I am working on a small relational database for school and having trouble with a simple query. I am supposed to find all records in a table that have a field with the word 'OMG' somewhere in the text.
I have tried a few other and I can't figure it out. I am getting an invalid relational operator error.
my attempts:
select * from comments where contains('omg');
select * from comments where text contains('omg');
select * from comments where about('omg');
and several more variations of the above. As you can see I am brand spanking new to SQL.
text is the name of the field in the comments table.
Thanks for any help.
Assuming that the column name is text:
select * from comments where text like '%omg%';
The percentage % is a wild-card which means that any text can come before/after omg
Pay attention, if you don't want the results to contain words in which omg` is a substring - you might want to consider adding whitespaces:
select * from comments where text like '% omg %';
Of course that it doesn't cover cases like:
omg is at the end of the sentence (followed by a ".")
omg has "!" right afterwards
etc
You may want to use the LIKE operator with wildcards (%):
SELECT * FROM comments WHERE text LIKE '%omg%';
Related
My database is Oracle 11g.
I want to do a in query in sql. The query criteria is to matched some of the characters of a field:
description
CWLV321900017391;EFHU3832239
CWLV321900017491;ERHU3832239
CWLV321900017591;ERHU3832239
CWLV321900017691;ERHU3832239
My query is like this:
select * from product where description in ('CWLV321900017391', 'CWLV321900017491');
It returns no records in result.
I expect the result like below:
description
CWLV321900017391;EFHU3832239
CWLV321900017491;ERHU3832239
How to get it by SQL?
thanks.
You are using IN to search for a partial string match in the description table. This is not going to return the results, as IN will only match exact values.
Instead, one way to achieve this is to use a LIKE operator with %:
select *
from product
where (description LIKE 'CWLV321900017391%' OR
description LIKE 'CWLV321900017491%');
The % at the end indicates that anything can follow after the specified text.
This will return any description that starts with CWLV321900017391 or CWLV321900017491.
Incidentally, if your search term occurs anywhere in the description field, you will need to use a % at each end of the search term:
description LIKE '%CWLV321900017391%' OR description LIKE '%CWLV321900017491%'
There are various ways to solve it. Here is one:
select * from product
where instr (description, 'CWLV321900017391') > 0
or instr (description, 'CWLV321900017491') > 0;
If you know you're always searching for the start of the description you can use:
select * from product
where substr (description, 1, 16) in ('CWLV321900017391','CWLV321900017491')
Also, there's LIKE or REGEX_LIKE solution. It depends really on what strings you're searching for.
Of course none of these solutions is truly satisfactory, and for large volumes of data may exhibit suck-y performance. The problem is the starting data model violates First Normal Form by storing non-atomic values. Poor data models engender clunky SQL.
You can try below way -
select * from product
where substr(description,1,instr(description,';',1,1)-1) in ('CWLV321900017391', 'CWLV321900017491')
I am opposed to storing multiple values in a delimited string format. There are many better alternatives, but let me assume that you don't have a choice on the data model.
If you are stuck with strings in this format, you should take the delimiters into account. You can do this using like:
where ';' || description || ';' like '%;CWLV321900017391;%'
You can also do something similar with regexp_like() if you want to look for one of several values:
where regexp_like(';' || description || ';',
'(;CWLV321900017391;)|(;CWLV321900017391;)'
)
I am using Netezza.
One of my columns (COMMENTS) in my table (TABLE1) has text.
COMMENTS column is a free text column and can have any text in it.
Here are a few samples of how the text looks:
OLD STUDENT:N(Y/N/U=UNKNOWN) Some other text could be here
Some other text could be here OLD STUDENT:Y(Y/N/U=UNKNOWN)
OLD STUDENT:Y(Y/N/U=UNKNOWN) Some other text could be here
Note: Some records have "" between OLD STUDENT: and Y
I want to filter records based on text search.
The idea is to isolate records having the text "STUDENT:Y"
Here is what I have so far but is not working:
SELECT COMMENTS
FROM
TABLE1
WHERE
LOWER(COMMENTS) LIKE '%student%'
AND LOWER(COMMENTS) NOT LIKE '%student:n%'
How do I handle this?
Thank you for your time.
Even though there are valid records, WHERE LOWER(COMMENTS) LIKE '%student:y% where not returning any results. Some of the text fields have weird characters between the student: and y
Adding another single wildcard character "_" solved the issue.
SELECT COMMENTS
FROM
TABLE1
WHERE
LOWER(COMMENTS) LIKE '%student%'
AND LOWER(COMMENTS) NOT LIKE '%student:_n%'
I have column store_name (varchar). In that column I have entries like prime sport, best buy... with a space. But when user typed concatenated string like primesport without space I need to show result prime sport. how can I achieve this? Please help me
SELECT *
FROM TABLE
WHERE replace(store_name, ' ', '') LIKE '%'+#SEARCH+'%' OR STORE_NAME LIKE '%'+#SEARCH +'%'
Well, I don't have much idea, and even I am searching for it. But may be what I know works for you, You can achieve this by performing different type of string operations:
Mike can be Myke or Myce or Mikke or so on.
