尝试将值传递给 ISSET 条件但出现错误

Question

我创建了一个 PHP 脚本，如果使用该脚本，它总是会转到 else 条件，我不确定为什么它不会转到 else 条件。

  <?php

   require_once  'db_functions.php';

   $db = new DB_Functions();

   $response = array();

   $phone="1234";

   $name="Test";

   $birthdate="1994-01-01";

   $address="123 M";

    if(isset($_POST['phone']) &&

    isset($_POST['name']) &&

    isset($_POST['birthdate']) &&

    isset($_POST['address']))

   {

    echo "Hello World 1";

    $phone = $_POST['phone'];

    $name = $_POST['name'];

    $birthdate = $_POST['birthdate'];

    $address = $_POST['address'];

    echo "Hello World 2";

   }

   else{

    echo "Hello";

    $response["error_msg"] = "Required parameter 

    (phone,name,birthdate,address) is missing!";

    echo json_encode($response);

    }

    ?>

输出：

_msg":"缺少必需的参数（电话、姓名、生日、地址）！"}

如果该值被传递，那么它应该转到 if 条件而不是 else 条件。

尝试的选项

尝试了以下选项，但我得到了空值：

$test=$_POST['手机']; echo "嘿......".$test;

echo isset($_POST['phone']);

使用的网址 https://www.aaa.ccc/php/register.php?phone=232&name=test&birthdate=1954-04-04&address=232

Answer 1

您正在使用GET方法而不是传递 url 中的参数POST，因此您需要：&nbsp;<?php&nbsp; &nbsp;require_once&nbsp; 'db_functions.php';&nbsp; &nbsp;$db = new DB_Functions();&nbsp; &nbsp;$response = array();&nbsp; &nbsp;$phone="1234";&nbsp; &nbsp;$name="Test";&nbsp; &nbsp;$birthdate="1994-01-01";&nbsp; &nbsp;$address="123 M";&nbsp; &nbsp; if(isset($_GET['phone']) &&&nbsp; &nbsp; isset($_GET['name']) &&&nbsp; &nbsp; isset($_GET['birthdate']) &&&nbsp; &nbsp; isset($_GET['address']))&nbsp; &nbsp;{&nbsp; &nbsp; echo "Hello World 1";&nbsp; &nbsp; $phone = $_GET['phone'];&nbsp; &nbsp; $name = $_GET['name'];&nbsp; &nbsp; $birthdate = $_GET['birthdate'];&nbsp; &nbsp; $address = $_GET['address'];&nbsp; &nbsp; echo "Hello World 2";&nbsp; &nbsp;}&nbsp; &nbsp;else{&nbsp; &nbsp; echo "Hello";&nbsp; &nbsp; $response["error_msg"] = "Required parameter&nbsp;&nbsp; &nbsp; (phone,name,birthdate,address) is missing!";&nbsp; &nbsp; echo json_encode($response);&nbsp; &nbsp; }&nbsp; &nbsp; ?>