What query could I use in sqlite to get the names of columns beginning with (for example) "thing" in a DB
Such as if they were formatted like this:
"thing_column1"
"thing_column2"
"thing_data"
etc.
You can use pragma_table_info() with the table's name:
SELECT name
FROM pragma_table_info('tablename')
WHERE name LIKE 'thing%'
You can use this query:
SELECT GROUP_CONCAT(name) AS columns
FROM pragma_table_info('tablename')
WHERE name LIKE 'thing%'
which returns only 1 column columns with a string value like 'thing_column1,thing_column2,thing_column3' and you can use it to construct a SELECT statement in your application.
Related
i have a table column looks like below.
what is the sql query statement i can use to have multiple partial match conditions?
search by ID or Name
if search abc then list the row A1 , row A2
if search test then list the row A1 , row A2, row 3
if search ghj then list the row A2
i was trying this but nothing return:
SELECT * FROM table where colB LIKE '"ID":"%abc%"'
updating data in text
{"ItemId":"123","IDs":[{"ID":"abc","CodingSystem":"cs1"}],"Name":"test itemgh"}
{"ItemId":"123","IDs":[{"ID":"ghj","CodingSystem":"cs1"}],"Name":"test abc"}
{"ItemId":"123","IDs":[{"ID":"defg","CodingSystem":"cs1"}],"Name":"test 111"}
JSON parsing
Oracle
Looked into the JSON parsing capabilities of Oracle and I managed to make running a query like this:
select * from table t where json_exists(t.colB, '$.IDs[?(#.ID=="abc")]') or json_exists(t.colB, '$.IDs?(#.name=="abc"')
And inside the same JSON query expression:
select * from table t where json_exists(t.colB, '$.IDs[?(#.ID=="abc" || #.name=="abc")]')
The call of function json_exists() is the key to this.
The first parameter can be a VARCHAR2, and I also tried with a BLOB containing text, and it works.
The second parameter is the path to your json object attribute that needs to be tested, with the condition.
I wrote two ORed conditions for the ID and for the Name, but maybe there is a better JSON query expression you can use to include them both.
More information about json_exists() function here.
Postgres
There is a JSON datatype in Postgres that supports parsing in queries.
So, if your colB column is declared as JSON you can do something like this:
select * from table where colB->>'Name' LIKE '%abc%';
And in order to have available the array elements of the IDs array, you should use the function json_array_elements().
select * from table, json_array_elements(colB->'IDs') e where colB->>'Name' LIKE '%abc%' or e->>'ID' = 'abc';
Check an example I created for you here.
Here is an online tool for online testing your JSON queries.
Check also this question in SO.
MSSQL Server 2017
I made a couple of tests also with MS SQL Server, and I managed to create an example searching for partial matching in the name field.
select * from table where JSON_VALUE(colB,'$.Name') LIKE '%abc%';
And finally I arrived to a working query that does partial match to the Name field and full match to the ID field like this:
select * from table t
CROSS APPLY OPENJSON(colB, '$.IDs') WITH (
ID VARCHAR(10),
CodingSystem VARCHAR(10)
) e
where JSON_VALUE(t.colB,'$.Name') LIKE '%abc%'
or e.ID = 'abc';
The problem is that we need to open the IDs array, and make something like a table from it, that can be queried also by accessing its columns.
The example I created is here.
LIKE text query
Your tries are good but you misplace the % symbols. They have to be first and last in your given string:
If you want the ID to be the given value:
SELECT * FROM table where colB LIKE '%"ID":"abc"%'
If the given value can be anywhere, then don't put the "ID" part:
SELECT * FROM table where colB LIKE '%abc%'
If the given value can be only on the ID or Name field then:
SELECT * FROM table where colB LIKE '%"ID":"abc"%' OR colB LIKE '%"Name":"abc"%'
And because you are giving hard-coded identifiers of fields (eg ID and Name) that can be in variable case:
SELECT * FROM table where lower(colB) LIKE '%"id":"abc"%' OR lower(colB) LIKE '%"name":"abc"%'
Assuming that the number of spaces do not vary between the : character and the value or the name of the properties.
