我有 2 个表
这里是表 1 表名是例如
项目清单
+-----+----------+-----+
| uid | username | age |
+-----+----------+-----+
| 1 | doe | 17 |
| 2 | smith | 18 |
| 3 | john | 30 |
+-----+----------+-----+
另一个是
最喜欢的
+-----+------+---------+
| uid | user | itemuid |
+-----+------+---------+
| 1 | alex | 2 |
+-----+------+---------+
这是我的mysql查询*不工作*当我运行php文件时解决这个问题的任何方法我在mysql语法中遇到错误
SELECT c.uid, c.username, c.age,i.uid,i.user,i.itemuid
from itemlist c
left join fav i on c.uid = i.itemuid
if (i.user = 'alex') THEN
SET @fav = 1;
ELSE
SET @fav = 0;
END IF
这是示例 php
while ($row = mysqli_fetch_array($res)){
if ($row['fav'] = '1'){
echo $row['username']." is exit in fav";
}else{
echo $row['username']." is not exit in fav";
}
}
我希望你能理解我的问题吗?