Question

如果用户的信息（登录信息）与存储在数据库中的信息相匹配，我想在php设备上生成一个php文件返回true或false。

到目前为止，我已经编写了以下代码：

<?php

require "init.php";

if($connection){

    $username = $_POST["username"];
    $password = $_POST["password"];

    $statement = mysqli_prepare($connection,"SELECT * FROM  `admin` WHERE username = ? AND password = ?");

    mysqli_stmt_bind_param($statement,"ss",$username,$password);
    mysqli_stmt_execute($statement);

    mysqli_stmt_store_result($statement);


    mysqli_close($connection);
}

Answer 1

简单的mysqli_stmt_execute返回一个布尔值只打印结果，或者你可以制作一个json。

<?php
    require "init.php";
    require "init.php";

    if($connection){
      $username = $_POST["username"];
   $password = $_POST["password"];

   $statement = mysqli_prepare($connection,"SELECT * FROM  `admin` WHERE username = ? AND password = ?");

   mysqli_stmt_bind_param($statement,"ss",$username,$password);
   if(mysqli_stmt_execute($statement)){
       echo "true";
     }else {
       echo "false";
      }

   mysqli_stmt_store_result($statement);


  mysqli_close($connection);
}

Answer 2

要返回一个json，你必须创建一个关联数组并将其编码为json。

<?php
require "init.php";
require "init.php";

if($connection){
  $username = $_POST["username"];
 $password = $_POST["password"];

 $statement = mysqli_prepare($connection,"SELECT * FROM  `admin` WHERE username = ? AND password = ?");

 mysqli_stmt_bind_param($statement,"ss",$username,$password);

if(mysqli_stmt_execute($statement)){
   $response=array('response'=>true);
 }else {
   $response=array('response'=>false);
  }
   echo json_encode($response);
mysqli_stmt_store_result($statement);


 mysqli_close($connection);
  }

如果用户信息与数据库信息匹配，如何返回true或false？

2 个答案: