I am querying a Google Fusion table as follows:
SELECT * FROM {mytableIDhere} WHERE col1 LIKE '%{mystringhere}%'
This shows all row data for entries containing {mystringhere} in column 1. Is it possible to use a * wildcard on the column field, so that the search applies to ANY column? I tried that, but I get a parse error when doing:
SELECT * FROM {mytableIDhere} WHERE * LIKE '%{mystringhere}%'
It seems that the location parameter cannot be open-ended. Is this correct? If so, is there a workaround?
EDIT: In terms of desired functionality, I am trying to create a global search of a table so that any rows containing the search query will be returned. The columns include e-mail addresses, ID numbers, commentary, and other values.
Related
I have a table named Posts I would like to count and profile in Snowflake using the current Snowsight UI.
When I return the results via EXPLAIN using TABLULAR I am able to return the set with the combination of TABLE, RESULT_SCAN, and LAST_QUERY_ID functions, but any predicate or filter or column reference seems to fail.
Is there a valid way to do this in Snowflake with the TABLE function or is there another way to query the output of the EXPLAIN using TABLULAR?
-- Works
EXPLAIN using TABULAR SELECT COUNT(*) from Posts;
-- Works
SELECT t.* FROM TABLE(RESULT_SCAN(LAST_QUERY_ID())) as t;
-- Does not work
SELECT t.* FROM TABLE(RESULT_SCAN(LAST_QUERY_ID())) as t where operation = 'GlobalStats';
-- invalid identifier 'OPERATION', the column does not seem recognized.
Tried the third example and expected the predicate to apply to the function output. I don't understand why the filter works on some TABLE() results and not others.
You need to double quote the column name
where "operation"=
From the Documentation
Note that because the output column names from the DESC USER command
were generated in lowercase, the commands use delimited identifier
notation (double quotes) around the column names in the query to
ensure that the column names in the query match the column names in
the output that was scanned
I want to return records based on a wildcard search from a table of keywords
Current Table
Message
Some Message
Keyword Table
Words
SomeWord1
SomeWord2
Currently I am writing a query on something like this
SELECT * FROM table WHERE message LIKE '%SomeWord1%' OR message LIKE '%SomeWord2%'
Is there a more elegant way to do this wild card search using sort of subquery
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
I have a Redshift table column that contains 1 to many hashtags (e.g. #a, #b, etc.). I want to write a query that finds rows where all tags from a given set exist (e.g. #a and #b) while not picking up other rows that have some but not all of the tags (e.g. only #a or only #b).
I can see how to do this with multiple LIKE statements (e.g. LIKE '%#a %' AND LIKE '%#b%') but I would really like to do it with a single statement. I can see how to do this with SIMILAR TO but not in a way that ignores ordering. The following would work but only if I include all possible combinations of ordering.
SELECT * FROM table WHERE field SIMILAR TO '(%#a%)(%#b%)|(%#b%)(%#a%)'
This works but having to list all combinations of the tags I'm looking for would be a royal pain and prone to error. Is there a way to express 'AND' in SIMLAR TO (or another function) in Redshift that ignores order?
Make sure to capture the whole tag in any position and not match on incomplete tags:
SELECT *
FROM table
WHERE (field LIKE '#a#%' OR field LIKE '%#a') AND
(field LIKE '#b#%' OR field LIKE '%#b')
This avoids matching data such as #ac#b
Use AND and LIKE:
SELECT t.*
FROM table t
WHERE field LIKE '%#a%' AND
field LIKE '%#b%';
My overall goal is to create new tables by selecting columns in existing tables with certain patterns/tags in their column names. This is in SQL Server.
For example, I have a common need to get all contact information out of a company table, and into its own contact table.
I haven't been able to find a programmatic approach in SQL to express excluding columns from a SELECT statement based on string.
When looking to options like the COL_NAME function, those require an ID arg, which kills that option for me.
Wishing there was something built in that could work like the following:
SELECT *
FROM TABLE
WHERE COL_NAME() LIKE 'FLAG%'
Any ideas? Open to anything! Thanks!!
One trick could be to firstly using the following code to get the column names you desire:
select * from information_schema.columns
where table_name='tbl' and column_name like 'FLAG%'
Then concatenate them into a comma-delimited string and finally create a dynamic sql query using the created string as the column names.