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Syed Ali Raza's questions

Martin Hope
Syed Ali Raza
Asked: 2018-04-19 22:12:18 +0800 CST
  • 0

这是我的站点链接，当我尝试在网络中登录时，我得到的响应是整个 login.php 代码，而不是执行 php 文件后的响应。我的 app.yaml 文件是正确的，我已经仔细检查过了。我的 php 文件无法在谷歌云服务中执行。
这是我的 app.yaml 代码

runtime: php55
api_version: 1
threadsafe: yes

handlers:
- url: /favicon\.ico
  static_files: favicon.ico
  upload: favicon\.ico
- url: /css
  static_dir: css
- url: /images
  static_dir: images
- url: /fonts
  static_dir: fonts
- url: /utils
  static_dir: utils
- url: /js
  static_dir: js  
- url: /index.html
  script: /index.html
- url: /signup.html
  script: /signup.html
- url: /.*
  script: index.html 
- url: /utils/login.php
  script: /utils/login.php
- url: /js/login.js
  script: /js/login.js
- url: /js/script.js
  script: /js/script.js
- url: /js/jquery.min
  script: /js/jquery.min
- url: /js/bootstrap.min
  script: /js/bootstrap.min
- url: .*
  script: main.php

这是我的 login.php 代码

<?php


require 'dblinker.php';

function login(){
    try {
        $username = $_POST['username'];
        $password = $_POST['password'];

        // login function
        $handle=$conn->prepare("SELECT * FROM `users` WHERE `username`= :username AND `password`= :password");
        $handle->bindParam('username', $username, PDO::PARAM_STR);
        $handle->bindParam('password', $password, PDO::PARAM_STR);
        $handle->execute();
        if($handle->rowCount() == 1){
            $result=$handle->fetch(PDO::FETCH_ASSOC);
            $_SESSION["username"] = $result['username'];
            $_SESSION["role_id"] = $result['role_id'];
            echo "success";
        }else{
            echo "fail";
        }
    }
    catch(Exception $e){
        echo "F";
    }
}
session_start();
echo login();

?>
php5