);
$qSQL->bind_param("s", $UEml);
$qSQL->execute();
$result = $qSQL->get_result();
while ($row = $result->fetch_row()) {
return $row[0];
}
//  return $row[0];
// I do not know why you wrote this code. If you get an user this code will not be executed
if ($qSQL) {
return array(status => true);
} else {
return array(status => false, message => 'Not Found');
}
}

和你的echo:

// ...
$UsID = get_userID($DB, "[e-mail]");
echo "UserID: <span style="color: blue">{$UsID}</span>";