For partial matching you can use more % in between like '%"name":"%abc%"%':
SELECT * FROM table where lower(colB) LIKE '%"id":"abc"%' OR lower(colB) LIKE '%"name":"%abc%"%'
Regular Expressions
A different option would be to test with regular expressions.
Consider checking this: Oracle extract json fields using regular expression with oracle regexp_substr
So I want to do a simple select like so:
select * from table
But this table has, for example, 100 columns. I'll only need the columns where "substring" is in the name, like say table has 3 columns: "firstPoints", "secondPoints", "thirdPoints", and then 97 columns without "Points" in the name, is there a way to do a select that only selects the columns that have the substring "Points" in it's name? Like:
select * like '%Points%' from table
Or something like that, so that, without explicitly passing the columns I want, I still get the select but only of those 3 columns that has "Points" in their name.
The Answer is like this:
SELECT TOP 1000 COLUMN_NAME
FROM {databasename}.INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME = '{tablename}' and COLUMN_NAME like '%Datum%'
This will return a column "COLUMN_NAME" with records of my column names on that table containing "Datum" for example, in my example on my DB it returns 6 column names.
I am trying to have the user search for a value in a SQL table, and the user is returned with any row that contains that value. At the moment, I can make it work such that the code is:
SELECT * FROM table WHERE lower('foo') in (lower('col1'),lower('col2'),etc)
However, I would like it to be able to search every column and return any row LIKE 'foo'. For instance,
SELECT * FROM table WHERE (lower('col1'), lower('col2'), etc) like lower('%foo%')
But that doesn't work.
Any suggestions?
I believe you need to use multiple WHERE clauses instead of grouping them all into one statement. Try this:
SELECT * FROM table
WHERE lower(col1) like lower('%foo%')
OR lower(col2) like lower('%foo%')
OR etc like lower('%foo%')
You can convert the whole row to a string and then use LIKE on the result of that:
select *
from the_table
where lower(the_table::text) like '%foo%';
the_table::text returns all columns of each row as a comma separated list enclosed with parentheses, e.g. (42,Arthur,Dent). So the above is not 100 identical to a LIKE condition applied on each column - but probably does what you want.
I have written an sqlite3 query to execute the below,
cur.execute("select * from table where column like '%s'"% (variable1))
which gives the strings the completely match with the variable1.
But I want an sqlite3 query that can show:
cur.execute("select * from table where column *contains* '%s'"% (variable1))
What query should I write find any string that contains 'variable1'.
Any help would be appreciable.
To ignore characters, you have to put percent signs into the LIKE pattern (and double them so that the % operator does not try to replace them):
cur.execute("select * from table where column like '%%%s%%'" % (variable1))
But direct string interpolation brings formatting problems and possibly SQL injections. Better use a parameter:
cur.execute("select * from table where column like ?", ['%' + variable1 + '%'])
Try
cur.execute("""select * from table where column like '%{}%' """.format(variable))
Refer this link for sql
SQL SELECT WHERE field contains words
See, LIKE functionality is simple.
If you want to search a string, you will have to place it in single quotes:
search all names starting with s
select name from names where name like 's%';
search all names ending with s
select name from names where name like '%s';
search all names that contain s
select name from names where name like '%s%';
But if you have stored your string in a variable var
You just have to do this:
select name from names where name like var;
Hello i have a table with colums called NAME, in this colum i can have this type of name : First name Uppercase and other lower (Jack), all name uppercase (JACK), and name with space (Jack ) or (JACK ).
How can show all name than have jack in all type ?
i need to search multiple name with IN statement
i use ORACLE
Assuming SQL Server:
SELECT Name FROM Table WHERE Name LIKE '%jack%'
SELECT Name
FROM Table
WHERE UPPER(Name) LIKE '%JACK%'
will work in all SQL. In SQL Server I believe LIKE is case insensitive but it might depend on configuration settings and collation.
or maybe you need
SELECT Name
FROM Table
WHERE UPPER(TRIM(Name)) IN ('JACK','HOGAN','VICTOR')
(replace TRIM(Name) with RTRIM(LTRIM(Name)) if not using SQL Server.)
If you have a table with the list of names you can do it like this:
SELECT Name
FROM Table
JOIN NameTable ON UPPER(RTRIM(LTRIM(Table.Name))) = NameTable.Name