Question

我正在尝试将一些JSON解析为Android中的单独对象。对象由末端括号分隔，没有空格，然后是起始大括号（“结束第一个} {第二个开始”）。到目前为止，我的代码允许我找到这些点在JSON中的位置，但我无法弄清楚使用哪些方法来拆分对象然后存储它们。这是我的代码获取JSON，然后尝试解析：

public ArrayList<String> getQuestionJSONFromUrl(String url, List<NameValuePair> params) {

    // Making HTTP request
try {
        // defaultHttpClient
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);
        httpPost.setEntity(new UrlEncodedFormEntity(params));

        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        ArrayList<String> questionsList = new ArrayList<String>();
        int count = 0;
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        String newLine = null;
        String newLine2 = null;
        int previousJ = 0;
        if ((line = reader.readLine()) != null) {
            for(int j = 0; j < line.length(); j++) {
                while (line.charAt(j) == '}' && line.charAt(j+1) == '{') {
                    //everything before j = 1 string
                    //everything after gets deleted because I can't think
                    //of anything better, yet                       
                    sb.append(line);
                    sb.delete(j+1, sb.length());
                    line = sb.toString();
                    questionsList.add(count, line);
                    //line = line.split("\\}\\{");
                    Log.v("forintbs", questionsList.toString());
                    count++;
                }
            }
        }
        is.close();
        json = sb.toString();            
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }        
    // try parse the string to a JSON object
    try {
        jObj = new JSONObject(json);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    } 
    // return JSON String
    return null; 
}

显然，该方法现在返回null，因为无论如何它都无关紧要。到目前为止，当我查看日志以查看打印输出时，我的arraylist是一个很长的并且按预期正确获取了字符串 - 但原始数据的其余部分也按预期删除。我完全按原样保留了我的代码，以便您可以看到我尝试过的内容。有没有人有什么建议？以下是一些示例JSON：

{"category":"elections","id":"0","title":"Who will you vote for in November's Presidential election?","published":"2012-04-02","enddate":"2012-04-30","responsetype":"0"}

{"category":"elections","id":"2","title":"Question title, ladies and gents","published":"2012-04-02","enddate":"2012-04-30","responsetype":"1"}

我没有使用ArrayList的问题，而是使用JSON对象数组。我只是按照我昨天提出的一个问题的建议尝试。

编辑〜这是我的PHP脚本：

<?php

define("DB_HOST", "localhost");
define("DB_USER", "*");
define("DB_PASSWORD", "*");
define("DB_DATABASE", "*");

mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
mysql_select_db(DB_DATABASE);
$sql=mysql_query("select * from QUESTIONS where CATEGORY like 'elections'");
while($row=mysql_fetch_assoc($sql)) {
$output[]=$row;
}
foreach($output as $value) {
echo json_encode($value);
}
?>

Answer 1

正如Marc B所说，我需要更强大/更正确的PHP代码。另一个Stack Overflow用户帮我改进了Marc B帮助我理解的代码。现在JSON作为正确编码的数据返回，但我仍然没有确定如何将其解析为单独的对象。我很快就会到达那里。

以下是部分解决方案：https://stackoverflow.com/a/10019165/1231943

试图单独存储JSON

1 个答案: