CODEWITHSUNDEEP
mysqlselectjoinif-statement
Ken Tang
Ken Tang

Reputation: 529

MySQL SELECT + LEFT JOIN + IF query

I have one of simple DB tables named category:

category.id category.name category.lang
1           'apple'       'en'
1           'manzana'     'es'
1           ''            'de'
1           ''            'it'

Variable:

$default_lang = 'en';
$current_lang = 'it';

I would like to make query that returns category_name field according to $current_lang, but if category.name is empty then should use category.name from $default_lang like this:

SELECT * FROM cat_list
    LEFT JOIN category ON cal_list.id = category.id
          AND IF(category.name = $current_lang = '', $default_lang, 'No Lang')

Please let me know, how could i fix this query to make it work?

===

Ok now I have a good Idea:

SELECT * FROM cat_list LEFT JOIN category 
           ON cal_list.id = category.id 
          AND category_lang IN ($curreng_lang, $default_lang)
          AND category_name <> ''

it works fine but should I worry about the "IN" order in case if both languages has name values?

Upvotes: 4

Views: 18469

Answers (3)

Patriks
Patriks

Reputation: 1012

user1664869, what about this?

SELECT catname=CASE c1.name
       WHEN '' THEN c2.name
       ELSE c1.name
       END
FROM
Category c1
INNER JOIN Category c2 ON c1.lang=$current_lang and c2.lang=$default_lang;

Upvotes: 0

Patriks
Patriks

Reputation: 1012

SELECT catname=
CASE name
         WHEN '' THEN (select name from Category where lang=$default_lang limit 1)
         ELSE name
      END
FROM Category where id=$current_lang

Explaination: here Query is using case statement inside select statment, where outer query is using condition "id=$current_lang"

so in select statement it will have name='', so for

catname=
    CASE name
             WHEN '' THEN (select name from Category where lang=$default_lang limit 1)
             ELSE name
          END

name is '', so first condition with when should execute here it will execute query

(select name from Category where lang=$default_lang limit 1) name apple is returned (because language is 'en') and that will be sotred to catname variable. which will be returned in result

If it will get value with something for name first time (id=$current_lang) it will return with it.

Upvotes: 3

John Woo
John Woo

Reputation: 263933

Is the category.name NULL or empty string?

If NULL, use COALESCE (Returns the first non-NULL value in the list)

SELECT category.id, 
       COALESCE(category.name, $default_lang) name,
       COALESCE(category.lang, $current_lang) lang
FROM cat_list
     LEFT JOIN category 
           ON cal_list.id = category.id

If Empty String, Use IF

SELECT category.id, 
       IF(category.name = '', $default_lang,category.name) name,
       IF(category.lang = '', $current_lang,category.lang) lang
FROM cat_list
     LEFT JOIN category 
           ON cal_list.id = category.id

Upvotes: 2

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