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JSON data type for like column
How do you set a default value for a MySQL Datetime column?Which MySQL data type to use for storing boolean valuesShould I use the datetime or timestamp data type in MySQL?How do I change the data type for a column in MySQL?SQL select only rows with max value on a columnReference - What does this error mean in PHP?Storing JSON in database vs. having a new column for each keySELECT all and ORDER by Column A without duplicate of Column B and then ORDER by Column CINSERT INTO table SELECT & Append a columnIs the MySQL JSON data type bad for performance for data retrieval?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom: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
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
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
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
mysql
asked Mar 25 at 9:34
JillAndMeJillAndMe
1481 silver badge10 bronze badges
1481 silver badge10 bronze badges
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1 Answer
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votes
@JillAndMe! you can use the JSON_ARRAY_APPEND() method. I hope this will help you JSON_ARRAY_APPEND()
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
@JillAndMe! you can use the JSON_ARRAY_APPEND() method. I hope this will help you JSON_ARRAY_APPEND()
add a comment |
@JillAndMe! you can use the JSON_ARRAY_APPEND() method. I hope this will help you JSON_ARRAY_APPEND()
add a comment |
@JillAndMe! you can use the JSON_ARRAY_APPEND() method. I hope this will help you JSON_ARRAY_APPEND()
@JillAndMe! you can use the JSON_ARRAY_APPEND() method. I hope this will help you JSON_ARRAY_APPEND()
answered Mar 25 at 9:40
MKJMKJ
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7681 gold badge11 silver badges27 bronze badges
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