Mysql Nested Subqueries Problem » History » Version 2
Amber Herold, 03/04/2010 01:29 PM
1 | 1 | Amber Herold | h1. Mysql Nested Subqueries Problem |
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3 | // This is the query that I wanted to do because it is easy to understand (= easier to maintain). |
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4 | // Unfortunatly there is a bug in mysql (http://bugs.mysql.com/bug.php?id=10312) which makes this |
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5 | // so slow that no one has seen it complete. The subsequent query is a bit more difficult to follow |
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6 | // but gets around this problem. |
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7 | // |
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8 | // NOTE: shareexperiments.experimentId and projectexperiments.projectexperimentId match but are not |
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9 | // the same as DB_LEGINON.SessionData.DEF_id. For finding the ProjectIds that correspond to a session, |
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10 | // we need to use the SessionData id. |
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12 | // $sharedProjectsQuery = "SELECT ".DB_PROJECT.".projectexperiments.projectId " |
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13 | // ." FROM ".DB_PROJECT.".projectexperiments " |
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14 | // ." WHERE ".DB_PROJECT.".projectexperiments.name IN " |
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15 | // ." ( " |
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16 | // ." SELECT ".DB_LEGINON.".SessionData.name " |
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17 | // ." FROM ".DB_LEGINON.".SessionData " |
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18 | // ." WHERE ".DB_LEGINON.".SessionData.`DEF_id` IN " |
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19 | // ." ( " |
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20 | // ." SELECT ".DB_PROJECT.".shareexperiments.`REF|leginondata|SessionData|experiment` " |
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21 | // ." FROM ".DB_PROJECT.".shareexperiments " |
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22 | // ." WHERE ".DB_PROJECT.".shareexperiments.`REF|leginondata|UserData|user` = ".$userId." " |
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23 | // ." ) " |
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24 | // ." )"; |
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26 | $sharedProjectQuery = " SELECT amberproject.projectexperiments.projectId " |
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27 | ." FROM amberproject.projectexperiments " |
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28 | ." INNER JOIN ( " |
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29 | ." SELECT amberdbemdata.SessionData.name AS SessionName " |
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30 | ." FROM amberdbemdata.SessionData " |
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31 | ." INNER JOIN ( " |
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32 | ." SELECT amberproject.shareexperiments.`REF|leginondata|SessionData|experiment` AS SessionId " |
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33 | ." FROM amberproject.shareexperiments " |
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34 | ." WHERE amberproject.shareexperiments.`REF|leginondata|UserData|user` = ".$userId." " |
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35 | ." ) AS SessionIds ON amberdbemdata.SessionData.`DEF_id` = SessionIds.SessionId " |
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36 | ." ) AS SessionNames ON amberproject.projectexperiments.name = SessionNames.SessionName "; |
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39 | 2 | Amber Herold | See another example "here":http://forums.mysql.com/read.php?24,54721,54721#msg-54721. |