[英]Sending JSON array using PHP to Android Volley
I am sending an array of data from MySQL query, and I am using json_encode to do that:我正在从 MySQL 查询发送一组数据,我正在使用 json_encode 来做到这一点:
public function getFriends($id){
$query = mysql_query(" SELECT uid, name, email
FROM users
RIGHT JOIN friendships
ON friendid2 = uid") or die (mysql_error());
$jsonData = '{"tag":"friends",error":"false",';
$jsonData .= '"friends":[';
while($row = mysql_fetch_array($query)){
$i++;
$id = $row["uid"];
$name = $row["name"];
$email = $row["email"];
$jsonData .= '{"id":"'.$id.'","name":"'.$name.'","email":"'.$email.'"},';
}
$jsonData = chop($jsonData, ",");
$jsonData .= ']}';
echo json_encode($jsonData);
In the Android LogCat, when I'm creating JSON object I see this:在 Android LogCat 中,当我创建 JSON 对象时,我看到:
org.json.JSONException: Value {"tag":"friends",error":"false","friends":[{"id":"4","name":"GrubyJohny2","email":"gruby@gmail.com"},{"id":"243000000","name":"Wariacik","email":"karol@wp.com"}]} of type java.lang.String cannot be converted to JSONObject
Can anyone tell me what am I doing wrong ?谁能告诉我我做错了什么? Becouse I searched bounch of toutorials and I think my json syntax is correct.因为我搜索了一堆教程,我认为我的 json 语法是正确的。
The way I am receiving message from server:我从服务器接收消息的方式:
private void getFriendships(final String id) {
String tag_string_req = "req_friendships";
pDialog.setMessage("Sending Request for list of friends");
showDialog();
final String TAG = "List of friends request";
StringRequest strReq = new StringRequest(Request.Method.POST, AppConfig.URL_REGISTER, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
Log.d(TAG, "Friendship request Response: " + response.toString());
hideDialog();
try {
JSONObject jObj = new JSONObject(response);
boolean error = jObj.getBoolean("error");
if (!error) {
Toast.makeText(getApplicationContext(), "Friends uploaded", Toast.LENGTH_LONG).show();
} else {
String errorMsg = jObj.getString("error_msg");
Toast.makeText(getApplicationContext(), errorMsg, Toast.LENGTH_LONG).show();
}
} catch (JSONException e) {
Toast.makeText(getApplicationContext(), e.getMessage(), Toast.LENGTH_LONG).show();
e.printStackTrace();
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.e(TAG, "List of friends request Error: " + error.toString());
Toast.makeText(getApplicationContext(), error.getMessage(), Toast.LENGTH_LONG).show();
hideDialog();
}
}) {
@Override
protected Map<String, String> getParams() {
Map<String, String> params = new HashMap<String, String>();
params.put("tag", "friends");
params.put("sender", id);
return params;
}
};
AppController.getInstance().addToRequestQueue(strReq, tag_string_req);
}
"friends",error":"false"
The error
key needs quotes on both sides. error
键需要两边引号。
You can copy / paste the JSON in many free services, such as jsonlint.org that will parse your string for you and tell you if and where the errors are.您可以在许多免费服务中复制/粘贴 JSON,例如jsonlint.org ,它会为您解析您的字符串并告诉您是否存在错误以及错误在哪里。
As @Noman pointed out in the comments, you can also just build out the entirety of the data as a PHP array (and not just portions of it), and then simply json_encode
all of it at the very end:正如@Noman 在评论中指出的那样,您也可以将整个数据构建为 PHP 数组(而不仅仅是其中的一部分),然后在最后简单地对所有数据进行json_encode
:
$jsonData = array(
'tag' => 'friends',
'error' => 'false',
'friends' => array(),
);
while($row = mysql_fetch_array($query)) {
$jsonData['friends'][] = array(
'id' => $row['uid'],
'name' => $row['name'],
'email' => $row['email'],
);
}
echo json_encode($jsonData);
Also, it should be noted that mysql_query
as well as all mysql_*
functions are now deprecated .另外,应该注意的是mysql_query
以及所有mysql_*
函数现在都已弃用。 You should not write new code that uses these functions.您不应编写使用这些函数的新代码。 Instead you should learn to use either PDO or mysqli extensions for communicating with your database.相反,您应该学习使用PDO或mysqli扩展与您的数据库进行通信。
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