Question

我在这里做错了什么。我已经遵循了很多例子，但似乎无法让这个工作。我有2张桌子

表=＆gt;用户

user_id
user_name
user_email
user_password
user_country
user_dobdate
user_company
user_code
user_status
user_type

表=＆gt;应用

apply_id
apply_from
apply_leave_type
apply_priority
apply_start_date
apply_end_date
apply_halfday
apply_contact
apply_reason
apply_status
apply_comment
apply_dated
apply_action_date

SQLI QUERY

$query = $db->select("SELECT users.user_id, app.apply_from FROM users INNER JOIN applications ON  users.user_id = app.apply_from WHERE users.user_code='1'");
$rows = $db->rows();
foreach ($rows as $apply){
$apply_id = $apply['apply_id'];
$apply_from = $apply['apply_from'];

错误消息

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in xxxxxxxxxxxxxxx line 26

Answer 1

您的查询;

SELECT users.user_id, app.apply_from 
FROM users 
INNER JOIN applications 
  ON  users.user_id = app.apply_from 
WHERE users.user_code='1'

...对表app使用别名application，但不声明它。

INNER JOIN applications app

Answer 2

您已将applications表的别名错过app中的join。请尝试以下方法：

SELECT users.user_id,app.apply_from 
FROM users 
INNER JOIN applications app ON users.user_id = app.apply_from 
WHERE users.user_code='1'

Answer 3

将applications表格缩写为“app”：

SELECT 
    users.user_id, 
    app.apply_from 
FROM
    users 
INNER JOIN
    applications AS app
ON
    users.user_id = app.apply_from
WHERE
    users.user_code='1'

MySQLI Inner Join 2表

3 个答案: