Question

我的问题是这样

后端的

（1）我在MySQL中有数据库。它包含7个表。（2）在前端，我有一个HTML表单。（3）从用户输入的数据必须填充这7个表。（4）我使用PHP并使用mysqli将表单链接到数据库。

由于某种原因，该表单仅填充了第一个表，而其他6个没有填充数据。知道如何将表单一次连接到所有表吗？我不需要一个sql表，因为HTML表单有300多个输入字段。谢谢。

已编辑

这是代码

<?php
$host="localhost";
$user="root";
$pass="";
$db="ao db";

$conn = new mysqli ($host, $user, $pass, $db);

if ($conn->connect_error) { 
die("Connection failed: " . $conn->connect_error);
}

$firstname = $_POST["first_name"];
$lastname = $_POST["name"]; 
$female = $_POST["female"];

$stmt = $conn->prepare("INSERT INTO table_demo (Vorname, Nachname, female) VALUES (?, ?, ?)");
$stmt->bind_param("ssi", $firstname, $lastname, $female);

$stmt->execute();
$stmt->close();
echo "New records created in Demo";
$conn->close();

$link = new mysqli ($host, $user, $pass, $db);

if ($link->connect_error) { 
die("Connection failed: " . $link->connect_error);
}

$height = $_POST["height"];
$weight = $_POST["weight"];

$stmt=$link->prepare("INSERT INTO table_pre (height, weight) VALUE (?, ?)");
$stmt->bind_param("ii", $height, $weight);

$stmt->execute();

$stmt->close();
echo "New records created in Pre";
$link->close();

>

Answer 1

对于您来说，首先必须创建数据库连接对象。

`

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}`

用于插入表。有了帮助，您可以在所有七个表中插入详细信息。

$sql = "INSERT INTO MyGuests (firstname, lastname, email)VALUES ('John', 'Doe', 'john@example.com')";
使用此从表中获取详细信息。借助此功能，您可以从所有七个表中进行选择。

$sql = "SELECT id, firstname, lastname FROM MyGuests"; $result = $conn->query($sql);

Answer 2

我知道了！那里有2个问题：

问题一：表创建不正确，并且外键连接错误。我通过phpmyadmin菜单使用外键创建表时遇到问题。也许最好使用SQL数据库和表创建语法：

CREATE DATABASE IF NOT EXISTS database_name;

USE database_name;

CREATE TABLE table_1
(
    table_1_id int not null auto_increment primary key,
    variable_1 varchar(255) not null,
    variable_2 int(11),
    ...
    variable_to_be_used_as_foreign_key text,
    variable_n text,
) ENGINE=InnoDB;

CREATE TABLE table_2
(
    table_2_id int not null auto_increment primary key,
    variable_1 varchar(255) not null,
    variable_2 int(11),
    ...
    variable_n text,
    variable_to_be_used_as_foreign_key text,

    FOREIGN KEY fk_table_2(variable_to_be_used_as_foreign_key)
         REFERENCES table_1(variable_to_be_used_as_foreign_key)
              ON UPDATE CASCADE
              ON DELETE RESTRICT
) ENGINE=InnoDB;

请注意：必须在两个表中都添加外键变量（我知道这很明显，但仍然可以）。

问题＃2 ：php代码失败。我在准备好的语句中错过了第二个表的外键列中的输入，即：

$stmt=$link->prepare("INSERT INTO table_pre (height, weight) VALUE (?, ?)");
$stmt->bind_param("ii", $height, $weight);

代替：

$stmt=$link->prepare("INSERT INTO table_pre (height, weight, female) VALUE (?, ?, ?)");
$stmt->bind_param("iii", $height, $weight, $female);

PHP与MySQL表的连接问题

2 个答案: