【发布时间】:2017-04-24 06:31:04
【问题描述】:
我一直在努力尝试成功地将数据插入 MySQL 数据库(使用 Volley),这不是问题,因为数据已插入,但我一直遇到此错误 W/System.err: org.json.JSONException: Value
or an JSONEception, 这个错误停止做任何其他事情,如果我添加更多代码 UI 冻结,我试图从我的 php 文件中删除日期代码,一切正常,只是日期时间值不是我要查找的值,当我添加回来时,它再次向我显示错误。
这是我的代码:
Response.Listener<String> responseListener1 = new Response.Listener<String>() {
@Override
public void onResponse(String response) {
try {
JSONObject jsonResponse1 = new JSONObject(response);
boolean success = jsonResponse1.getBoolean("success");
if (success) {
Toast.makeText(MapActivity.this, "SUCCESS",Toast.LENGTH_LONG).show();
} else {
Toast.makeText(MapActivity.this, "INSERTION FAILED",Toast.LENGTH_LONG).show();
}
} catch (JSONException e) {
e.printStackTrace();
Toast.makeText(MapActivity.this, "EXCEPTION",Toast.LENGTH_LONG).show();
}
}
};
SendBookingRequest bookingRequest = new SendBookingRequest(idd,em,Adresse_source,duree,dist, responseListener1);
RequestQueue queue1 = Volley.newRequestQueue(MapActivity.this);
queue1.add(bookingRequest);
这是我的 php 文件 SendBookingRequest
<?php
require("password.php");
$connect = mysqli_connect("localhost", "XXXXX", "XXXXX", "XXXXX");
$driver_id = $_POST["driver_id"];
$email = $_POST["email"];
$adresse_source = $_POST["adresse_source"];
$duree = $_POST["duree"];
$distance = $_POST["distance"];
$response = array();
$dt_obj = new DateTime($response['send_moment'], new
DateTimeZone('America/Chicago'));
$dt_obj->setTimezone(new DateTimeZone('Europe/London'));
$send_time = $dt_obj->format('Y-m-d H:i:s');
echo $send_time;
function AddRequest() {
global $connect, $driver_id, $email, $adresse_source, $duree, $distance, $send_time ;
$statement = mysqli_prepare($connect, "INSERT INTO demande (driver_id, pass_id, adresse_source, duree, distance, send_moment) VALUES (?, (SELECT user_id FROM passager WHERE email = ?),?, ?, ?, ?)");
mysqli_stmt_bind_param($statement, "issdis", $driver_id, $email, $adresse_source, $duree, $distance, $send_time);
mysqli_stmt_execute($statement);
mysqli_stmt_close($statement);
}
$response["success"] = false;
AddRequest();
$response["success"] = true;
echo json_encode($response);
?>
任何帮助将不胜感激。
【问题讨论】:
-
展示数据样本的尴尬想法如何?
-
对不起,你能解释一下你的意思吗?
-
显示您尝试解析的导致您面临的问题的字符串
-
怎么弄!!,对不起,我是新手
-
通过记录?设置断点并使用调试器运行?
标签: php android mysql datetime android-volley