我一直在努力尝试将数据成功插入MySQL数据库(使用Volley),好吧,这不是问题因为数据是插入的,但是我一直有这个错误W/System.err:org.json.JSONException:值< br的java.lang.String类型不能转换为JSONObject或JSONEception,这个错误是停止做的其他任何东西,如果我添加更多的代码的UI冻结,我试图从我的PHP文件中删除日期代码,一切都很好,只有日期时间值不是我正在寻找的,当我加回来,它显示我再次出现错误。 这是我的代码:由于DATETIME数据,无法将java.lang.String类型的值转换为JSONObject
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);
这里是我的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);
?>
任何帮助将非常感激。
和怎么样显示,数据的样本的尴尬想法? –
对不起,你能解释一下你的意思吗? –
显示你正试图解析的字符串,导致你正面临的问题 –