检查局部变量是否等于局部变量的正确方法

Question

这不会计算，即使值应该相等，它会直接进入 else 事件。 table 列名与数据库匹配，所以...

我需要在 ($recruitcheck==$recruitpass) 中使用某种类型的引号吗？

$recruiter = $_POST["recruiter"];
$recruitpass = $_POST["recruitpass"];
$recruitcheck = mysqli_query($maindb, "select aurapass from auras where auraname='$recruiter'");
if($recruitcheck == $recruitpass) {
    $badgecheck = mysqli_query($maindb, "select recruitbadge from auras where auraname='$recruiter'");
    if($badgecheck == "0") {
        echo "<script>alert('Recruiter information correct, but there is no recruit badge.')</script>";
        exit();
    } else {
        echo "<script>alert('Recruiter badge accepted.')</script>";
        $emailcheck = mysqli_query($maindb, "select * from auras where email='$email'");
        $namecheck = mysqli_query($maindb, "select * from auras where auraname='$auraname'");
        if(mysqli_num_rows($emailcheck) > 0 || mysqli_num_rows($namecheck) > 0){
            echo "<script>alert('New auraname/email is already taken.')</script>";
            exit();
        } else {
            exit();                        
        }
    }
} else {
    echo "<script>alert('Aura information of recruiter is incorrect. Please, use preexisting aura information to confirm recruiter identity.')</script>";
    exit();
}

Answer 1

$recruitpass 是一个字符串，而 $recruitcheck 是 mysqli_result 对象。在将其与 $recruitpass

进行比较之前，您需要获取对象 inside 中的字符串值

检查局部变量是否等于局部变量的正确方法

Correct way to check if local variable equals local variable

php

syntax

mysqli

function

local-variables