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使用json字符串将数据从java发送到php不会';行不通_Java_Php_Json - Fatal编程技术网

使用json字符串将数据从java发送到php不会';行不通

使用json字符串将数据从java发送到php不会';行不通,java,php,json,Java,Php,Json,我的java应用程序中有一些数据,我希望使用json将它们发送到php,以便稍后将它们插入数据库。问题在于json字符串 Main.java String args = "{\"nom\":\""+hostName+"\",\"host_name\":\""+hostName+"\", \"os_name\":\""+nameOS+"\",\"os_type\":\""+osType+"\",\"os_version\":\""+osVersion+"\"}"; Cpu.main(args);

我的java应用程序中有一些数据,我希望使用json将它们发送到php,以便稍后将它们插入数据库。问题在于json字符串

Main.java

String args = "{\"nom\":\""+hostName+"\",\"host_name\":\""+hostName+"\", \"os_name\":\""+nameOS+"\",\"os_type\":\""+osType+"\",\"os_version\":\""+osVersion+"\"}";
Cpu.main(args);
java

package webservice;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintStream;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLConnection;

public class Cpu {

    public static void main(String args) {

        try {
            // make json string, try also hamburger

            // send as http get request
            URL url = new URL("http://localhost:8080/parc/index.php?order=" + args);
            URLConnection conn = url.openConnection();

            // Get the response
            BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            String line;
            while ((line = rd.readLine()) != null) {
                System.out.println(line);
            }
            rd.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}
PHPFILE.php

    <?php
$order = $_GET["order"];
$obj = json_decode($order);
$nom = $obj->{"nom"};
$host_name = $obj->{"host_name"};
$os_name = $obj->{"os_name"};
$os_type = $obj->{"os_type"};
$os_version = $obj->{"os_version"};
                echo $host_name;
            echo json_last_error(); // 4 (JSON_ERROR_SYNTAX)
echo json_last_error_msg(); // unexpected character 
$array = array("nom" => $nom, "host_name" => $host_name, "os_name" => $os_name, "os_type" => $os_type, "os_version" => $os_version);
echo json_encode($array);

?>


现在我认为json字符串格式的问题是“args”,因为当我更改变量
hostname
os\u type
。。。如下所示:(
String json=“{\'name\':\'Frank\',\'food\':\'pizza\',\'quantity\':3}”;
)(来自教程)它工作正常。

尝试在org.json包中创建JsonObject类的对象。然后使用它的toString()方法在url中输入。请您详细解释一下?我是初学者,你可以参考这些例子