php - 提交后添加表单数据并从另一个表获取数据形成新记录 - Php MySQL

Question

表 1 中的 5 个字段来自 Php 表单。从表2、表3、表4、表5中提取出5个字段。

这可以通过 Php 实现吗？

尝试了多个 INSERT SELECT 和 VIEWS；然而，与一些新手相比。

<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("site", "user", "password", "database");

// Check connection
if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

$new_record_id = trim($_GET["id"]); 

// Attempt insert query execution
$sql = "INSERT INTO persons (first_name, last_name, email, city, state) VALUES ('Peter', 'Parker', 'peterparker@mail.com', 'williamsburg', 'new york')";
if(mysqli_query($link, $sql)){
    echo "Records inserted successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// Attempt insert query1 execution
$sql1 = "INSERT INTO persons (SELECT cust_type, cust_revenues, cust_since FROM customer) VALUES ('Existing', '3029', '2016') WHERE id = $new_record_id";
if(mysqli_query($link, $sql)){
    echo "Records inserted successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// Attempt insert query2 execution
$sql2 = "INSERT INTO persons (SELECT order_no, order_date FROM orders) VALUES ('293048', '11/26/2016') WHERE id = $new_record_id";
if(mysqli_query($link, $sql)){
    echo "Records inserted successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// Close connection
mysqli_close($link);
?>

提交表单时，应使用所有 10 个字段中的数据创建一条新记录。目前，还没有发生。

Answer 1

我想我会用 PDO 强化的 PHP 语句和 MySQLi 替换代码。由于没有给出错误代码(它返回一个干净的页面)，我认为结果集太大。我将查询替换为仅从 Form 执行 INSERT。其他表将聚合到 View 中。