Question

这是我的java代码，它发出一个http请求并将json对象发送到php脚本。

login.java

    String username = jTextField1.getText();
    String password = jPasswordField1.getText();

    JSONObject obj = new JSONObject();

    obj.put("username", username);
    obj.put("password", password);

    //JSONArray list = new JSONArray();
    //list.add(username);
    //list.add(password);


    //obj.put("logindata", list);
    try {
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response;
        HttpPost httppost = new HttpPost("http://localhost/kolitha/json_test/index.php");
        StringEntity se = new StringEntity("myjson" + obj.toString());
        httppost.setEntity(se);
        System.out.print(se);
        httppost.setHeader("Accept", "application/json");
        httppost.setHeader("Content-type", "application/json");
        System.out.println(obj.toString());

        //response = httpclient.execute(httppost);

        HttpGet httpget = new HttpGet("http://localhost/kolitha/json_test/index.php");
        response = httpclient.execute(httpget);
        HttpEntity entity = response.getEntity();
        System.out.println(EntityUtils.toString(entity));



    } catch (Exception e) {
        e.printStackTrace();
        System.out.print("Cannot establish connection!");

    }

这是我的index.php脚本，我无法从json对象中提取用户名和密码。

的index.php

$obj = json_decode(file_get_contents('php://input'));

$username=$obj->{'username'};
$password=$obj->{'password'};


$connect=mysql_connect('localhost', 'root', '');
IF (!$connect){
die ('Failed Connecting to Database: ' . mysql_error());}

$d = mysql_select_db("kolitha_json_test");
if(!$d){ echo "db not selected";}

$sql="SELECT * FROM login WHERE username='$username' AND password='$password' ";
$result=mysql_query($sql) or die (mysql_error());
$count=mysql_num_rows($result);

 // If result matched $myusername and $mypassword, table row must be 1 row
if($count==1)
{
echo "true";
}
else
 {
echo "false";
}
?>

Answer 1

好的，我发表了评论，但似乎没有人注意评论。错误发生在Java代码

中

StringEntity se = new StringEntity("myjson" + obj.toString());

为名为myjson

的json字符串添加前缀

Answer 2

为什么同时使用 HttpPost 和 HttpGet。您正在将字符串实体添加到 HttpPost 请求并不使用它 ..

而您正在使用 Httpget ，其中未设置您的用户名和密码（显然没有与httpget请求关联的json数据）

在此处进行更改

 response = httpclient.execute(httppost);//don comment this line

如果您使用的是httppost

，请在此处添加评论

//HttpGet httpget = new HttpGet("http://localhost/kolitha/json_test/index.php");
//response = httpclient.execute(httpget);

我无法在php中解码json对象

2 个答案: