php MySQLI 内连接 2 个表
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MySQLI Inner Join 2 Tables
提问by user1772571
What am I doing wrong here. I have followed many examples but can't seem to get this working. I have 2 tables
我在这里做错了什么。我已经遵循了许多示例，但似乎无法使其正常工作。我有2张桌子
Table => users
表 => 用户
user_id
user_name
user_email
user_password
user_country
user_dobdate
user_company
user_code
user_status
user_type
Table => applications
表 => 应用程序
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
SQLI查询
$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'];
Error Message
错误信息
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in xxxxxxxxxxxxxxx line 26
采纳答案by Joachim Isaksson
Your query;
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'
...uses an alias appfor the table application, but does not declare it.
...app为 table使用别名application，但不声明它。
INNER JOIN applications app
回答by Bharathi
You have missed the alias name for table applicationsas appin join. Try the following:
你已经错过了表的别名applications为app在join。请尝试以下操作：
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'
回答by user4035
Put abbreviation 'app' for applicationstable:
将缩写“app”放在applications表中：
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'