Cat an be Kat or katt or catt or so on.
For this you should write a function to generate number of possible strings and then form a SQL Query using all these, and query the database.
A similar kind of search in known as Soundex Search from Oracle and Soundex Search from Microsoft. Have a look of it. this may work.
And overall make use of functions like upper and lower.
Have you tried using replace()
You can replace the white space in the query then use like
SELECT * FROM table WHERE replace(store_name, ' ', '') LIKE '%primesport%'
It will work for entries like 'prime soft' querying with 'primesoft'
Or you can use regex.
Does anyone have a LIKE pattern that matches whole words only?
It needs to account for spaces, punctuation, and start/end of string as word boundaries.
I am not using SQL Full Text Search as that is not available. I don't think it would be necessary for a simple keyword search when LIKE should be able to do the trick. However if anyone has tested performance of Full Text Search against LIKE patterns, I would be interested to hear.
Edit:
I got it to this stage, but it does not match start/end of string as a word boundary.
where DealTitle like '%[^a-zA-Z]pit[^a-zA-Z]%'
I want this to match "pit" but not "spit" in a sentence or as a single word.
E.g. DealTitle might contain "a pit of despair" or "pit your wits" or "a pit" or "a pit." or "pit!" or just "pit".
Full text indexes is the answer.
The poor cousin alternative is
'.' + column + '.' LIKE '%[^a-z]pit[^a-z]%'
FYI unless you are using _CS collation, there is no need for a-zA-Z
you can just use below condition for whitespace delimiters:
(' '+YOUR_FIELD_NAME+' ') like '% doc %'
it works faster and better than other solutions. so in your case it works fine with "a pit of despair" or "pit your wits" or "a pit" or "a pit." or just "pit", but not works for "pit!".
I think the recommended patterns exclude words with do not have any character at the beginning or at the end. I would use the following additional criteria.
where DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like 'pit[^a-z]%' OR
DealTitle like '%[^a-z]pit'
I hope it helps you guys!
Surround your string with spaces and create a test column like this:
SELECT t.DealTitle
FROM yourtable t
CROSS APPLY (SELECT testDeal = ' ' + ISNULL(t.DealTitle,'') + ' ') fx1
WHERE fx1.testDeal LIKE '%[^a-z]pit[^a-z]%'
If you can use regexp operator in your SQL query..
For finding any combination of spaces, punctuation and start/end of string as word boundaries:
where DealTitle regexp '(^|[[:punct:]]|[[:space:]])pit([[:space:]]|[[:punct:]]|$)'
Another simple alternative:
WHERE DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like '[^a-z]pit[^a-z]%' OR
DealTitle like '%[^a-z]pit[^a-z]'
This is a good topic and I want to complement this to someone how needs to find some word in some string passing this as element of a query.
SELECT
ST.WORD, ND.TEXT_STRING
FROM
[ST_TABLE] ST
LEFT JOIN
[ND_TABLE] ND ON ND.TEXT_STRING LIKE '%[^a-z]' + ST.WORD + '[^a-z]%'
WHERE
ST.WORD = 'STACK_OVERFLOW' -- OPTIONAL
With this you can list all the incidences of the ST.WORD in the ND.TEXT_STRING and you can use the WHERE clausule to filter this using some word.
You could search for the entire string in SQL:
select * from YourTable where col1 like '%TheWord%'
Then you could filter the returned rows client site, adding the extra condition that it must be a whole word. For example, if it matches the regex:
\bTheWord\b
Another option is to use a CLR function, available in SQL Server 2005 and higher. That would allow you to search for the regex server-side. This MSDN artcile has the details of how to set up a dbo.RegexMatch function.
Try using charindex to find the match:
Select *
from table
where charindex( 'Whole word to be searched', columnname) > 0
I have a field that has values like this...
s:10:"03/16/1983";
s:4:"Male";
s:2:"No";
I'd like to parse out the quoted values.
its going to be some sort of combination of substr and instr
its the doublequote i have issues finding its position.
i have tried things like select substr(field_value, instr(field_value,'"'),instr(field_value,'"',null,2)) from table where etc
apologies a noob question...
Here's something that should work (unable to test at the moment):
select substr(substr(field_value, instr(field_value,':')+1, CHAR_LENGTH(field_value)-1),
instr(substr(field_value, instr(field_value,':')+1, CHAR_LENGTH(field_value)-1),':')+1)
Edit: Putting my comment in the answer:
select substr(field_value, instr(field_value,'\"'),CHAR_LENGTH(field_value)-1)