您好，我想在我的安卓应用程序中访问我的网络记录。通过这样做，我使用JSON和PHP来做到这一点。这是我的json文件的网址：这里
但问题是它只是显示 php 代码。我需要能够访问/读取该文件。:(知道我在这里做什么吗？我非常感谢帮助。谢谢。
这是我到目前为止尝试过的：
<?php
include('connectdb.php');
$sql = "SELECT salesordercard_code, location_from, location_to, salesmancode FROM salesorderingcard";
$result = mysql_query($sql);
if($result === FALSE) {
die(mysql_error()); // TODO: better error handling
}
$set = array();
while($row1 = mysql_fetch_assoc($result)) {
$set[] = $row1;
}
header('Content-type: application/json');
echo json_encode($set);
?>
主要活动.class
@Override
protected Void doInBackground(Void... params) {
// Create the array
arraylist = new ArrayList<HashMap<String, String>>();
// Retrive JSON Objects from the given website URL in JSONfunctions.class
jsonobject = JSONFunctions.getJSONfromURL("http://www.shoppersgroup.net/vanmanagement/results.php");
try {
// Locate the array name
jsonarray = jsonobject.getJSONArray("posts");
for (int i = 0; i < jsonarray.length(); i++) {
HashMap<String, String> map = new HashMap<String, String>();
jsonobject = jsonarray.getJSONObject(i);
//Log.i(MainActivity.class.getName(), jsonobject.getString("movie_name"));
// Retrive JSON Objects
map.put(TAG_CODE, jsonobject.getString("salesordercard_code"));
map.put(TAG_LOCATION_FROM, jsonobject.getString("location_from"));
map.put(TAG_LOCATION_TO, jsonobject.getString("location_to"));
// Set the JSON Objects into the array
arraylist.add(map);
}
} catch (JSONException e) {
Log.e("Error", e.getMessage());
e.printStackTrace();
}
return null;
}
日志猫：
11-18 02:35:08.521: E/log_tag(1047): Error parsing data org.json.JSONException: Value [{"salesmancode":"SLMAN001","location_to":"MAIN","location_from":"IN-TRANSIT","salesordercard_code":"SLESO0001"},{"salesmancode":"SLMAN001","location_to":"MAIN","location_from":"IN-TRANSIT","salesordercard_code":"SLESO0002"}] of type org.json.JSONArray cannot be converted to JSONObject
首先作为 json 字符串
[ // Its an Array
"posts",// Its not an Array and its index is 0 so we will not use index 0
{//Index1
"salesordercard_code": "SLESO0001",
"location_from": "IN-TRANSIT",
"location_to": "MAIN",
"salesmancode": "SLMAN001"
},
{//index2
"salesordercard_code": "SLESO0002",
"location_from": "IN-TRANSIT",
"location_to": "MAIN",
"salesmancode": "SLMAN001"
}
]
更改您的 doInBackground（） 方法，如下所示，
@Override
protected Void doInBackground(Void... params) {
// Create the array
arraylist = new ArrayList<HashMap<String, String>>();
// Retrive JSON Objects from the given website URL in JSONfunctions.class
jsonobject = JSONFunctions.getJSONfromURL("http://www.shoppersgroup.net/vanmanagement/results.php");
try {
// Locate the array name
jsonarray = new JSONArray(jsonobject.toString());
for (int i = 1; i < jsonarray.length(); i++) {// strat from index1
HashMap<String, String> map = new HashMap<String, String>();
JSONObject jsonobj = jsonarray.getJSONObject(i);
//Log.i(MainActivity.class.getName(), jsonobj.getString("movie_name"));
// Retrive JSON Objects
map.put(TAG_CODE, jsonobj.getString("salesordercard_code"));
map.put(TAG_LOCATION_FROM, jsonobj.getString("location_from"));
map.put(TAG_LOCATION_TO, jsonobj.getString("location_to"));
// Set the JSON Objects into the array
arraylist.add(map);
}
} catch (JSONException e) {
Log.e("Error", e.getMessage());
e.printStackTrace();
}
return null;
}
希望这能解决您的问题。
您的results.json文件无法运行 php，因为它没有被处理为 PHP，而是作为 JSON 文件处理，因此它将其中的文本显示为文本。
将其重命名为 results.php 并尝试，然后，它应该打印出带有所需结果的 json 编码数据。