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Sakot ar kuru My SQL versiju var izmantot sub selektus

 

piem

 

"delete from ratings where wallpaperid in (select wallpaperid from wallpaper where galleryid = $gallid)"

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Sakot ar kuru My SQL versiju var izmantot sub selektus

 

piem

 

"delete from ratings where wallpaperid in (select wallpaperid from wallpaper where galleryid = $gallid)"

19102[/snapback]

Subqueries Add Advanced Querying Capabilities

 

MySQL Version 4.1 will include support for SQL subqueries, also called subselects or nested queries, a powerful feature that lets users search complex data with ease and efficiency.

http://www.mysql.com/news-and-events/press...se_2003_07.html

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Tam kverijam jau nevajag subselektu!

DELETE ratings 
USING wallpaper
WHERE 
 wallpaper.wallpaperid = ratings.wallpaperid AND 
 wallpaper.galleryid = $gallid

19107[/snapback]

 

hmm, shitaadu delete versiju liidz shim nebiju izmantojis. kaadreiz iespeejams noderees. thanx :)

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Tam kverijam jau nevajag subselektu!

DELETE ratings 
USING wallpaper
WHERE 
 wallpaper.wallpaperid = ratings.wallpaperid AND 
 wallpaper.galleryid = $gallid

19107[/snapback]

tieshi taa :)

apmeeram 50-80% gadijumu var izmantot shadu pierakstu pat aizvietojot

3 un vairaak iekseejos subselectus :)

papildus vienai un tai pashai tabulai var pieskirt dazaadus nosaukumus un

izmantot taas kaa divas dazaadas tabulas :)

P.S. teiksim kad vajag selekteet no tabulas ailes kaadu kaudziiti datu no tiem

atlasiit vel dalju un tad tikai dabuut gala rez (teoretiski 2 iekseejie subselekti)...

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Varēji vismaz patiekt, kas nestrādā! (mysql_error).

Pamēģini šādi, laikam aizmirsu FROM uzlikt:

DELETE FROM ratings
USING wallpaper
WHERE
wallpaper.wallpaperid = ratings.wallpaperid AND
wallpaper.galleryid = $gallid

(USING vietā var lietot arī FROM keywordu)

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ahh, nu pareiz, tur ar tabulu vajdzēj :)

Šis jau ļoti normāls veids. Bez subselektiem daudzkur var iztikt, tikai labi jāpārzin JOIN'u veidošana, tad daudz lietas var izdarīt (nu bet protams ne visas, kuras ar subselektiem).

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