select * from table where column like '%a|b%'
The above query matches all rows with the column having either 'a' OR 'b' as a substring.
What if I want to match the substring "a|b"?
Using the query,
select * from table where column like '%a\|b%'
yields the same result.
Can I get the complete reference for the LIKE operator in hive? The UDF manual seems insufficient.
You can use RLIKE (regular expression) : select * from table where column rlike '.*a\|b.*'
You can use select * form table where column like '%a[|]b%'
Related
I am looking a way to select all columns name that "startwith" a specific character. My data contains the same column name multiple time with a digit number at the end and I want the code to always select all the columns regardless the last digit numbers.
For example, if I have 3 kinds of apple in my column names, the dataset will contains the column: "apple_1", "apple_2" and "apple_3". Therefore, I want to select all columns that startwith "apple_" in a proc sql statement.
Thanks you
In regular SAS code you can use : as a wildcard to create a variable list. You normally cannot use variable lists in SQL code, but you can use them in dataset options.
proc sql ;
create table want as
select *
from mydata(keep= id apple_: )
;
quit;
Use like:
proc sql;
select t.*
from t
where col like 'apple%';
If you want the _ character as well, you need to use the ESCAPE clause, because _ is a wildcard character for LIKE:
proc sql;
select t.*
from t
where col like 'apple$_%' escape '$';
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 table called table with column called column with datatype text with values like '["1","2"]'.
I need to get all records which has "1" as one of the element.
select *
from table
where column.....?
How should the where clause be?
Simply use LIKE. Keep the double quotes to pass 1, but avoid other numbers containing that digit.
select *
from table
where column like '%"1"%'
I think you can use ? operator on jsonb type:
select *
from (
select '["1","2"]' union all
select '["0"]'
) as a(data)
where
a.data::jsonb ? '1'
In general, I'd consider storing your data as jsonb instead of string.
db<>fiddle example
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 want to put a condition in my query where I have a column that should contain second position as an alphabet.
How to achieve this?
I've tried with _[A-Z]% in where clause but is not working. I've also tried [A-Z]%.
Any inputs please?
I think you want mysql query. like this
SELECT * FROM table WHERE column REGEXP '^.[A-Za-z]+$'
or sql server
select * from table where column like '_[a-zA-Z]%'
You can use regular expression matching in your query. For example:
SELECT * FROM `test` WHERE `name` REGEXP '^.[a-zA-Z].*';
That would match the name column from the test table against a regex that verifies if the second character is either a lowercase or uppercase alphabet letter.
Also see this SQL Fiddle for an example of data it does and doesn't match.
agree with #Gordon Linoff, your ('_[A-Z]%') should work.
if not work, kindly add some sample data with your question.
Declare #Table Table
(
TextCol Varchar(20)
)
Insert Into #Table(TextCol) Values
('23423cvxc43f')
,('2eD97S9')
,('sAgsdsf')
,('3Ss08008')
Select *
From #Table As t
Where t.TextCol Like '_[A-Z]%'
The use of '%[A-Z]%' suggests that you are using SQL Server. If so, you can do this using LIKE:
where col like '_[A-Z]%'
For LIKE patterns, _ represents any character. If the first character needs to be a digit:
where col like '[0-9][A-Z]%'
EDIT:
The above doesn't work in DB2. Instead:
where substr(col, 2, 1) between 'A' and 'Z'