Android PostResponseAsyncTask无法从android获取字符串

Question

我有这个应用程序，我在Android工作室使用PHP和Mysql查询但是当我使用更新它不工作时，php应该给我$result给我的字符串，但它没有给它任何空白。

这是我的代码。

的Android

HashMap postData = new HashMap();

    if (TextUtils.isEmpty(etCarModel.getText().toString())) {
        Toast.makeText(UpdateClick.this, "Car model is empty", Toast.LENGTH_SHORT).show();
        return;
    }
    if (TextUtils.isEmpty(etCarType.getText().toString())) {
        Toast.makeText(UpdateClick.this, "Car type is empty", Toast.LENGTH_SHORT).show();
        return;
    }
    if (TextUtils.isEmpty(etCapacity.getText().toString())) {
        Toast.makeText(UpdateClick.this, "Capacity is empty", Toast.LENGTH_SHORT).show();
        return;
    }
    if (TextUtils.isEmpty(etPlateNumber.getText().toString())) {
        Toast.makeText(UpdateClick.this, "Plate Number is empty", Toast.LENGTH_SHORT).show();
        return;
    }
    if (TextUtils.isEmpty(etPrice.getText().toString())) {
        Toast.makeText(UpdateClick.this, "Price is empty", Toast.LENGTH_SHORT).show();
        return;
    }
    postData.put("txtCar_No", tvCar_No.getText().toString());
    postData.put("txtCarModel", etCarModel.getText().toString());
    postData.put("txtCarType", etCarType.getText().toString());
    postData.put("txtCapacity", etCapacity.getText().toString());
    postData.put("txtPlateNumber", etPlateNumber.getText().toString());
    postData.put("txtCarPrice", etPrice.getText().toString());
    postData.put("image", toString());
    postData.put("txtFuelType", spFuelType.getSelectedItem().toString());

    PostResponseAsyncTask taskUpdate = new PostResponseAsyncTask(UpdateClick.this, postData, new AsyncResponse() {
        @Override
        public void processFinish(String q) {
            Log.d(TAG,q);
            if(q.contains("success")){
                Toast.makeText(UpdateClick.this, "Car details updated!", Toast.LENGTH_SHORT).show();
                Intent in = new Intent(UpdateClick.this, OwnerTabs.class);
                startActivity(in);
                finish();
            }else{
                Toast.makeText(getApplicationContext(), "Error ? " + q.toString(), Toast.LENGTH_SHORT).show();
            }
        }
    });
    taskUpdate.execute("http://carkila.esy.es/update.php");
}

PHP

    <?php  
include_once("connection.php");

if(isset($_POST['txtCar_No']) &&  isset($_POST['txtCarModel']) && 
isset($_POST['txtCarType']) && isset($_POST['txtCapacity']) && 
isset($_POST['image']) && isset($_POST['txtFuelType']) && 
isset($_POST['txtPlateNumber']) &&  isset($_POST['txtcarPrice']))
{
$now = DateTime::createFromFormat('U.u', microtime(true));
$id = $now->format('YmdHis');

$upload_folder = "upload";
$path = "$upload_folder/$id.jpeg";
$fullpath = "http://carkila.esy.es/$path";

$image = $_POST['image'];
$Car_No = $_POST['txtCar_No'];
$Car_Model = $_POST['txtCarModel'];
$Car_Type = $_POST['txtCarType'];
$Capacity = $_POST['txtCapacity'];
$Fuel_Type = $_POST['txtFuelType'];
$PlateNumber = $_POST['txtPlateNumber'];
$carPrice = $_POST['carPrice'];

$query = "UPDATE tbl_cars SET Car_Model='$Car_Model', Car_Type='$Car_Type', Capacity='$Capacity', Image='$fullpath', fuelType='$Fuel_Type',carPlatenuNumber='$PlateNumber' ,carPrice= '$carPrice' WHERE 'Car_No'=$Car_No";

$result = mysqli_query($conn,$query);
$count = mysqli_affected_rows($conn);

if($result == TRUE && $count > 0){
echo "success";
exit();
}else{
print_r (mysqli_error($conn));
echo "failed";
exit();
}
}
?>

Answer 1

我来自移动设备，但尝试更改 $ result ：

$result = mysqli_query($conn,$query);

有了这个：

$result = mysqli_query($conn,$query) or die(mysqli_error($conn));

修改

您必须检查信息是否已发送，因此请尝试将其插入您的代码：

$filename   =   "log.txt";
$fh         =   fopen($filename, "a") or die("Could not open log file.");
$post       =   var_dump($post);
fwrite($fh, date("d-m-Y, H:i")." - $post\n") or die("Could not write file!");
fclose($fh);

不要忘记使用chmod 777

创建log.txt文件