我有一些麻烦来完成我的Mysql查询以返回我需要的东西。我是MYSQL中长期查询的新手。
SELECT
lang_rel_a_id,
lang_rel_b_id,
lang_rel_id,
tla.text_lang_t AS atext,
lald.lang_data_lang_id AS laid,
lald.lang_data_position AS lapp,
lald.lang_data_font_weight AS lafw,
lald.lang_data_font_size AS lafs,
lald.lang_data_font_color AS lafc,
lald.lang_data_bg_color AS labg,
lasdf.funca AS lafunc,
lang_ship,
lbld.lang_data_lang_id AS lbid,
lbld.lang_data_position AS lbpp,
lbld.lang_data_font_weight AS lbfw,
lbld.lang_data_font_size AS lbfs,
lbld.lang_data_font_color AS lbfc,
lbld.lang_data_bg_color AS lbbg,
tlb.text_lang_t AS btext,
lbsdf.funcb AS lbfunc
FROM lang_relation
LEFT JOIN
(SELECT *, GROUP_CONCAT(text_func_t SEPARATOR ', ') AS funca
FROM synt_data_func
LEFT JOIN text_func ON text_func_id = synt_df_func
GROUP BY synt_df_lang_data
)
lasdf ON lang_rel_a_id = lasdf.synt_df_lang_data
LEFT JOIN lang_data lald ON lald.lang_data_id = lang_rel_a_id
LEFT JOIN text_lang tla ON lald.lang_data_lang_id = tla.text_lang_id
LEFT JOIN
(SELECT *, GROUP_CONCAT(text_func_t SEPARATOR ', ') AS funcb
FROM synt_data_func
LEFT JOIN text_func ON text_func_id = synt_df_func
GROUP BY synt_df_lang_data
)
lbsdf ON lang_rel_b_id = lbsdf.synt_df_lang_data
LEFT JOIN lang_data lbld ON lbld.lang_data_id = lang_rel_b_id
LEFT JOIN text_lang tlb ON lbld.lang_data_lang_id = tlb.text_lang_id
WHERE lang_rel_a_id < lang_rel_b_id
GROUP BY lang_rel_id
我的lang_relation表中有两种语言的关系。我需要查询它们中的每一个2个子表，但其中一个是一个关系表，其中包含lang_data_id（= lang_rel_a_id或lang_rel_b_id，= synt_df_lang_data）与不同语言函数的文本之间的关系，其中可能有多个值。
我不明白为什么这个子查询中的group_concat只返回一行。如果我只做这个查询，我会得到所有结果。但是当我把它放到这个更大的查询中时，一切都很好，但是这个......不是。
我的language_relation表
CREATE TABLE `lang_relation`
(
`lang_rel_id` int(11) NOT NULL,
`lang_rel_a_id` int(11) NOT NULL,
`lang_rel_b_id` int(11) NOT NULL,
`lang_ship` tinyint(1) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
已加入的lang_data
CREATE TABLE `lang_data` (
`lang_data_id` int(11) NOT NULL,
`lang_data_pic_key` int(11) NOT NULL,
`lang_data_position` tinyint(1) NOT NULL,
`lang_data_lang_id` int(11) NOT NULL,
`lang_data_font_weight` tinyint(2) NOT NULL,
`lang_data_font_size` tinyint(2) NOT NULL,
`lang_data_font_color` tinyint(2) NOT NULL,
`lang_data_bg_color` tinyint(2) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
还有synt_data_func。 text_table是一个简单的2列表，其中包含id + text。
CREATE TABLE `synt_data_func` (
`synt_df_id` int(11) NOT NULL,
`synt_df_lang_data` int(11) NOT NULL,
`synt_df_func` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin;
我尝试了不同的方法。这似乎是我所需要的最接近的那个。我不知道我改变GROUP BY子句的次数，我甚至试图在父SELECT中执行CONCAT_GROUP ..
