I'm trying to filter the String value, but it's not supported in VisualVM. How can I do that?
You can use OQL to filter the strings. For example the following query will show strings containing 'AWT'
select s from java.lang.String s where s.toString().contains("AWT")
More information about OQL can found in this document.
Related
I am currently working on building a dataware house in snowflake for the business that i work for and i have encounter some problems. I used to apply the function Json_value in TSQL for extracting certain key/value pair from json format field inside my original MSSQL DB.
All the other field are in the regular SQL format but there is this one field that i really need that is formated in JSON and i can't seems to exact the key/value pair that i need.
I'm new to SnowSQL and i can't seems to find a way to extract this within a regular query. Does anyone knows a way around my problem ?
* ID /// TYPE /// Name (JSON_FORMAT)/// Amount *
1 5 {En: "lunch, fr: "diner"} 10.00
I would like to extract this line (for exemple) and be able to only retrieve the EN: "lunch" part from my JSON format field.
Thank you !
Almost any time you use JSON in Snowflake, it's advisable to use the VARIANT data type. You can use the parse_json function to convert a string into a variant with JSON.
select
parse_json('{En: "lunch", fr: "diner"}') as VARIANT_COLUMN,
VARIANT_COLUMN:En::string as ENGLISH_WORD;
In this sample, the first column converts your JSON into a variant named VARIANT_COLUMN. The second column uses the variant, extracting the "En" property and casting it to a string data type.
You can define columns as variant and store JSON natively. That's going to improve performance and allow parsing using dot notation in SQL.
For anyone else who also stumbles upon this question:
You can also use JSON_EXTRACT_PATH_TEXT. Here is an example, if you wanted to create a new column called meal.
select json_extract_path_text(Name,'En') as meal from ...
I need to extract a field from a JSON string with MariaDB and search for specific patterns in that field.
This field is just a property of all the properties the JSON object has. I had read the documentation and I saw the JSON_EXTRACT function. I am still a newbie with databases so I would like some help in this matter.
{"user_id":"1","status_id":"1","text":"Hello, world"}
Lets say I want to get all the "text" values that have the "world" in the database table. I can extract with JSON_EXTRACT. But I want patterns, not absolute values.
How can I do that?
You can extract the value with json_extract(), and then do pattern matching with like:
select t.*
from mytable t
where json_extract(my_json_col, '$.text') like '%world%'
I have totally rewritten my question because of inaccurate description of the problem!
We have to store a lot of different informations about a specific region. For this we need a flexible data structure which does not limit the possibilities for the user.
So we've create a key-value table for this additional data which is described through a meta table which contains the datatype of the value.
We already use this information for queries over our rest api. We then automatically wrap the requested field with into a cast.
SQL Fiddle
We return this data together with information form other tables as a JSON object. We convert the corresponding rows from the data-table with array_agg and json_object into a JSON object:
...
CASE
WHEN count(prop.name) = 0 THEN '{}'::json
ELSE json_object(array_agg(prop.name), array_agg(prop.value))
END AS data
...
This works very well. Now the problem we have is if we store data like a floating point number into this field, we then get returned a string representation of this number:
e.g. 5.231 returns as "5.231"
Now we would like to CAST this number during our select statement into the right data-format so the JSON result would be correctly formatted. We have all the information we need so we tried following:
SELECT
json_object(array_agg(data.name),
-- here I cast the value into the right datatype!
-- results in an error
array_agg(CAST(value AS datatype))) AS data
FROM data
JOIN (
SELECT name, datatype
FROM meta)
AS info
ON info.name = data.name
The error message is following:
ERROR: type "datatype" does not exist
LINE 3: array_agg(CAST(value AS datatype))) AS data
^
Query failed
PostgreSQL said: type "datatype" does not exist
So is it possible to dynamically cast the text of the data_type column to a postgresql type to return a well-formatted JSON object?
First, that's a terrible abuse of SQL, and ought to be avoided in practically all scenarios. If you have a scenario where this is legitimate, you probably already know your RDBMS so intimately, that you're writing custom indexing plugins, and wouldn't even think of asking this question...
If you tell us what you're actually trying to do, there's about a 99.9% chance we can tell you a better way to do it.
Now with that disclaimer aside:
This is not possible, without using dynamic SQL. With a sufficiently recent version of PostgreSQL, you can accomplish this with the use of 'EXECUTE IMMEDIATE', which you can read about in the manual. It basically boils down to using EXEC.
Note, however, that even using this method, the result for every row fetched in the same query must have the same data type. In other words, you can't expect that row 1 will have a data type of VARCHAR, and row 2 will have INT. That is completely impossible.
The problem you have is, that json_object does create an object out of a string array for the keys and another string array for the values. So if you feed your JSON objects into this method, it will always return an error.
So the first problem is, that you have to use a JSON or JSONB column for the values. Or you can convert the values from string to json with to_json().
Now the second problem is that you need to use another method to create your json object because you want to feed it with a string array for the keys and a json-object array for the values. For this there is a method called json_object_agg.
Then your output should be like the one you expected! Here the full query:
SELECT
json_object_agg(data.name, to_json(data.value)) AS data
FROM data
I put some fields like manufacturer, group, description, num.
SimpleFaceted works ok if I use query like sometext* with QueryParser.
Im trying to use num:[100 TO 200],
num is NumericField with SetIntValue(150).
I got nothing returned.
Am I missing something?
You can't use a normal query parser for numeric range queries. However, assuming you know at query time which fields are numeric, it isn't too hard to derive a class from the Lucene query parser and create numeric range queries as necessary.
Support for numeric queries in the standard query parser looks like it should be available in Lucene.Net when a port of v3.4 is available. (See Java Lucene issue 1768)
Good luck,
Using nHibernate, I would like to query on an integer datatype, but its always returning the exact match.
How could I write an expression that returns a list starting with the number entered?
right now I am using it as:
(clientNum is a long)
crit.Add(Expression.Like("ClientNumber", clientNum)); //this always gives me exact matches only
so I tried the following, but its complainging of a wroing type (its only expecting a string)
crit.Add(Expression.Like("ClientNumber", clientNum, MatchMode.Start));
Update: Also I tried the clientNum.ToString() but I get a db exception saying invalid type.
I can use the sql as follows to get what I want, but how do I do this in nHibernate??
SELECT * FROM ClientTable
WHERE clientNum LIKE '3%' --incase I wanted a list that starts with 3...
I am not sure about the .NET version, but hybernate supported limited casting (e.g. cast (... as ...)) in HQL. I think you would need to cast the value to a string and then apply your 'Like' clause. Should be able to do this all in HQL. Alternatively, you can do it using the SQL Expressions (ex. Expression.Sql(...)) and do the same thing using T-SQL.