[英]MySQL SUM json values grouped by json keys
是否可以計算按json鍵分組的json值之和?
Mysql版本在Google雲sql上為5.7.17。
Example_1:我的觀點的一個簡短例子:
col1 | col2
-----|-----------------------
aaa | {"key1": 1, "key2": 3}
-----|-----------------------
bbb | {"key1": 0, "key2": 2}
-----|-----------------------
aaa | {"key1": 50, "key2": 0}
SQL查詢應該產生:
col1 | col2
-----|-----------------------
aaa | {"key1": 51, "key2": 3}
-----|-----------------------
bbb | {"key1": 0, "key2": 2}
要么
是否可以使用以下任何架構?
Example_2:
col1 | col2
-----|-----------------------
aaa | {{"key_name" : "key1", "key_value" : 1}, {"key_name" : "key2", "key_value" : 3}}
-----|-----------------------
bbb | {{"key_name" : "key1", "key_value" : 0}, {"key_name" : "key2", "key_value" : 2}}
-----|-----------------------
aaa | {{"key_name" : "key1", "key_value" : 50}, {"key_name" : "key2", "key_value" : 0}}
Example_3:
col1 | col2
-----|-----------------------
aaa | [{"key_name" : "key1", "key_value" : 1}, {"key_name" : "key2", "key_value" : 3}]
-----|-----------------------
bbb | [{"key_name" : "key1", "key_value" : 0}, {"key_name" : "key2", "key_value" : 2}]
-----|-----------------------
aaa | [{"key_name" : "key1", "key_value" : 50}, {"key_name" : "key2", "key_value" : 0}]
Example_4:
col1 | col2
-----|-----------------------
aaa | {"key1": {"key_name" : "key1", "key_value" : 1}, "key2": {"key_name" : "key2", "key_value" : 3}}
-----|-----------------------
bbb | {"key1": {"key_name" : "key1", "key_value" : 0}, "key2": {"key_name" : "key2", "key_value" : 2}}
-----|-----------------------
aaa | {"key1": {"key_name" : "key1", "key_value" : 50}, "key2": {"key_name" : "key2", "key_value" : 0}}
TL; DR:是的,它可以在不事先知道密鑰名的情況下完成,並且沒有任何備用數據格式比原始格式具有任何優勢。
這可以在不事先知道鍵名的情況下完成,但是很痛苦......基本上你必須查看表中的每個值,以便在你可以求和之前確定表中不同鍵的集合。 由於這個要求,以及備用數據格式每個條目都可以有多個密鑰的事實,使用它們中的任何一個都沒有任何優勢。
由於您必須查找所有不同的鍵,因此在查找它們時可以輕松完成總和。 這個功能和程序將一起做到這一點。 函數json_merge_sum
接受兩個JSON值並合並它們,對兩個值中出現鍵的值求和,例如
SELECT json_sum_merge('{"key1": 1, "key2": 3}', '{"key3": 1, "key2": 2}')
輸出:
{"key1": 1, "key2": 5, "key3": 1}
功能代碼:
DELIMITER //
DROP FUNCTION IF EXISTS json_merge_sum //
CREATE FUNCTION json_sum_merge(IN j1 JSON, IN total JSON) RETURNS JSON
BEGIN
DECLARE knum INT DEFAULT 0;
DECLARE jkeys JSON DEFAULT JSON_KEYS(j1);
DECLARE kpath VARCHAR(20);
DECLARE v INT;
DECLARE l INT DEFAULT JSON_LENGTH(jkeys);
kloop: LOOP
IF knum >= l THEN
LEAVE kloop;
END IF;
SET kpath = CONCAT('$.', JSON_EXTRACT(jkeys, CONCAT('$[', knum, ']')));
SET v = JSON_EXTRACT(j1, kpath);
IF JSON_CONTAINS_PATH(total, 'one', kpath) THEN
SET total = JSON_REPLACE(total, kpath, JSON_EXTRACT(total, kpath) + v);
ELSE
SET total = JSON_SET(total, kpath, v);
END IF;
SET knum = knum + 1;
END LOOP kloop;
RETURN total;
END
過程count_keys
執行GROUP BY
子句的等效操作。 它在表中找到col1
所有不同值,然后為具有該值col1
每一行調用json_sum_merge
。 請注意,行選擇查詢執行SELECT ... INTO
虛擬變量,因此不生成輸出,並使用MIN()
確保只有一個結果(以便可以將其分配給變量)。
步驟:
DELIMITER //
DROP PROCEDURE IF EXISTS count_keys //
CREATE PROCEDURE count_keys()
BEGIN
DECLARE finished INT DEFAULT 0;
DECLARE col1val VARCHAR(20);
DECLARE col1_cursor CURSOR FOR SELECT DISTINCT col1 FROM table2;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET finished=1;
OPEN col1_cursor;
col1_loop: LOOP
FETCH col1_cursor INTO col1val;
IF finished=1 THEN
LEAVE col1_loop;
END IF;
SET @total = '{}';
SET @query = CONCAT("SELECT MIN(@total:=json_sum_merge(col2, @total)) INTO @json FROM table2 WHERE col1='", col1val, "'");
PREPARE stmt FROM @query;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
SELECT col1val AS col1, @total AS col2;
END LOOP col1_loop;
END
稍微大一點的例子:
col1 col2
aaa {"key1": 1, "key2": 3}
bbb {"key1": 4, "key2": 2}
aaa {"key1": 50, "key3": 0}
ccc {"key2": 5, "key3": 1, "key4": 3}
bbb {"key1": 5, "key2": 1, "key5": 3}
CALL count_keys()
產生:
col1 col2
aaa {"key1": 51, "key2": 3, "key3": 0}
bbb {"key1": 9, "key2": 3, "key5": 3}
ccc {"key2": 5, "key3": 1, "key4": 3}
注意我在程序中調用了表table2
,你需要編輯它(在兩個查詢中)以適應。
我相信這樣的事情可行。
SELECT SUM(col2->>"$.key1"), SUM(col2->>"$.key2") FROM your_table GROUP BY col1
用於“簡短示例”的SQL:
SELECT col1,
JSON_OBJECT('key1', SUM(value1), 'key2', SUM(value2)) AS col2
FROM
(SELECT col1,
JSON_EXTRACT(col2, '$.key1') AS value1,
JSON_EXTRACT(col2, '$.key2') AS value2
FROM tbl) subq
GROUP BY col1;
Example_3的解決方案:
DROP TABLE IF EXISTS jsondata;
CREATE TABLE jsondata (json JSON, col varchar(11));
INSERT INTO jsondata VALUES
('[{"key_name" : "key1", "key_value" : 1}, {"key_name" : "key2", "key_value" : 3}]', 'aaa'),
('[{"key_name" : "key1", "key_value" : 0}, {"key_name" : "key3", "key_value" : 2}]', 'bbb'),
('[{"key_name" : "key1", "key_value" : 50}, {"key_name" : "key2", "key_value" : 0}]', 'aaa');
DROP FUNCTION IF EXISTS json_sum_by_col;
CREATE FUNCTION json_sum_by_col(col varchar(100)) RETURNS JSON
BEGIN
DECLARE i INT DEFAULT 0;
DECLARE done INT DEFAULT FALSE;
DECLARE select_values JSON;
DECLARE temp_result JSON;
DECLARE json_result JSON DEFAULT '[]';
DECLARE temp_key varchar(11);
DECLARE temp_value int;
DECLARE curs CURSOR FOR SELECT json FROM jsondata WHERE jsondata.col = col;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = TRUE;
OPEN curs;
read_loop: LOOP
SET i = 0;
FETCH curs INTO select_values;
IF done THEN
LEAVE read_loop;
END IF;
WHILE i < JSON_LENGTH(select_values) DO
-- extract key and value for i element
SET temp_key = JSON_EXTRACT(JSON_EXTRACT(select_values, CONCAT('$[',i,']')), '$.key_name');
SET temp_value = JSON_EXTRACT(JSON_EXTRACT(select_values, CONCAT('$[',i,']')), '$.key_value');
-- search json_result for key
SET @search = JSON_SEARCH(json_result, 'one', JSON_UNQUOTE(temp_key));
IF @search IS NOT NULL THEN
-- if exists add to existing value
SET @value_path = JSON_UNQUOTE(REPLACE(@search, 'name', 'value'));
SET temp_value = temp_value + JSON_EXTRACT(json_result, @value_path);
SET json_result = JSON_REPLACE(json_result, @value_path, temp_value);
ELSE
-- else attach it to json_result
SET temp_result = JSON_OBJECT("key_name", JSON_UNQUOTE(temp_key), "key_value", temp_value);
SET json_result = JSON_INSERT(json_result, CONCAT('$[',JSON_LENGTH(json_result),']'), temp_result);
END IF;
SELECT i + 1 INTO i;
END WHILE;
END LOOP;
CLOSE curs;
RETURN json_result;
END;
SELECT col, json_sum_by_col(col) FROM jsondata GROUP BY col;
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