Question

我有2张桌子：

Category主键ID和列Name
Employee主键ID和列Category_id

注意：Category_id现在正确显示ID

我想为Name的输出显示ID而不是Employee。

尝试：

$categ = mysql_query("SELECT * FROM employee WHERE id = '" . $_GET['id'] . "'");
$rows = array();

while ($row = mysql_fetch_assoc($categ)) {
  $website_cat = $row;
}

Category表：

+----+----------------+
| ID | Name           |
+----+----------------+
| 23 | Manager        |
| 10 | Boss           |
| 14 | Worker         |
| 41 | Another        |
+----+----------------+

Employee表：

+----+----------------+
| ID | Category_id    |
+----+----------------+
|  1 | Manager        |
|  2 | Boss           |
|  3 | Worker         |
|  4 | Another        |
+----+----------------+

输出：

echo $website_cat['category_id'];

Answer 1

您正在寻找的SQL关键字是JOIN。您的查询可能是这样的：

SELECT * FROM employee INNER JOIN category ON employee.category_id = category.id WHERE id = ...

或者，更可读：

SELECT
  *
FROM
  employee
  INNER JOIN category
    ON employee.category_id = category.id
WHERE
  id = ...

（注意：我故意删除了WHERE子句的最后一位，因为我不习惯将SQL注入漏洞放在答案中。Please read this要正确学习一些正确的基础知识执行涉及用户输入的SQL查询。目前，您的代码对一种非常常见的攻击形式持开放态度。）

由于您的某些列共享相同的名称，您甚至可能希望更明确地请求它们：

SELECT
  employee.id AS employee_id,
  category.id AS category_id,
  category.name AS category_name
FROM
  employee
  INNER JOIN category
    ON employee.category_id = category.id
WHERE
  id = ...

然后在您的代码中，您可以访问这些字段：

employee_id, category_id, category_name

所以你可以输出你想要的值：

echo $website_cat['category_name'];

Answer 2

您需要join类别表

$categ = mysql_query("
SELECT employee.*, category.name as category_name FROM employee 
INNER JOIN category on category.id = employee.category_id
WHERE id = '" . $_GET['id'] . "'");

然后使用$website_cat['category_name']

输出

显示名称而不是来自不同表的ID

2 个答案: