client.post("http://10.0.2.2/project/process/selectalluser.php", new AsyncHttpResponseHandler() {
@Override
public void onSuccess(String response) {
Integer contacts = controller.getContactsCount();
Log.d("Reading contacts: ", contacts+"");
System.out.println("onSuccess");
JSONArray jsonArray = new JSONArray(response);
Log.d("Reading response: ", response.length()+"");
System.out.println(response);
Toast.makeText(getApplicationContext(), "MySQL DB has been informed about Sync activity", Toast.LENGTH_LONG).show();
}
@Override
public void onFailure(int statusCode, Throwable error, String content) {
System.out.println("onFailure");
System.out.println(statusCode);
System.out.println(error);
System.out.println(content);
Toast.makeText(getApplicationContext(), "Error Occured", Toast.LENGTH_LONG).show();
}
});
然后我的PHP:字符串转换成JSON在Android的
$stmt = $dbh->prepare("SELECT * FROM `admin_tbl`");
if ($stmt->execute()) {
$basicinfo = array();
if ($stmt->rowCount() > 0) {
while ($selected_row = $stmt->fetch(PDO::FETCH_ASSOC)) {
$basicinfo[] = array(
'email' => $selected_row['email'],
'username' => $selected_row['username'],
'password' => $selected_row['password'],
'fname' => $selected_row['fname'],
'mname' => $selected_row['mname'],
'lname' => $selected_row['lname'],
'suffix' => $selected_row['suffix'],
'status' => $selected_row['status']);
}
echo json_encode($basicinfo, JSON_UNESCAPED_UNICODE);
//return $basicinfo;
} else {
echo json_encode(array(
'error' => false,
'message' => "No record found!"
));
exit;
}
}
这是我的代码,我想这是字符串类型,以JSON的,所以我可以算多少条记录那里的响应转换。但我得到的Unhandled Exception: org.json.JSONException
错误行JSONObject json = new JSONObject(response);
线System.out.println(response);
将导致如下:
I /的System.out:[{ “电子邮件”: “[email protected]”, “用户名”: “管理员”, “密码”: “管理”,其中 “fname”: “布拉德利”, “MNAME”: “Buenafe”, “L-NAME”: “Dalina”, “后缀”: “”, “状态”: “1” }]
如何字符串格式正确转换成JSON
我需要有一个特定的进口对于我不是能提到这一点,但我想这已经 –
'org.json.JSONArray'? –
我认为不会有像上面那样的错误 –