java - JSON 编码字符串的键是什么

Question

我有一个 Java 应用程序通过 PHP 脚本使用 JSON 与 SQL 数据库通信。我之前使用 JSON 对数组进行 json_encode，效果很好，您可以使用每个条目的键调用 Java 中的值。但是现在，我在 PHP 中对一个字符串进行了 json_encoded，如下所示: json_encode("成功!"); 现在，我如何在业务的 Java 端请求该值？我需要将一些东西作为 key 放入 jsonResponse.getString('key'); 中。关键是什么...

我希望你能理解我的问题...

public String send(String username, String password, String database){
    //Create a HTTPClient as the form container
    HttpClient httpclient;

//Use HTTP POST method
HttpPost httppost;

//Create an array list for the input data to be sent
ArrayList<NameValuePair> nameValuePairs;

//Create a HTTP Response and HTTP Entity
HttpResponse response;
HttpEntity entity;

//Create new default HTTPClient
httpclient = new DefaultHttpClient();

//Create new HTTP POST with URL to php file as parameter
httppost = new HttpPost("http://192.168.144.150/login/add.php");

String pass = md5(password);

//Next block of code needs to be surrounded by try/catch block for it to work
try {
    //Create new Array List
    nameValuePairs = new ArrayList<NameValuePair>();

    //place them in an array list
    nameValuePairs.add(new BasicNameValuePair("username", username));
    nameValuePairs.add(new BasicNameValuePair("password", pass));
    nameValuePairs.add(new BasicNameValuePair("database", database));

    //Add array list to http post
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

    //assign executed form container to response
    response = httpclient.execute(httppost);

    //check status code, need to check status code 200
    if(response.getStatusLine().getStatusCode()== 200){

        //assign response entity to http entity
        entity = response.getEntity();

        //check if entity is not null
        if(entity != null){


            //Create new input stream with received data assigned
            InputStream instream = entity.getContent();

            //Create new JSON Object. assign converted data as parameter.
            //JSONObject jsonResponse = new JSONObject(convertStreamToString(instream));
            JSONArray a = new JSONArray(convertStreamToString(instream));

            //assign json responses to local strings
            String res = a.getString(0);

            //Return the json response to the gui
            return res;

        } else {
            //Toast.makeText(this, "kapot", Toast.LENGTH_SHORT).show();
            return new String("De entiteit is leeg. Kortom: kapot");
        }


    } else {
        //Toast.makeText(this, "statuscode was niet 200", Toast.LENGTH_SHORT).show();
        return new String("De statuscode was niet 200.");
    }


} catch(Exception e){
    e.printStackTrace();
    //Display toast when there is a connection error
    //Change message to something more friendly
    //Toast.makeText(this, "verbindingsproblemem", Toast.LENGTH_SHORT).show();
    return new String("Er is een fout opgetreden. Controleer a.u.b. uw gegevens en de internetverbinding.");
}

不要介意荷兰语部分，这不重要。

Answer 1

json_encode("Success!") 的输出是 "Success"，它本身不是有效的 JSON document .有效的 JSON 文档的顶级实体(根)必须是对象或数组。

如果你只想返回一个字符串，你可以这样做:

1. 使其成为数组中的唯一条目:

   json_encode(array("Success!"))
   

   ...结果是 ["Success!"]，您可以在您的 Java 代码中像这样使用它(因为您说您正在使用 org.java 库):

   JSONArray a = new JSONArray(jsonStringFromPHP);
   String s = a.getString(0);
   

   或者当然

   String s = new JSONArray(jsonStringFromPHP).getString(0);
   

2. 或者让它成为一个对象的唯一属性:

   json_encode(array('result' => "Success!")
   

   ...结果是 {"result":"Success!"}，您可以在 Java 代码中像这样使用它:

   JSONObject o = new JSONObject(jsonStringFromPHP);
   String s = o.getString("result");
   

或者当然

    String s = new JSONObject(jsonStringFromPHP).getString("result");

您可能会使用 JSONTokenizer读取您当前的 JSON 片段(“成功!”)，但我不建议这样做。坚持交换有效的 JSON 文档。