GROUP_CONCAT AND ORDER BY FIELD in MySQL 查询出现问题
Issue with GROUP_CONCAT AND ORDER BY FIELD in MySQL query
我在查询时遇到了一些问题。我想通过多个 id 来排序查询。这是我的查询:
这个查询没有return我想要的结果:(
SELECT *
FROM question q
INNER JOIN answer a ON a.question_id = q.question_id
WHERE q.lesson_id = 1
AND q.question_id IN (
SELECT e.question_id
FROM exame e
WHERE e.inscription_id = 1
AND e.lesson_id = 1
AND e.attempt = 1
ORDER BY e.question_id
) /*this subquery returns (10,2,1,8,3,12,4,11,14,7)*/
ORDER BY FIELD(q.question_id,(
SELECT GROUP_CONCAT(DISTINCT ee.question_id ORDER BY ee.order_of_appearance SEPARATOR ',') AS final_order
FROM exame ee
WHERE ee.inscription_id = 1
AND ee.lesson_id = 1
AND ee.attempt = 1)
) /*this subquery returns (10,2,1,8,3,12,4,11,14,7)*/
可以看到,两个子查询return结果相同(10,2,1,8,3,12,4,11,14,7)。如您所知,它们之间的区别在于第一个 return 是一个结果集,而第二个只是一个将所有 id 连接起来的字段。
问题 #1: 如果我复制第一个子查询并将其写入第二个子查询所在的位置,我会收到此错误:
1242 - Subquery returns more than 1 row
所以我创建了第二个子查询(使用 GROUP_CONCAT 函数),但结果不是我所期望的。结果按 "question_id" 排序,我希望它按 "order_of_appearance" 字段排序。
问题 #2: 如果我在 ORDER BY 子句中编写子查询,我不会得到按 "order_of_appearance" 字段排序的结果,但是如果我删除子查询和我手动写 id 的数组,结果按 "order_of_appearance" 排序!!为什么???
这个查询return结果如我所愿!! :)
SELECT *
FROM question q
INNER JOIN answer a ON a.question_id = q.question_id
WHERE q.lesson_id = 1
AND q.question_id IN (
SELECT e.question_id
FROM exame e
WHERE e.inscription_id = 1
AND e.lesson_id = 1
AND e.attempt = 1
ORDER BY e.question_id
)
ORDER BY FIELD(q.question_id,10,2,1,8,3,12,4,11,14,7)
最后一个问题:是否可以在不必手动编写 id 数组的情况下实现我想要的?我需要动态地做。
提前致谢! (我希望你能听懂我的英语!)
我认为您需要使用 FIND_IN_SET()
而不是 FIELD()
:
来自参考手册:
Returns the index (position) of str
in the str1
, str2
, str3
, ... list. Returns 0 if str is not found.
If all arguments to FIELD()
are strings, all arguments are compared as strings. If all arguments are numbers, they are compared as numbers. Otherwise, the arguments are compared as double.
If str
is NULL
, the return value is 0 because NULL
fails equality comparison with any value. FIELD()
is the complement of ELT()
.
mysql> SELECT FIELD('ej', 'Hej', 'ej', 'Heja', 'hej', 'foo');
-> 2
mysql> SELECT FIELD('fo', 'Hej', 'ej', 'Heja', 'hej', 'foo');
-> 0
Returns a value in the range of 1 to N if the string str is in the string list strlist consisting of N substrings. A string list is a string composed of substrings separated by “,” characters. If the first argument is a constant string and the second is a column of type SET
, the FIND_IN_SET()
function is optimized to use bit arithmetic. Returns 0 if str
is not in strlist
or if strlist
is the empty string. Returns NULL
if either argument is NULL
. This function does not work properly if the first argument contains a comma (“,”) character.
mysql> SELECT FIND_IN_SET('b','a,b,c,d');
-> 2
我在查询时遇到了一些问题。我想通过多个 id 来排序查询。这是我的查询:
这个查询没有return我想要的结果:(
SELECT *
FROM question q
INNER JOIN answer a ON a.question_id = q.question_id
WHERE q.lesson_id = 1
AND q.question_id IN (
SELECT e.question_id
FROM exame e
WHERE e.inscription_id = 1
AND e.lesson_id = 1
AND e.attempt = 1
ORDER BY e.question_id
) /*this subquery returns (10,2,1,8,3,12,4,11,14,7)*/
ORDER BY FIELD(q.question_id,(
SELECT GROUP_CONCAT(DISTINCT ee.question_id ORDER BY ee.order_of_appearance SEPARATOR ',') AS final_order
FROM exame ee
WHERE ee.inscription_id = 1
AND ee.lesson_id = 1
AND ee.attempt = 1)
) /*this subquery returns (10,2,1,8,3,12,4,11,14,7)*/
可以看到,两个子查询return结果相同(10,2,1,8,3,12,4,11,14,7)。如您所知,它们之间的区别在于第一个 return 是一个结果集,而第二个只是一个将所有 id 连接起来的字段。
问题 #1: 如果我复制第一个子查询并将其写入第二个子查询所在的位置,我会收到此错误:
1242 - Subquery returns more than 1 row
所以我创建了第二个子查询(使用 GROUP_CONCAT 函数),但结果不是我所期望的。结果按 "question_id" 排序,我希望它按 "order_of_appearance" 字段排序。
问题 #2: 如果我在 ORDER BY 子句中编写子查询,我不会得到按 "order_of_appearance" 字段排序的结果,但是如果我删除子查询和我手动写 id 的数组,结果按 "order_of_appearance" 排序!!为什么???
这个查询return结果如我所愿!! :)
SELECT *
FROM question q
INNER JOIN answer a ON a.question_id = q.question_id
WHERE q.lesson_id = 1
AND q.question_id IN (
SELECT e.question_id
FROM exame e
WHERE e.inscription_id = 1
AND e.lesson_id = 1
AND e.attempt = 1
ORDER BY e.question_id
)
ORDER BY FIELD(q.question_id,10,2,1,8,3,12,4,11,14,7)
最后一个问题:是否可以在不必手动编写 id 数组的情况下实现我想要的?我需要动态地做。
提前致谢! (我希望你能听懂我的英语!)
我认为您需要使用 FIND_IN_SET()
而不是 FIELD()
:
来自参考手册:
Returns the index (position) of
str
in thestr1
,str2
,str3
, ... list. Returns 0 if str is not found.If all arguments to
FIELD()
are strings, all arguments are compared as strings. If all arguments are numbers, they are compared as numbers. Otherwise, the arguments are compared as double.If
str
isNULL
, the return value is 0 becauseNULL
fails equality comparison with any value.FIELD()
is the complement ofELT()
.mysql> SELECT FIELD('ej', 'Hej', 'ej', 'Heja', 'hej', 'foo'); -> 2 mysql> SELECT FIELD('fo', 'Hej', 'ej', 'Heja', 'hej', 'foo'); -> 0
Returns a value in the range of 1 to N if the string str is in the string list strlist consisting of N substrings. A string list is a string composed of substrings separated by “,” characters. If the first argument is a constant string and the second is a column of type
SET
, theFIND_IN_SET()
function is optimized to use bit arithmetic. Returns 0 ifstr
is not instrlist
or ifstrlist
is the empty string. ReturnsNULL
if either argument isNULL
. This function does not work properly if the first argument contains a comma (“,”) character.mysql> SELECT FIND_IN_SET('b','a,b,c,d'); -> 2