php - 在单个属性中插入多个值

Question

我在一个数据库中有两个表，

  tbl_language:                           tbl_users
  lang_id    lang_name               user_id          user_name        user_lang
  1          English                                 
  2          Spanish                                 
  3          German
  4          Portuguese

I have PHP code to select languages from tbl_lang, Now I have a multiple select in a form where users can choose more than one languages. While inserting the values in database, I want to store multiple language values in one column/attribute. Where user_lang attribute would only store lang_ids of the tbl_language separated by comma(,).

Like

  tbl_users:
  user_id          user_name        user_lang
  1                 ABC              1,2
  2                 XYZ              1,4

How to accomplish this using PHP and MySQL ?

I have sample code here:

<?php

$user_name = $_POST['user_name'];
$user_lang = $_POST['user_lang']; // user_lang as array from multiple select
$qry = ""; // I need help with this query

Answer 1

根据您的设置，考虑到 user_id 是一个自动递增的 PK:

$user_name = mysql_real_escape_string($_POST['user_name']);
$user_lang = mysql_real_escape_string(implode(',', $_POST['user_lang']);
$sql = "INSERT INTO tbl_users (`user_name`, `user_lang`) VALUES ('{$user_name}', '{$user_lang}');

但是，正如其他人所提到的，您可能应该寻找第 3^rd 级表:

tbl_users_lang
user_id    lang_id

并插入like(例如，没有错误处理，只是一个建议)

$user_name = mysql_real_escape_string($_POST['user_name']);
$sql = "INSERT INTO tbl_users (`user_name`) VALUES ('{$user_name}');"
mysql_query($sql);

$user_id = mysql_insert_id();
foreach ($user_lang as $lang) {
   $lang = mysql_real_escape_string($lang);
   $sql = "INSERT INTO tbl_users_lang (`user_id`, `lang_id`) VALUES ({$user_id}, '{$lang}');"
   mysql_query($sql);       
}

** 编辑 **

作为奖励，使用此查询，您可以获取任何给定用户的语言列表的逗号分隔行。我想你可能会感兴趣:

SELECT DISTINCT u.user_id, u.user_name, GROUP_CONCAT(l.lang_name) 
  FROM `tbl_users` as u
  LEFT JOIN `tbl_users_lang` as ul ON u.user_id = ul.user_id 
  LEFT JOIN `tbl_language` as l on ul.lang_id = l.lang_id
  -- add WHERE clause here to filter (ex : WHERE u.user_name LIKE 'John%')