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JSON data type for like column

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0

I'm trying to make like column.

like column is used when users like the content they push like button and users can do 1 like for 1 content.

So I decided to use JSON data type for like in order to prevent duplication of like.

but, here is my problem: I've read document, functions are seemed to be required original data as first parameter.

ex:)

mysql> SET @j = '"a": 1'; //this is original data
mysql> SELECT JSON_ARRAY_APPEND(@j, '$', 'z') as result;
+---------------------------------+
| result |
+---------------------------------+
| ["a": 1, "z"] |
+---------------------------------+

If I want to add "b":2 then I need to fetch all json data, all I want to do is just push a data to it not fetching all json data from like column.

I want to do this :

mysql> SELECT database.table.`like`.JSON_ARRAY_APPEND('$', '"b":2') as result;
+---------------------------------+
| result |
+---------------------------------+
| ["a": 1, "z"] |
+---------------------------------+

Let me know if you need more information.
Thanks.

mysql

share|improve this question

asked Mar 25 at 9:34

JillAndMeJillAndMe

14811 silver badge1010 bronze badges

add a comment | 

0

I'm trying to make like column.

like column is used when users like the content they push like button and users can do 1 like for 1 content.

So I decided to use JSON data type for like in order to prevent duplication of like.

but, here is my problem: I've read document, functions are seemed to be required original data as first parameter.

ex:)

mysql> SET @j = '"a": 1'; //this is original data
mysql> SELECT JSON_ARRAY_APPEND(@j, '$', 'z') as result;
+---------------------------------+
| result |
+---------------------------------+
| ["a": 1, "z"] |
+---------------------------------+

If I want to add "b":2 then I need to fetch all json data, all I want to do is just push a data to it not fetching all json data from like column.

I want to do this :

mysql> SELECT database.table.`like`.JSON_ARRAY_APPEND('$', '"b":2') as result;
+---------------------------------+
| result |
+---------------------------------+
| ["a": 1, "z"] |
+---------------------------------+

Let me know if you need more information.
Thanks.

mysql

share|improve this question

asked Mar 25 at 9:34

JillAndMeJillAndMe

14811 silver badge1010 bronze badges

add a comment | 

I'm trying to make like column.

like column is used when users like the content they push like button and users can do 1 like for 1 content.

So I decided to use JSON data type for like in order to prevent duplication of like.

but, here is my problem: I've read document, functions are seemed to be required original data as first parameter.

ex:)

mysql> SET @j = '"a": 1'; //this is original data
mysql> SELECT JSON_ARRAY_APPEND(@j, '$', 'z') as result;
+---------------------------------+
| result |
+---------------------------------+
| ["a": 1, "z"] |
+---------------------------------+

If I want to add "b":2 then I need to fetch all json data, all I want to do is just push a data to it not fetching all json data from like column.

I want to do this :

mysql> SELECT database.table.`like`.JSON_ARRAY_APPEND('$', '"b":2') as result;
+---------------------------------+
| result |
+---------------------------------+
| ["a": 1, "z"] |
+---------------------------------+

Let me know if you need more information.
Thanks.

mysql

share|improve this question

asked Mar 25 at 9:34

JillAndMeJillAndMe

14811 silver badge1010 bronze badges

mysql> SET @j = '"a": 1'; //this is original data
mysql> SELECT JSON_ARRAY_APPEND(@j, '$', 'z') as result;
+---------------------------------+
| result |
+---------------------------------+
| ["a": 1, "z"] |
+---------------------------------+

mysql> SELECT database.table.`like`.JSON_ARRAY_APPEND('$', '"b":2') as result;
+---------------------------------+
| result |
+---------------------------------+
| ["a": 1, "z"] |
+---------------------------------+

share|improve this question

asked Mar 25 at 9:34

JillAndMeJillAndMe

14811 silver badge1010 bronze badges

share|improve this question

asked Mar 25 at 9:34

JillAndMeJillAndMe

14811 silver badge1010 bronze badges

asked Mar 25 at 9:34

JillAndMeJillAndMe

14811 silver badge1010 bronze badges

asked Mar 25 at 9:34

JillAndMeJillAndMe

14811 silver badge1010 bronze badges

1 Answer
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1

@JillAndMe! you can use the JSON_ARRAY_APPEND() method. I hope this will help you JSON_ARRAY_APPEND()

share|improve this answer

answered Mar 25 at 9:40

MKJMKJ

76811 gold badge1111 silver badges2727 bronze badges

add a comment | 

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1

@JillAndMe! you can use the JSON_ARRAY_APPEND() method. I hope this will help you JSON_ARRAY_APPEND()

share|improve this answer

answered Mar 25 at 9:40

MKJMKJ

76811 gold badge1111 silver badges2727 bronze badges

add a comment | 

1

@JillAndMe! you can use the JSON_ARRAY_APPEND() method. I hope this will help you JSON_ARRAY_APPEND()

share|improve this answer

answered Mar 25 at 9:40

MKJMKJ

76811 gold badge1111 silver badges2727 bronze badges

add a comment | 

@JillAndMe! you can use the JSON_ARRAY_APPEND() method. I hope this will help you JSON_ARRAY_APPEND()

share|improve this answer

answered Mar 25 at 9:40

MKJMKJ

76811 gold badge1111 silver badges2727 bronze badges

share|improve this answer

answered Mar 25 at 9:40

MKJMKJ

76811 gold badge1111 silver badges2727 bronze badges

answered Mar 25 at 9:40

MKJMKJ

76811 gold badge1111 silver badges2727 bronze badges

answered Mar 25 at 9:40

MKJMKJ

76811 gold badge1111 silver badges2727 bronze badges