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
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
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
I need to find the '&' in a string.
SELECT * FROM TABLE WHERE FIELD LIKE ..&...
Things we have tried :
SELECT * FROM TABLE WHERE FIELD LIKE '&&&'
SELECT * FROM TABLE WHERE FIELD LIKE '&\&&'
SELECT * FROM TABLE WHERE FIELD LIKE '&|&&' escape '|'
SELECT * FROM TABLE WHERE FIELD LIKE '&[&]&'
None of these give any results in SQLServer.
Well some give all rows, some give none.
Similar questions that didn't work or were not specific enough.
Find the % character in a LIKE query
How to detect if a string contains special characters?
some old reference Server 2000
http://web.archive.org/web/20150519072547/http://sqlserver2000.databases.aspfaq.com:80/how-do-i-search-for-special-characters-e-g-in-sql-server.html
& isn't a wildcard in SQL, therefore no escaping is needed.
Use % around the value your looking for.
SELECT * FROM TABLE WHERE FIELD LIKE '%&%'
Your statement contains no wildcards, thus is equivalent to WHERE FIELD = '&'.
& isn't a special character in SQL so it doesn't need to be escaped. Just write
WHERE FIELD LIKE '%&%'
to search for entries that contain & somewhere in the field
Be aware though, that this will result in a full table scan as the server can't use any indexes. Had you typed WHERE FIELD LIKE '&%' the server could do a range seek to find all entries starting with &.
If you have a lot of data and can't add any more constraints, you should consider using SQL Server's full-text search to create and use and FTS index, with predicates like CONTAINS or FREETEXT
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.
In phpMyAdmin, there is a search in database feature by which I can input a word and find it in any table(s).
How to implement it by SQL statement? I know the LIKE operation, but it's syntax is:
WHERE column_name LIKE pattern
How to search in all columns? Any how to specify it's a exact keyword or regular express?
SELECT * FROM your_table_name WHERE your_column_name LIKE 'search_box_text';
Where search_box_text is what you enter in the search. It will also say in the result page what kind of query it made. The same query with regular expressions is:
SELECT * FROM your_table_name WHERE your_column_name REGEXP 'search_box_text';
Remember that the wildcard in mysql is %. Eg. "LIKE '%partial_search_text%'
If you want to search in multiple columns, you can check which columns are in table with:
DESCRIBE TABLE your_table_name;
Or if you already know your columns:
SELECT * FROM your_table_name
WHERE your_column_1 LIKE '%search%'
AND your_column_2 LIKE '%search%'
AND your_column_3 LIKE '%search%';