Returing单列多次

Question

我有两个表这样的：

summary(id, status, value, time, aperson) and long_summary(id, who, comment, alltext) 

对于我有试过下面的查询，如：

SELECT l.id,summary.status 
FROM long_summary l INNER JOIN summary ON l.id = summary.id INNER JOIN long_summary ON summary.aperson = long_summary.who 
WHERE summary.status IN('old','new','waiting') 
AND summary.value IN ('') 
AND summary.time >= DATE_SUB(NOW(),INTERVAL 2 MONTH) 
AND l.alltext LIKE '% relational database management system %' ORDER BY FIELD(summary.status,'old','new','waiting'),summary.time DESC 

在运行它给正确的输出，但retured行是多次出现，如：

(1,'old') 
(1,'old') 
(1,'old') 
(2,'new') 
(2,'new') 
(3,'new') 
(4,'waiting') 
(4,'waiting') 

对于那i have used SELECT DISTINCT and after that it is giving correct output without repetition of the single row。但我没有得到我做错了什么？ 你会建议吗？

Answer 1

看看发生了什么的方法是（a）创建一个精简的测试用例，并（b）做一个SELECT *，这样你就可以看到你在其他列返回的东西，这就是重复。

所以，我这样做：

CREATE TABLE summary(id, status, value, time, aperson); 
INSERT INTO "summary" VALUES(1,'old',23,'time0','joe'); 
INSERT INTO "summary" VALUES(2,'new',42,'time1','bob'); 
INSERT INTO "summary" VALUES(3,'new',32,'time2','mike'); 
CREATE TABLE long_summary(id, who, comment, alltext); 
INSERT INTO "long_summary" VALUES(1,'someone','i say!','some text'); 
INSERT INTO "long_summary" VALUES(1,'joe','joe likes','some text'); 
INSERT INTO "long_summary" VALUES(2,'joe','joe likes bob','some text'); 
INSERT INTO "long_summary" VALUES(3,'joe','joe likes mike','some text'); 
INSERT INTO "long_summary" VALUES(1,'bob','nice one, joe','some text'); 
INSERT INTO "long_summary" VALUES(2,'bob','nice one, me','some text'); 
INSERT INTO "long_summary" VALUES(2,'bob','double nice one, me','some text'); 
INSERT INTO "long_summary" VALUES(3,'bob','nice one, mike','some text'); 
COMMIT; 

然后，我把您的查询的一个简化版本：

SELECT l.id,summary.status 
FROM long_summary l INNER JOIN summary ON l.id = summary.id INNER JOIN long_summary ON summary.aperson = long_summary.who 
WHERE summary.status IN('old','new','waiting') 

的其他的东西都不能使事情更糟，对吗？所以这是无关紧要的。我运行这个时会得到什么？ 9份1|old和2|new 12份。

所以，让我们改变它看到整排：

SELECT * 
FROM long_summary l INNER JOIN summary ON l.id = summary.id INNER JOIN long_summary ON summary.aperson = long_summary.who 
WHERE summary.status IN('old','new','waiting') 

1|bob|nice one, joe|some text|1|old|23|time0|joe|1|joe|joe likes|some text 
1|bob|nice one, joe|some text|1|old|23|time0|joe|2|joe|joe likes bob|some text 
1|bob|nice one, joe|some text|1|old|23|time0|joe|3|joe|joe likes mike|some text 
1|joe|joe likes|some text|1|old|23|time0|joe|1|joe|joe likes|some text 
1|joe|joe likes|some text|1|old|23|time0|joe|2|joe|joe likes bob|some text 
1|joe|joe likes|some text|1|old|23|time0|joe|3|joe|joe likes mike|some text 
1|someone|i say!|some text|1|old|23|time0|joe|1|joe|joe likes|some text 
1|someone|i say!|some text|1|old|23|time0|joe|2|joe|joe likes bob|some text 
1|someone|i say!|some text|1|old|23|time0|joe|3|joe|joe likes mike|some text 
2|bob|double nice one, me|some text|2|new|42|time1|bob|1|bob|nice one, joe|some text 
2|bob|double nice one, me|some text|2|new|42|time1|bob|2|bob|double nice one, me|some text 
2|bob|double nice one, me|some text|2|new|42|time1|bob|2|bob|nice one, me|some text 
2|bob|double nice one, me|some text|2|new|42|time1|bob|3|bob|nice one, mike|some text 
2|bob|nice one, me|some text|2|new|42|time1|bob|1|bob|nice one, joe|some text 
2|bob|nice one, me|some text|2|new|42|time1|bob|2|bob|double nice one, me|some text 
2|bob|nice one, me|some text|2|new|42|time1|bob|2|bob|nice one, me|some text 
2|bob|nice one, me|some text|2|new|42|time1|bob|3|bob|nice one, mike|some text 
2|joe|joe likes bob|some text|2|new|42|time1|bob|1|bob|nice one, joe|some text 
2|joe|joe likes bob|some text|2|new|42|time1|bob|2|bob|double nice one, me|some text 
2|joe|joe likes bob|some text|2|new|42|time1|bob|2|bob|nice one, me|some text 
2|joe|joe likes bob|some text|2|new|42|time1|bob|3|bob|nice one, mike|some text 

OK，现在你可以看到这个问题，让我们来看看为什么它的发生。是否这些行中的每一个都假设在那里？换句话说，如果前三个组合中至少有一个在那里，你是否应该看到1|old？

如果不是，哪个不应该导致它？您需要在WHERE或JOIN中筛选出某些内容。

如果是这样，那么你需要一个GROUP BY来合并相关的字段，或一个OR某处，更可能是第一个。

逐步通过所有组询问相同的问题。如果您在实际显示的列上达到需要GROUP BY的点，则只需使用SELECT DISTINCT即可。

您也可以退后一步，询问您是否真的想要long_summary的完整JOIN以及带有long_summary的摘要。这是8 * 3 * 8 = 192行，您已经过滤到21行。这是否有意义，或者你只期望，例如，24行来过滤？如果是后者，你得到了JOIN错误。这些JOIN中的任何一个都不应该存在，或者它应该是一对一的而不是一对多的JOIN，或者其他问题是错误的。

顺便说一下，从上面的测试中可以看出，我使用sqlite3而不是mysql，只是因为它的获取和运行要简单得多。我怀疑这有什么不同，但如果你在mysql进行测试并看到不同的结果，请通过任何方式告诉我。