我面临 json 解析器错误,我不知道是什么原因导致它,因为有时它工作正常,有时错误只是弹出并且我的数据没有更新到数据库中。数据将被发布到 php 文件以加密并更新到数据库。有时它可以将数据插入数据库,当它不能时,它会显示 json 解析错误。
这是我的错误:
08-31 01:10:55.598: E/JSON Parser(360): Error parsing data org.json.JSONException: End of input at character 0 of
这是我的代码:
String email = db.getEmail();
String currentPassword = cpCurrentPassword.getText().toString();
String newPassword = cpNewPassword.getText().toString();
// Building Parameters
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("email", email));
params.add(new BasicNameValuePair(
"currentpassword", currentPassword));
params.add(new BasicNameValuePair("newpassword",
newPassword));
JSONObject json = jsonParser
.makeHttpRequest(
newpassword_url, "POST",
params);
try {
int success1 = json1
.getInt(TAG_SUCCESS);
if (success1 == 1) {
showToast("Password successfully changed!");
这是我的PHP:
<?php
// array for JSON response
$response = array();
// check for required fields
if (isset($_POST['email']) && isset($_POST['newpassword'])){
$email = $_POST['email'];
$newpassword = $_POST['newpassword'];
}
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
$result = mysql_query("UPDATE users SET encrypted_password = '$newpassword' WHERE email = '$email'");
// check if row inserted or not
if ($result) {
// successfully updated
$response["success"] = 1;
$response["message"] = "Password successfully changed";
// echoing JSON response
echo json_encode($response);
}
?>
即使数据没有更新到数据库,我的成功标签仍然会响应 1。这是我不明白的地方。