php - Android PHP 查询不起作用

Question

我的php代码可以工作，它说结果是成功的，在我的android中也是如此，但它没有执行我的查询来更新一行，它只是说成功但没有更新。 我非常确定 CarStatus 和 Car_No 在我的数据库中。

PHP 查询

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

if(isset($_POST['txtCarNo']) && isset($_POST['txtCarStatus']))
{
    $carNo = $_POST['txtCarNo'];
    $carStatus = $_POST['txtCarStatus'];

    $query = "UPDATE tbl_cars SET carStatus = '$carStatus' WHERE Car_No = '$carNo'";
    $result = mysqli_query($conn,$query);
    if($result > 0){
        echo "success";
        exit;
    }else{
        echo "failed";
        exit;
    }
}
?>

移动

HashMap post = new HashMap();

    post.put("txtCarNo",tvCarID.toString());
    post.put("txtCarStatus","for approval");

    PostResponseAsyncTask task1 = new PostResponseAsyncTask(DetailActivity.this, post, new AsyncResponse() {
        @Override
        public void processFinish(String s) {
            if (s.contains("success")) {
                Log.d(TAG, s);
                Toast.makeText(DetailActivity.this, "Wait for owners approval", Toast.LENGTH_SHORT).show();
                Intent in = new Intent(DetailActivity.this, RenterTabs.class);
                startActivity(in);
                finish();
            } else {
                Toast.makeText(getApplicationContext(), "Error", Toast.LENGTH_SHORT).show();
            }
        }
    });
    task1.execute("http://carkila.esy.es/carkila/rent2.php");
}

EDIT

连接.php

<?php
$servername = "***";
$username = "***";
$password = "***"; 
$dbname = "***";

$conn = mysqli_connect($servername, $username, $password, $dbname);

if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
?>

Answer 1

在 post.put("txtCarNo",tvCarID.toString()); 上，tvCarID 需要 .getText 所以改为 postData.put("txtCarNo",tvCarID.getText().toString());