MYSQL left join if then statement favorite table procedure - php

Question

i have 2 tables
here is table 1 table name is
forexample
itemlist
+-----+----------+-----+
| uid | username | age |
+-----+----------+-----+
| 1 | doe | 17 |
| 2 | smith | 18 |
| 3 | john | 30 |
+-----+----------+-----+
and other one is
fav
+-----+------+---------+
| uid | user | itemuid |
+-----+------+---------+
| 1 | alex | 2 |
+-----+------+---------+
Here is my mysql query *NOT Working * any way to fix this problem when i run php file i got error in mysql syntax
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
this is sample 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";
}
}
i hope you understand my question right ?

Answer 1

To get a column named fav returned in the resultset, you would need to include an expression in the SELECT list, and give it an alias fav.
It's not at all clear why you would need a MySQL user-defined variable; if you don't know why you'd need one, then you probably don't need one.
Given that your PHP code is looking for a column named fav in the resultset, likely you want something like this:
SELECT c.uid
, c.username
, c.age
, i.uid AS i_uid
, i.user
, i.itemuid
, IF(i.user='alex',1,0) AS fav
FROM itemlist c
LEFT
JOIN fav i ON i.itemuid = c.uid
Note that the original query had two columns named uid; if you want to return both, and be able to reference both of those by column name, you need to have distinct names for each. In the query above, I've assigned an alias to the i.uid column so that both uid columns will be available by distinct column name.