JSON_EXTRACT作爲值的INSERT INTO？

Question

我期待下面的MySQL代碼：

SET @json= '{"1":{"name":"Name","value":"James Bowery","id":1,"type":"name","first":"James","middle":"","last":"Bowery"},"2":{"name":"Birthdate","value":"06\/23\/2017","id":2,"type":"date-time","date":"06\/23\/2017","time":"","unix":1498176000},"3":{"name":"Gender","value":"Male","value_raw":"Male","id":3,"type":"radio"},"4":{"name":"Ethnicity","value":"European","value_raw":"European","id":4,"type":"radio"},"5":{"name":"Email","value":"jabowery@emailservice.com","id":5,"type":"text"}}'; 
SET @array = JSON_EXTRACT(@json, '$."1".first','$."1".last','$."5".value','$."2".value','$."3".value','$."4".value'); 
INSERT INTO user (firstname,lastname,birthdate,gender,ethnicity,email) VALUES (@array); 

要插入行從JSON中提取的命名字段填充user表。但是，INSERT會產生語法錯誤。什麼是正確的語法？

Answer 1

也許試試這個。請注意，我已將日期格式更改爲yyyy-MM-dd。您將需要努力將此編程轉換爲適合MySQL。另外，當我運行這個時，數組在出生日期之前有電子郵件，所以我改變了插入的順序。

SET @json= '{"1":{"name":"Name","value":"James Bowery","id":1,"type":"name","first":"James","middle":"","last":"Bowery"},"2":{"name":"Birthdate","value":"06\/23\/2017","id":2,"type":"date-time","date":"2017-06-23","time":"","unix":1498176000},"3":{"name":"Gender","value":"Male","value_raw":"Male","id":3,"type":"radio"},"4":{"name":"Ethnicity","value":"European","value_raw":"European","id":4,"type":"radio"},"5":{"name":"Email","value":"jabowery@emailservice.com","id":5,"type":"text"}}'; 
SET @array = JSON_EXTRACT(@json, '$."1".first','$."1".last','$."5".value','$."2".value','$."3".value','$."4".value'); 
SET @queryString= CONCAT('INSERT INTO user (firstname,lastname,email,birthdate,gender,ethnicity) VALUES (',REPLACE(REPLACE(@array, '[',''),']',''),')'); 
PREPARE stmt FROM @queryString; 
EXECUTE stmt; 
DEALLOCATE PREPARE stmt;