如何自动将Users表的usersId插入Parents表的user_id外键字段？

我有Users和Parents两个独立数据表，表单数据能正常存入phpMyAdmin数据库，用过程式PHP和预处理语句处理表单，两个表单都能成功插入数据，但Users表的usersId作为外键插入Parents表的user_id字段时，数据库中始终显示NULL，只能通过phpMyAdmin的下拉列表手动更新该字段。

我需要实现：提交表单时，Users表的usersId自动填入Parents表的user_id字段。以下是我当前使用的预处理语句代码：

<?php

if (isset($_POST["submit"])) {
$mFname = $_POST["mFname"];
$mLname = $_POST["mLname"];
$mEmail = $_POST["mEmail"];
$mPhone = $_POST["mPhone"];

$fFname = $_POST["fFname"];
$fLname = $_POST["fLname"];
$fEmail = $_POST["fEmail"];
$fPhone = $_POST["fPhone"];

$addressL1 = $_POST["addressL1"];
$addressL2 = $_POST["addressL2"];
$city = $_POST["city"];
$stateAbbr = $_POST["stateAbbr"];
$zip = $_POST["zip"];
$user_Id = $_POST['user_Id'];


include_once 'dbh.inc.php';
include_once 'functions.inc.php';
 
    $sql1 = "UPDATE parent INNER JOIN users ON users.usersId = parent.user_id SET parent.user_id = users.usersId";

    if (mysqli_query($conn, $sql1)) {
    echo "Record updated successfully";
    } else {
    echo "Error updating record: " . mysqli_error($conn);
    }


    $sql = "INSERT INTO parent (mFname, mLname, mEmail, mPhone, fFname, fLname, fEmail, fPhone, addressL1, addressL2, city, stateAbbr, zip, user_id) VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?,?);";

    $stmt = mysqli_stmt_init($conn);
    if (!mysqli_stmt_prepare($stmt, $sql)) {
        echo "SQL error";
    } else {
        mysqli_stmt_bind_param($stmt, "ssssssssssssis", $mFname, $mLname, $mEmail, $mPhone, $fFname, $fLname, $fEmail, $fPhone, $addressL1, $addressL2, $city, $stateAbbr, $zip, $user_id);
        mysqli_stmt_execute($stmt);
 
    }
    header("Location: ../mf2.php?parentinfo=success");
}

截图显示其他数据均正常插入，但user_id字段为NULL。我曾尝试用内连接更新语句，提示更新成功但数据库无变化；也曾尝试使用usersId会话变量，均未成功。作为新手，恳请详细的解决方案。

内容的提问来源于stack exchange，提问作者Joshua Lee