php - 获取在 PHP 函数中声明的变量的值

Question

我有 2 个页面，“signup.php”和“globalfunctions.php”。在 signup.php 上，我从表单提交中获取所有信息，我对密码进行哈希处理(通过附加在 globalfunctions.php 中生成的随机字符串)，并使用我定义的函数 executeSQL。

signup.php:

include('/home/www/portaldev.samgoodman.co/processes/globalfunctions.php');
$singleAppendString = generateRandomAppend(16);
$form_email = $_POST['email'];
$form_password = $_POST['password'];
$form_name = $_POST['name'];
$form_school = $_POST['schoolid'];
$form_grad = $_POST['gradyear'];
$form_ip = $_SERVER['REMOTE_ADDR'];
$password_with_hash = $form_password.$singleAppendString;
$hashedPassword = sha1($password_with_hash);

executeSQL("$nextUserQuery", "SELECT id FROM users ORDER BY id DESC LIMIT 0 , 1");

这里是我想从数据库中获取 $nextUserQuery 值的地方，但我需要在函数中返回该值。

executeSQL("$insertUser", "INSERT INTO users (id, name, email, password, school_id, grad_year, lvl, signup_ip) VALUES ('".$calc_userid."', '".$form_name."', '".$form_email."', '".$hashedPassword."', '".$form_school."', '".$form_grad."', '0','".$form_ip."')");
executeSQL("$insertHash", "INSERT INTO vault (id, hash) VALUES ('".$calc_userid."', '".$singleAppendString."')");

globalfunctions.php

function generateRandomAppend($length) {
    $characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
    $randomString = '';
    for ($i = 0; $i < $length; $i++) {
        $randomString .= $characters[rand(0, strlen($characters) - 1)];
    }
    return $randomString;
}

function executeSQL($varName, $query) {
    global $varName;
    $con=mysqli_connect("localhost", "hugopak1_spm", "Massavailable1", "hugopak1_spm");
    $varName = mysqli_query($con, $query);
    return $varName;
}

Answer 1

您目前所做的不是获取 executeSQL 函数返回值的正确方法。从 executeSQL 函数中移除 $varName 参数

function executeSQL($query) {
    $con = mysqli_connect("localhost", "hugopak1_spm", "Massavailable1", "hugopak1_spm");
    $varName = mysqli_query($con, $query);
    return $varName;
}

并声明一个新变量来保存返回值

$nextUserQuery = executeSQL("SELECT id FROM users ORDER BY id DESC LIMIT 0 , 1");

在上面的例子中，$nextUserQuery的值是executeSQL("SELECT id FROM users ORDER BY id DESC LIMIT 0 , 1")返回的值。您应该对其他两行应用相同的内容，如下所示

$insertUser = executeSQL("INSERT INTO users (id, name, email, password, school_id, grad_year, lvl, signup_ip) VALUES ('".$calc_userid."', '".$form_name."', '".$form_email."', '".$hashedPassword."', '".$form_school."', '".$form_grad."', '0','".$form_ip."')");
$insertHash = executeSQL("INSERT INTO vault (id, hash) VALUES ('".$calc_userid."', '".$singleAppendString."')");