英文:
How to change ISO date format to string format("DD/MM/YYYY") in Couchbase query?
问题
以下是已翻译的内容:
我在文档中有一个日期属性,格式如下:
2016-09-23T18:48:11.000+00:00
我需要将此格式转换为以下格式:
“DD/MM/YYYY”
是否有任何方法可以进行转换?
我尝试了以下查询:
SELECT date_format_str("2016-09-23T18:48:11.000+00:00", '11-11-1111' ) ;
结果:
[
{
"$1": "2016-09-23T18:48:11Z"
}
]
预期结果:
"23/09/2016"
提前感谢。
英文:
I have a date attribute in document in the below format
2016-09-23T18:48:11.000+00:00
I need to convert this format into below format
“DD/MM/YYYY”
Is there any method to convert so?
I tried the below query :-
SELECT date_format_str("2016-09-23T18:48:11.000+00:00", '11-11-1111' ) ;
Result :-
[
{
"$1": "2016-09-23T18:48:11Z"
}
]
Expected Result :-
"23/09/2016"
Thanks in advance.
答案1
得分: 3
可以使用 DATE_FORMAT_STR
完成。
对于您的示例:
SELECT DATE_FORMAT_STR("2016-09-23T18:48:11.000+0000", "DD/MM/YYYY") AS formatted_date
更多信息请参考:https://docs.couchbase.com/server/current/n1ql/n1ql-language-reference/datefun.html#fn-date-format-str
英文:
Can be accomplished with DATE_FORMAT_STR
For your example;
SELECT DATE_FORMAT_STR("2016-09-23T18:48:11.000+0000", "DD/MM/YYYY") AS formatted_date
For more information; https://docs.couchbase.com/server/current/n1ql/n1ql-language-reference/datefun.html#fn-date-format-str
答案2
得分: 3
依赖于服务器版本
检查各种选项
https://docs.couchbase.com/server/current/n1ql/n1ql-language-reference/datefun.html#date-string
SELECT DATE_FORMAT_STR("2016-09-23T18:48:11.000+0000", "%d/%m/%Y") AS formatted_date;
SELECT DATE_FORMAT_STR("2016-09-23T18:48:11.000+0000", "DD/MM/YYYY") AS formatted_date;
如果您收到版本不受支持的错误。您可以使用以下方式
SELECT CONCAT2("/",SUBSTR(date,8,2), SUBSTR(date,5,2), SUBSTR(date,0,4)) AS newdate LET date = "2016-09-23T18:48:11.000+0000";
英文:
It depends on the Server version
Check various tabs
https://docs.couchbase.com/server/current/n1ql/n1ql-language-reference/datefun.html#date-string
SELECT DATE_FORMAT_STR("2016-09-23T18:48:11.000+0000", "%d/%m/%Y") AS formatted_date;
SELECT DATE_FORMAT_STR("2016-09-23T18:48:11.000+0000", "DD/MM/YYYY") AS formatted_date;
If you get error that your version may not supported. You can use as follows
SELECT CONCAT2("/",SUBSTR(date,8,2), SUBSTR(date,5,2), SUBSTR(date,0,4)) AS newdate LET date = "2016-09-23T18:48:11.000+0000";
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