I need help for sql query LIKE.
Value for column in database is same below:
record 1 : "3,13,15,20"
record 2 : "13,23,14,19"
record 3 : "3,14,15,19,20"......
for now I want to get the most accurate record with a value of 3
This is my query :
SELECT * FROM accounts where type like '%3%'
This query will find all record with value exist is '3' eg: 13,23 ....
And It does not solve my problem.
Try this:
SELECT *
FROM accounts
WHERE CONCAT(',', type, ',') LIKE '%,3,%';
Demo
This trick places commas around the end of the type CSV string, so that we all we have to do is then check for ,3, anywhere in that string.
By the way, it is generally not desirable to store CSV data like this in your SQL tables. Instead, consider normalizing your data and storing those CSV values across separate rows.
Related
I have table with data like this:
Id | StringValue
----+-------------
1 | 4,50
2 | 90,40
I will get input StringValue like 4. I need to fetch the data exact matched record. When I am using LIKE operator, select query is returning two rows, but I need exact matched data record only.
Can anybody please help me with this?
SELECT *
FROM Table1
WHERE StringValue like '%4%'
But that returns two rows - both ID 1 and 2.
My expectation is I need to get ID = 1 row only
Storing delimited data like this is a well documented anti-pattern, violates basic normalisation principles and prevents the database engine from fully utilising an index.
What you can do is delimit your search value and also ensure the expression to search is correctly delimited; this is an unsargable expression however and the strorage engine will have to scan all rows every time -
declare #valueToFind varchar(10) = '4';
select *
from t
where Concat(',', t.StringValue, ',') like Concat('%,' #valueToFind, ',%');
for SQL Server 2016 and later you can use STRING_SPLIT or earlier version of SQL Server, there are many alternative, just do a search for it.
Or, you can simply do
SELECT * FROM Table1 where ',' + StringValue + ',' like '%,4,%'
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 numbers column in my sql table and I want to pull all the numbers that start with 3 specific digits; how would I query this?
Use the "LIKE" query:
123 is the prefix. I think you have to store the numbers as strings though, just try it out on your data set :-)
SELECT * from TableName Where ColumnName LIKE '123%'
See also this Q&A: In MySql, find strings with a given prefix
I would like to ask if it is possible to do this:
For example the search string is '009' -> (consider the digits as string)
is it possible to have a query that will return any occurrences of this on the database not considering the order.
for this example it will return
'009'
'090'
'900'
given these exists on the database. thanks!!!!
Use the Like operator.
For Example :-
SELECT Marks FROM Report WHERE Marks LIKE '%009%' OR '%090%' OR '%900%'
Split the string into individual characters, select all rows containing the first character and put them in a temporary table, then select all rows from the temporary table that contain the second character and put these in a temporary table, then select all rows from that temporary table that contain the third character.
Of course, there are probably many ways to optimize this, but I see no reason why it would not be possible to make a query like that work.
It can not be achieved in a straight forward way as there is no sort() function for a particular value like there is lower(), upper() functions.
But there is some workarounds like -
Suppose you are running query for COL A, maintain another column SORTED_A where from application level you keep the sorted value of COL A
Then when you execute query - sort the searchToken and run select query with matching sorted searchToken with the SORTED_A column
Does SQLite offer a way to search every column of a table for a searchkey?
SELECT * FROM table WHERE id LIKE ...
Selects all rows where ... was found in the column id. But instead to only search in the column id, I want to search in every column if the searchstring was found. I believe this does not work:
SELECT * FROM table WHERE * LIKE ...
Is that possible? Or what would be the next easy way?
I use Python 3 to query the SQLite database. Should I go the route to search through the dictionary after the query was executed and data returned?
A simple trick you can do is:
SELECT *
FROM table
WHERE ((col1+col2+col3+col4) LIKE '%something%')
This will select the record if any of these 4 columns contain the word "something".
No; you would have to list or concatenate every column in the query, or reorganize your database so that you have fewer columns.
SQLite has full-text search tables where you can search all columns at once, but such tables do not work efficiently with any other queries.
I could not comment on #raging-bull answer. So I had to write a new one. My problem was, that I have columns with null values and got no results because the "search string" was null.
Using coalesce I could solve that problem. Here sqlite chooses the column content, or if it is null an empty string (""). So there is an actual search string available.
SELECT *
FROM table
WHERE (coalesce(col1,"") || coalesce(col2,"") || coalesce(col3,"") || coalesce(col4,"")) LIKE '%something%')
I'm not quite sure, if I understood your question.
If you want the whole row returned, when id=searchkey, then:
select * from table where id=searchkey;
If you want to have specific columns from the row with the correct searchkey:
select col1, col2, col3 from table where id=searchkey;
If you want to search multiple columns for the "id": First narrow down which columns this could be found in - you don't want to search the whole table! Then:
select * from table where col1=searchkey or col2=searchkey or col3=searchkey;