9/2/2023 0 Comments Update mysql with json![]() ![]() ![]() ![]() It increases the complexity of any code you write to do it, and the developer who needs to take over maintenance of your code after you have moved on to another project will curse your name. Storing data in a JSON document when you eventually want to use SQL expressions to search or update individual fields within the JSON document is a costly mistake. You can replace data at multiple places in the document if required. You provide the JSON document as the first argument, followed by the path to replace data at, followed with the value to replace the existing value with. Now compare how difficult that is to: UPDATE content SET status = 1 WHERE name = 'Jack' In MySQL, the JSONREPLACE () function replaces values in a JSON document and returns the result. Using this in an UPDATE looks like this: mysql> update example set content = json_set(content, concat(json_unquote(json_search(json_extract(content, '$.name'), 'one', 'Jack')), '.status'), '1') How to UPDATE a mySQL table with JSON Object in Node. Now use it in a JSON_SET() call: mysql> select json_pretty(Ĭoncat(json_unquote(json_search(json_extract(content, '$.name'), 'one', 'Jack')), '.status'), status and you can get the full path to the field you want to update: mysql> select concat(json_unquote(json_search(json_extract(content, '$.name'), 'one', 'Jack')), '.status') as path from example Then you can search that array to get the array position: mysql> select json_search(json_extract(content, '$.name'), 'one', 'Jack') as root from example Here's a trick: You can extract the name field, and that turns into an array of those values: mysql> select json_extract(content, '$.name') as a from example mysql> select j.* from example cross join json_table(content, '$' columns( And that doesn't tell you the path to the element, it only allows you to return a specific row out of the array. To search for a key/value pair, you need to use JSON_TABLE() and that requires you upgrade to MySQL 8.0. But still, if its a URL and it allows to get only certain prams based on request query params, then it can be done. Only picking partial object may not be possible. mysql> select json_unquote(json_search(content, 'one', 'Jack')) as path from example 1,168 22 38 What is the source of your JSON A URL or a variable You need complete object to traverse upon. If "Jack" occurs in some other field, that would also be found. You can get the path to a JSON element with JSON_SEARCH(), but you can only search by value, not by a key/value pair. The solution in that case depends on you knowing which array element your pseudo-record exists in. The question like yours has come up before on Stack Overflow, for example JSON update single value in MySQL table. ![]() mysql> select json_pretty(json_replace(content, '$.status', '1')) as j So in this case, we can see that the array element is $ but if you didn't know that, you couldn't use this solution. You can use JSON_REPLACE() or JSON_SET(), but both require that you know the path to the field you want to change. The key column has a BIN2 collation, so you need to convert with collate option.This is very awkward, nearly impossible with MySQL's JSON functions. SELECT JSON_MODIFY(d.Data, N'$.data COLLATE Latin1_General_CI_AS) + N'].someData.sample1', N'NewValue') AS Data SELECT '$.data + '].someData.sample1', N'NewValue') AS JsonData The JSON array is splitted into elements using OPENJSON with default. Note, that this feature is available in SQL Server 2017 (14.x) and in Azure SQL Database. WHERE somecolumnsomevalue Notice the WHERE clause in the UPDATE syntax: The WHERE clause specifies which record or records that should be updated. In this case, one possible approach is to use an expression as path parameter for JSONMODIFY. The UPDATE statement is used to update existing records in a table: UPDATE tablename SET column1value, column2value2. The JSON values stored in that column, are just a way to differenciate a row from another, nothing more. But later, I was thinking about using index to speed up the queries, and because of that, change the JSON type to maybe a TEXT type to create a index with the 4 columns. How can I update the interests if there are multiple values to be updated Running the above update statement to update the Visual Arts interest results in an. The JSON array is splitted into elements using OPENJSON with default schema and in this case the returned columns are key, value and type. Example: Your JSON is an array of JSON objects, so you need an index to access each element. From the beginning I'm using the filter column as a JSON type of MySQL. UPDATE table SET meta JSONREPLACE (meta, JSONUNQUOTE (JSONSEARCH (meta, 'all', 'oldname')), 'newname') This works because the JSONSEARCH returns one path. In this case, one possible approach is to use an expression as path parameter for JSON_MODIFY. Your JSON is an array of JSON objects, so you need an index to access each element. ![]()
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