我甚至想知道这是否可行，因为子查询需要2个不同的ID ..这是问题吗？
提前感谢任何提示。
答案 0 :(得分：0)
我终于明白了。也许它会帮助有类似问题的人。我改变了我的查询方法。
SELECT
lrel.lang_rel_pic_key,
lrel.lang_rel_id,
langdata_a.lascore,
lasdf.func_a,
langdata_a.latext,
lasf.score_astyle,
SUM(lasf.score_astyle) + (langdata_a.lascore) AS atotal,
lang_ship,
langdata_b.lbtext,
langdata_b.lbscore,
lbsdf.func_b,
lbsf.bformat,
lbsf.score_bstyle,
SUM(lbsf.score_bstyle) + (langdata_b.lbscore) AS btotal
FROM lang_relation lrel
INNER JOIN
(
SELECT DISTINCT
lald.lang_data_id,
lafw.field_value AS lafweight,
lafs.field_value AS lafsize,
lafc.field_value AS laffc,
lafbg.field_value AS lafbg,
lapos.field_value AS laposa,
tla.text_lang_t AS latext,
SUM(lafw.field_value) + (lafs.field_value) + (lafc.field_value) + (lafbg.field_value) + (lapos.field_value) AS lascore
FROM lang_data lald
LEFT JOIN text_lang tla ON lald.lang_data_lang_id = tla.text_lang_id
LEFT JOIN `fields` lafw ON lald.lang_data_font_weight = lafw.field_id
LEFT JOIN `fields` lafs ON lald.lang_data_font_size = lafs.field_id
LEFT JOIN `fields` lafc ON lald.lang_data_font_color = lafc.field_id
LEFT JOIN `fields` lafbg ON lald.lang_data_bg_color = lafbg.field_id
LEFT JOIN `fields` lapos ON lald.lang_data_position = lapos.field_id
GROUP BY lald.lang_data_id
)
langdata_a ON langdata_a.lang_data_id = lrel.lang_rel_a_id
LEFT JOIN
(SELECT sdf.synt_df_lang_data, GROUP_CONCAT(latf.text_func_t) AS func_a
FROM synt_data_func sdf
INNER JOIN text_func latf ON latf.text_func_id = sdf.synt_df_func
GROUP BY sdf.synt_df_lang_data
)
lasdf ON lasdf.synt_df_lang_data = lrel.lang_rel_a_id
LEFT JOIN
(
SELECT sfb.synt_format_lang_data,
sfb.synt_format_fields_id,
GROUP_CONCAT(sfbf.field_text SEPARATOR ', ') AS aformat,
SUM(sfbf.field_value) AS score_astyle
FROM synt_format sfb
INNER JOIN `fields` sfbf ON sfbf.field_id = sfb.synt_format_fields_id
GROUP BY sfb.synt_format_lang_data
)
lasf ON lasf.synt_format_lang_data = lrel.lang_rel_a_id
INNER JOIN
(
SELECT DISTINCT
lbld.lang_data_id,
lbfw.field_value AS lbfweight,
lbfs.field_value AS lbfsize,
lbfc.field_value AS lbffc,
lbfbg.field_value AS lbfbg,
lbpos.field_value AS lbposa,
tlb.text_lang_t AS lbtext,
SUM(lbfw.field_value) + (lbfs.field_value) + (lbfc.field_value) + (lbfbg.field_value) + (lbpos.field_value) AS lbscore
FROM lang_data lbld
LEFT JOIN text_lang tlb ON lbld.lang_data_lang_id = tlb.text_lang_id
LEFT JOIN `fields` lbfw ON lbld.lang_data_font_weight = lbfw.field_id
LEFT JOIN `fields` lbfs ON lbld.lang_data_font_size = lbfs.field_id
LEFT JOIN `fields` lbfc ON lbld.lang_data_font_color = lbfc.field_id
LEFT JOIN `fields` lbfbg ON lbld.lang_data_bg_color = lbfbg.field_id
LEFT JOIN `fields` lbpos ON lbld.lang_data_position = lbpos.field_id
GROUP BY lbld.lang_data_id
)
langdata_b ON langdata_b.lang_data_id = lrel.lang_rel_b_id
LEFT JOIN
(SELECT sdfb.synt_df_lang_data, GROUP_CONCAT(lbtf.text_func_t) AS func_b
FROM synt_data_func sdfb
INNER JOIN text_func lbtf ON lbtf.text_func_id = sdfb.synt_df_func
GROUP BY sdfb.synt_df_lang_data
)
lbsdf ON lbsdf.synt_df_lang_data = lrel.lang_rel_b_id
LEFT JOIN
(
SELECT sfb.synt_format_lang_data,
sfb.synt_format_fields_id,
GROUP_CONCAT(sfbf.field_text SEPARATOR ', ') AS bformat,
SUM(sfbf.field_value) AS score_bstyle
FROM synt_format sfb
INNER JOIN `fields` sfbf ON sfbf.field_id = sfb.synt_format_fields_id
GROUP BY sfb.synt_format_lang_data
)
lbsf ON lbsf.synt_format_lang_data = lrel.lang_rel_b_id
GROUP BY lrel.lang_rel_id
可能有点长，但输出正是所需要的： - ）