Question

我试图通过URL将python文件中的JSON发送到PHP文件。正确调用并运行PHP文件。但它没有捕获我尝试发送的msg数据。但它显示为NULL。

myPython.py

import requests, json
url = 'http:/mywebsite.com/api/v1/myPhpFile.php?msg='
payload = {'firstname':'Olivia', 'city':'London', 'country':'UK'}
r = requests.post(url, data=payload)

也尝试过：

r = requests.post(url, data=json.dumps(payload))

myPHP.php

<?php
...

class AddUser extends Abstractor {
    public $log;

    public function __construct() {
        $this -> log = new MyLogPHP('../includes/debug.log.txt');       
    }

    public function processJob($workerMsg) {
        $this -> log -> info($workerMsg, 'W/ADD-USER');
    }
}

$addUser = new AddUser();
$response = $addUser -> processJob($_REQUEST['msg']);

在日志中，我得到NULL值。

我甚至尝试通过删除＆＃39;？msg =＆＃39;

来修改网址

myPython.py

import requests, json
url = 'http:/mywebsite.com/api/v1/myPhpFile.php'
payload = {'msg':{'firstname':'Olivia', 'city':'London', 'country':'UK'}}
r = requests.post(url, data=payload)

或

r = requests.post(url, data=json.dumps(payload))

我在这里做错了什么？

Answer 1

@Alex是对的。

您正在发送正文中的数据，您应该使用PHP请求正文。下面的代码将帮助您入门。

<强> myPython.py

import requests, json
url = 'http:/mywebsite.com/api/v1/myPhpFile.php'
payload = {'firstname':'Olivia', 'city':'London', 'country':'UK'}
r = requests.post(url, data=json.dumps(payload))

<强> myPHP.php

<?php
...
class AddUser extends Abstractor {
    public $log;

    public function __construct() {
        $this -> log = new MyLogPHP('../includes/debug.log.txt');       
    }

    public function processJob($workerMsg) {
        $this -> log -> info($workerMsg, 'W/ADD-USER');
    }
}

$addUser = new AddUser();
$entityBody = file_get_contents('php://input');
$response = $addUser -> processJob($entityBody);

通过URL将Python从MS发送到PHP文件的NULL值

1 个答案: