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"}]}
你能显示http://192.100.1.1:4000/news/news.php?latest_news=10 –
的输出我已经更新了这个问题并添加了JSON输出数据 –
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))。 –