Question

我有一个.php用于在表格中创建记录。我认为这是正确的，但我无法将参数从URL传递给它，有些事情是错误的。

<?php

/*
 * Following code will create a new product row
 * All product details are read from HTTP Post Request
 */

// array for JSON response
$response = array();

// check for required fields
if (isset($_POST['name']) && isset($_POST['price']) && isset($_POST['description'])) {

    $name = $_POST['name'];
    $price = $_POST['price'];
    $description = $_POST['description'];

    // include db connect class
    require_once __DIR__ . '/db_connect.php';

    // connecting to db
    $db = new DB_CONNECT();

    // mysql inserting a new row
    $result = mysql_query("INSERT INTO products(name, price, description) VALUES('$name', '$price', '$description')");

    // check if row inserted or not
    if ($result) {
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "Product successfully created.";

        // echoing JSON response
        echo json_encode($response);
    } else {
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        // echoing JSON response
        echo json_encode($response);
    }
} else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);
}
?>

我在网址中传递params：

localhost:81/android/create_product.php?name='test'&price='35000'&description='test'

是php代码错了还是别的什么？

Answer 1

将$_GET用于网址参数

if (isset($_GET['name']) && isset($_GET['price']) && isset($_GET['description']))

Answer 2

如果你在url中传递参数，那么你需要通过

来获取它

$_GET['name']

方法不是通过$ _POST ['name']喜欢

if (isset($_GET['name']) && isset($_GET['price']) && isset($_GET['description']))

在php上传递带URL的多个参数

2 个答案: