2017-08-16 78 views
0

我想要消耗Android端的PHP文件(news.php)生成的JSON数据。我有一个PHP文件,似乎可以正确生成JSON数据,而我的Android版本正常工作。问题是我得到了上述例外。JSONException:Value <! - ?php的java.lang.String类型无法转换为JSONObject

令我困惑的是,如果我将PHP文件生成的JSON细节输出复制到单独的文件并将其另存为JSON文件(例如news.json),我的Android应用程序就能够使用JSON数据,但如果我将它重新命名为php(news.php)文件,我会收到上述异常。

我看了其他可能的重复问题,但他们不太适合我自己的。任何帮助是极大的赞赏。

我有个PHP文件编码到JSON格式:

<?php 

include("includes/db_connection.php");   

if (isset($_GET['latest_news'])) { 
    $limit = $_GET['latest_news'];  
    $query = "SELECT * FROM tbl_news_category c, tbl_news n WHERE c.cid = n.cat_id ORDER BY n.nid DESC LIMIT $limit"; 
    $result = mysqli_query($con, $query); 
} 

$total_records = mysqli_num_rows($result); 
if ($total_records >= 1) { 
    while ($link[] = mysqli_fetch_array($result, MYSQLI_ASSOC)) { 
     $set['RecentNews'] = $link; 
    } 
}  

echo $val = str_replace('\\/', '/', json_encode($set,JSON_UNESCAPED_UNICODE)); 

//} 

?> 

然后,我有一个 'Android JSON类' 代码段:

try { 
    JSONObject mainJson = new JSONObject(result); 
    JSONArray jsonArray = mainJson.getJSONArray(Constant.CATEGORY_ARRAY_NAME); 
    JSONObject objJson = null; 
    for (int i = 0; i < jsonArray.length(); i++) { 
     objJson = jsonArray.getJSONObject(i); 
     ItemLatest objItem = new ItemLatest(); 
     objItem.setCId(objJson.getString(Constants.CATEGORY_ITEM_CID)); 
     objItem.setCategoryName(objJson.getString(Constants.CATEGORY_ITEM_NAME)); 
     //objItem.setCategoryImage(objJson.getString(Constants.CATEGORY_ITEM_IMAGE)); 
     objItem.setCatId(objJson.getString(Constants.CATEGORY_ITEM_CAT_ID)); 
     objItem.setNewsImage(objJson.getString(Constants.CATEGORY_ITEM_NEWSIMAGE)); 
     objItem.setNewsHeading(objJson.getString(Constants.CATEGORY_ITEM_NEWSHEADING)); 
     objItem.setNewsDescription(objJson.getString(Constants.CATEGORY_ITEM_NEWSDESCRI)); 
     objItem.setNewsDate(objJson.getString(Constants.CATEGORY_ITEM_NEWSDATE)); 
     arrayOfLatestnews.add(objItem); 
    } 
} 
catch (JSONException exc) { 
    exc.printStackTrace(); 
} 

Android的常量该类的代码片断:

public class Constants implements Serializable { 

    public static final String SERVER_URL = "http://192.100.1.1:4000/news"; 

    public static final String LATEST_URL = "http://192.100.1.1:4000/news/news.php?latest_news=10"; 

    public static final String CATEGORY_ARRAY_NAME = "RecentNews"; 
    public static final String CATEGORY_NAME = "category_name"; 
    public static final String CATEGORY_CID = "cid"; 
    public static final String CATEGORY_IMAGE = "category_image"; 
    ... 
} 

编辑:

JSON输出:

{"RecentNews":[{"cid":"7","category_name":"World","category_image":"91771_world.jpg","status":"1","nid":"9","cat_id":"7","news_heading":"World Sample News Heading","news_description":" 
World Sample News 

\r\n","news_image":"88702_IMG_5038.JPG","news_date":"08-16-2017","news_status":"1"},{"cid":"12","category_name":"IT","category_image":"98162_IMG_1303.JPG","status":"1","nid":"8","cat_id":"12","news_heading":"IT Sample News Heading","news_description":" 
IT Sample News 

\r\n","news_image":"24966_IMG_5018.JPG","news_date":"08-03-2017","news_status":"1"},{"cid":"5","category_name":"Business","category_image":"4591_download.jpg","status":"1","nid":"7","cat_id":"5","news_heading":"Business Sample News Heading","news_description":" 
Business Sample News 

\r\n","news_image":"13015_IMG_5017.JPG","news_date":"08-02-2017","news_status":"1"},{"cid":"4","category_name":"Sports","category_image":"22814_sports.jpg","status":"1","nid":"6","cat_id":"4","news_heading":"Sports Sample News Heading","news_description":" 
Sports Sample News 

\r\n","news_image":"72021_IMG_5016.JPG","news_date":"08-01-2017","news_status":"1"}]} 
+0

你能显示http://192.100.1.1:4000/news/news.php?latest_news=10 –

+0

的输出我已经更新了这个问题并添加了JSON输出数据 –

+1

你确定你发布的json输出是与应用程序收到的相同吗?你得到的错误消息说结果字符串以“<! - ?php”开头,它看起来像破损的php openning标签。也许尝试在Android代码中记录结果变量的内容(Log.d(“mytag”,result))。 –

回答

0

由于josef.adamcik,我用Log.d(“标签”,结果)来获取日志的异常,并强调它确实读PHP文件,以“<!--”开头。

的“db_connection.php”在news.php文件“包含”文件已经在它开始<!--comments -->和在去除这部分的一切运作良好的部分。

编辑:因为我用括号IDE

我在PHP中的多线使用<!--comments-->评论,而不是/* comments */

,反正变灰<!--comments-->部分,我忽略了,这是错误的。

+1

但是..在php中的评论是用'//'制作的。 –

+0

谢谢@DebanikDawn,我已经编辑了上面的答案,我确定它现在应该更清楚了 –

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