Question

我一直在尝试使用我的python代码将SQL查询数据从运行python的客户端传输到运行PHP的Web服务器，然后将该数据输入到MySQL数据库中。

以下是模拟单行的python测试代码：

    #!/usr/bin/python

import requests
import json


url = 'http://192.168.240.182/insert_from_json.php'
payload = {"device":"gabriel","data_type":"data","zone":1,"sample":5,"count":0,"time_stamp":"00:00"}
headers = {'content-type': 'application/json'}

response = requests.post(url, data=dict(payload=json.dumps(payload)), headers=headers)
print response

以下是服务器端的PHP脚本：

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "practice";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo "Connection made...";

$payload_dump = $_POST['payload']
//$payload = '{"device":"gabriel","data_type":"data","zone":1,"sample":4,"count":0,"time_stamp":"00:00"}';

$payload_array = json_decode($payload_dump,true);

//get the data_payload details
$device = $payload_array['device'];
$type = $payload_array['data_type'];
$zone = $payload_array['zone'];
$sample = $payload_array['sample'];
$count = $payload_array['count'];
$time = $payload_array['time_stamp'];

$sql = "INSERT INTO data(device, data_type, zone, sample, count, time_stamp) VALUES('$device', '$type', '$zone', '$sample', '$count', '$time')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}


$conn->close();
?>

如果我注释掉$ payload_dump而改为使用$ payload并运行脚本，则正确添加行。但是当我在客户端上调用python脚本时，我得到一个“响应200”但是没有添加记录。我在python中的JSON字符串格式不正确或者我没有正确传递值。

1）python代码的语法/过程是什么，可以让我看到PHP代码实际收到的内容。（即将“$ payload_dump”变量内容转换为字符串并发送回python脚本？

2）如何解决上述代码？

Answer 1

您的第一个问题是使用 json 标题，因为您没有发送原始json ，所以您不应该使用它：

您的数据看起来像

{payload = {'key1': 'value1',
            'key2': 'value2' }
         }

并且原始json数据应采用以下形式：

{'key1': 'value1', 
 'key2': 'value2'}

所以你需要从请求中删除标题

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

其次，您需要修复丢失的分号

$payload_dump = $_POST['payload']

到

$payload_dump = $_POST['payload'];

更好的解决方案

您可以使用请求直接发送json数据

response = requests.post(url, json=payload)

然后在php中用

抓住它

$payload_array = json_decode(file_get_contents('php://input'), true);

将JSON传递给PHP无法正常工作

1 个答案: