Question

<?php               
include 'connection.php';
session_start();
$noteinfo = array();
$noteinfo['note'] = $_POST['note'];
$_SESSION['noteinfo'] = $noteinfo;

if (isset($_POST['submit'])) {
    if (empty($_POST['note'])) {
        echo "Dobavete Komentar";
    }
    if (!empty($_SESSION['noteinfo'])) {
        $check = mysqli_escape_string($conn,$_SESSION['userinfo']['fname']);
        print_r($_SESSION['userinfo']);     
        $sql = "INSERT INTO users (user_fname,user_mname,user_lname,user_login,user_email,user_phone) 
                VALUES ('{$_SESSION['userinfo']['fname']}', '{$_SESSION['userinfo']['mname']}', '{$_SESSION['userinfo']['lname']}', '{$_SESSION['userinfo']['login']}', '{$_SESSION['userinfo']['email']}', '{$_SESSION['userinfo']['phone']}')";
        $sql1 = "INSERT INTO addresses (address_line_1,address_line_2,address_zip,address_city,address_province,address_country) 
                VALUES ('{$_SESSION['addressinfo']['adr1']}', '{$_SESSION['addressinfo']['adr2']}', '{$_SESSION['addressinfo']['zip']}', '{$_SESSION['addressinfo']['city']}', '{$_SESSION['addressinfo']['provinciq']}', '{$_SESSION['addressinfo']['durjava']}')"; 
        $sql2 = "INSERT INTO notes (note_text) 
                VALUES ('{$_SESSION['noteinfo']['note']}')";
        $sql3 = "INSERT INTO users_addresses (ua_user_id,ua_address_id) 
                SELECT users.user_id,addresses.address_id 
                FROM ( VALUES () )";

        if (mysqli_query($conn,$sql)) {
            echo "Added";
            if (mysqli_query($conn,$sql1)) {
                echo "Added";
                if (mysqli_query($conn,$sql2)) {
                    echo "Added";
                    if (mysqli_query($conn,$sql3)) {
                        echo "Mai stana toq put";
                        header("refresh:3 ; url=profile.php");
                    }
                }
            } 
        }
        else{
            echo "Error";
        }
    }
    else{
        header("refresh:1 ; url=zapiski.php");
    }
}
?>

我的问题是针对$ sql3：我想在表单填写完毕后用用户ID和地址表中的ID填充users_addresses，但是现在SQL并没有在表users_addresses中填充任何内容。这些是我的表格：

Answer 1

当用户在应用程序中填写值并且您需要为它们查找id时，编写查询的典型方法如下所示：

INSERT INTO users_addresses (ua_user_id, ua_address_id)
    VALUES ( (SELECT u.user_id FROM users u WHERE u.username = ?),
             (SELECT a.address_id FROM address a WHERE a.address = ?)
           );

也就是说，您要将参数传递到查询中，使用参数查找ID，然后使用这些ID插入联结表。

Answer 2

您似乎在VALUES()中缺少INSERT函数。

INSERT INTO users_addresses(ua_user_id,ua_address_id)

应该是

INSERT INTO users_addresses(ua_user_id,ua_address_id) VALUES(some_variable,some_variable)

您也可以这样做

INSERT INTO users_addresses SET ua_user_id=some_variable, ua_address_id=some_variable

在选择数据时，您需要JOIN表。我建议在此问题上使用LEFT JOIN。重要的是，您要在JOIN中相互匹配的id是一致的，以便能够将表链接在一起。然后，如果要解析user_id（即WHERE子句），则由您决定是否要满足特定条件。

选择特定匹配项的示例：

SELECT
  users_addresses.ua_id,
  addresses.address_id
FROM
  users_addresses
LEFT JOIN
  users ON users_addresses.ua_user_id = users.user_id
LEFT JOIN
  addresses ON users_addresses.ua_address_id = addresses.address_id
WHERE
  users_addresses.ua_id = $some_id

选择所有匹配项的示例：

SELECT
  users_addresses.ua_id,
  addresses.address_id
FROM
  users_addresses
LEFT JOIN
  users ON users_addresses.ua_user_id = users.user_id
LEFT JOIN
  addresses ON users_addresses.ua_address_id = addresses.address_id

让我知道这是否是您想要的，以及是否需要详细说明。

一个重要的注意事项，因为您正在使用mysqli_ *扩展名，所以应该研究准备好的语句。由于您现在可以使用SQL注入，因此它将大大提高安全性。

用LEFT JOIN插入

2 个答案: