Question

在MariaDB 10.2.19中，我有一个名为forms的表，该表的列template始终包含对象的JSON数组。其中一些对象将具有我想返回的属性：name（应始终存在），rule和parameters。我该如何从整个数组中仅返回这三个属性，而只返回存在rule的对象呢？

一个示例数组（格式化以便于查看）：

[{
    "label": "Employed?",
    "class": "select",
    "name": "employed",
    "parameters": "Yes",
    "rule": "in"
},
{
    "label": "Breed of dog?",
    "class": "select",
    "name": "breed",
    "parameters": "spaniel, collie, mix",
    "rule": "in"
},
{
    "label": "Number",
    "class": "text",
    "name": "breed"
}]

Answer 1

如果您使用的是MySQL 8.0.4或更高版本，则一种方法是使用JSON_TABLE：

mysql> SELECT * FROM foo;
+-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| data                                                                                                                                                                                                                                                                          |
+-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| [{"name": "employed", "rule": "in", "class": "select", "label": "Employed?", "parameters": "Yes"}, {"name": "breed", "rule": "in", "class": "select", "label": "Breed of dog?", "parameters": "spaniel, collie, mix"}, {"name": "breed", "class": "text", "label": "Number"}] |
+-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0,00 sec)

mysql> SELECT name, parameters 
         FROM foo,
              JSON_TABLE (
                  foo.data, 
                  "$[*]" COLUMNS (
                      name VARCHAR(100) PATH "$.name",
                      rule VARCHAR(100) PATH "$.rule",
                      parameters VARCHAR(100) PATH "$.parameters")
              ) AS t
       WHERE rule IS NOT NULL;
+----------+----------------------+
| name     | parameters           |
+----------+----------------------+
| employed | Yes                  |
| breed    | spaniel, collie, mix |
+----------+----------------------+
2 rows in set (0,00 sec)

https://dev.mysql.com/doc/refman/8.0/en/json-table-functions.html

Answer 2

您可以使用专用的numbers_table表来完成此操作：

SELECT
    JSON_VALUE(f.template,CONCAT('$[',n.number,'].name')) AS `name`,
    JSON_VALUE(f.template,CONCAT('$[',n.number,'].rule')) AS `rule`,
    JSON_VALUE(f.template,CONCAT('$[',n.number,'].parameters')) AS `parameters`
FROM forms AS f
JOIN numbers_table AS n
WHERE
    n.number < JSON_LENGTH(f.template) AND
    JSON_VALUE(f.template,CONCAT('$[',n.number,'].rule')) IS NOT NULL;

numbers_table表包含一个名为number的列，该列从0开始，并且可以根据您的用例需要而定（我有1000个值0到{ {1}}）。将999字段的每个元素提取到其自己的行中非常有用。

第一个JSON_ARRAY条件确保我们使用的数字与WHERE中的元素数量一样（在您的情况下为JSON_ARRAY）。

根据您的用例，第二个template条件消除了没有WHERE的情况。

如何使用MariaDB从JSON数组的多个对象中提取值？

2 个答案: