Hope one can help me and explain this query for me,
why the first query return result but the second does not:
EDIT:
first query:
select name from Items where name like '%abc%'
second Query:
select name from Items where name like substring('''%abc%''',1,10)
why the first return result but the second return nothing while
substring('''%abc%''',1,10)='%abc%'
If there are a logic behind that, Is there another approach to do something like the second query,
my porpuse is to transform a string like '''abc''' to 'abc' in order to use like statement,
You can concatenate strings to form your LIKE string. To trim the first 3 and last 3 characters from a string use the SUBSTRING and LEN functions. The following example assumes your match string is called #input and starts and ends with 3 quote marks that need to be removed to find a match:
select name from Items where name like '%' + SUBSTRING(#input, 3, LEN(#input) - 4) + '%'
Related
I need help with one of my current tasks wherein i am trying to pick only the table names from the query via Python
So basically lets say a query looks like this
Create table a.dummy_table1
as
select a.dummycolumn1,a.dummycolumn2,a.dummycolumn3 from dual
Now i am passing this query into Python using STRINGIO and then reading only the strings where it starts with "a" and has "_" in it like below
table_list = set(re.findall(r'\ba\.\w+', str(data)))
Here data is the dataframe in which i have parsed the query using StringIO
now in table_list i am getting the below output
a.dummy_table1
a.dummycolumn1
a.dummycolumn2
whereas the Expected output should have been like
a.dummy_table1
<Let me know how we can get this done , have tried the above regular expression but that is not working properly>
Any help on same would be highly appreciated
Your current regex string r"\ba.\w+" simply matches any string which:
Begins with "a" (the "\ba" part)
Followed by a period (the "." part)
Followed by 1 or more alphanumeric characters (the "\w+" part).
If I've understood your problem correctly, you are looking to extract from str(data) any string fragments which match this pattern instead:
Begins with "a"
Followed by a period
Followed by 1 or more alphanumeric characters
Followed by an underscore
Followed by 1 or more alphanumeric characters
Thus, the regular expression should have "_\w+" added to the end to match criteria 4 and 5:
table_list = set(re.findall(r"\ba\.\w+_\w+", str(data)))
If I have a string for example:
"this.is.a.string.and.I.need.the.last.part"
I am trying to get the last part of the string after the last ".", which in this case is "part"
How to I achieve this?
One way I tried was to split the string on ".", I get a array back, but then I don't know how to retrieve the last item in the array.
| extend ToSplitstring = split("this.is.a.string.and.I.need.the.last.part", ".")
gives me:
["this", "is","a","string","and","I","need","the","last", "part"]
and a second try I have tried this:
| extend ToSubstring = substring(myString, lastindexof(myString, ".")+1)
but Kusto do not have a function of lastindexof.
Anyone with tips?
you can access the last member of the array using a negative index -1.
e.g. this:
print split("this.is.a.string.and.I.need.the.last.part", ".")[-1]
returns a single table, with a single column and a single record, with the value part
You can try the code below, and feel free to change it to meet your need:
let lastIndexof = (input:string, lookup: string) {
indexof(input, lookup, 0, -1, countof(input,lookup))
};
your_table_name
| extend ToSubstring = substring("this.is.a.string.and.I.need.the.last.part", lastIndexof("this.is.a.string.and.I.need.the.last.part", ".")+1)
I have strings which looks like this [NAME LASTNAME/NAME.LAST#emailaddress/123456678]. What I want to do is parse strings which have the same format as shown above so I only get NAME LASTNAME. My psuedo idea is find the index of the first instance of /, then strip from index 1 to that index of / we found. I want this as a VBScript.
Your way should work. You can also Split() your string on / and just grab the first element of the resulting array:
Const SOME_STRING = "John Doe/John.Doe#example.com/12345678"
WScript.Echo Split(SOME_STRING, "/")(0)
Output:
John Doe
Edit, with respect to comments.
If your string contains the [, you can still Split(). Just use Mid() to grab the first element starting at character position 2:
Const SOME_STRING = "[John Doe/John.Doe#example.com/12345678]"
WScript.Echo Mid(Split(SOME_STRING, "/")(0), 2)
Your idea is good here, you should also need to grab index for "[".This will make script robust and flexible here.Below code will always return strings placed between first occurrence of "[" and "/".
var = "[John Doe/John.Doe#example.com/12345678]"
WScript.Echo Mid(var, (InStr(var,"[")+1),InStr(var,"/")-InStr(var,"[")-1)
I have a string to modify as per the requirements.
For example:
The given string is:
str1 varchar = '123,456,789';
I want to show the string as:
'456,789'
Note: The first part (delimited) with comma, I want to remove from string and show the rest of string.
In SQL Server I used STUFF() function.
SELECT STUFF('123,456,789',1,4,'');
Result:
456,789
Question: Is there any string function in PostgreSQL 9.3 version to do the same job?
you can use regular expressions:
select substring('123,456,789' from ',(.*)$');
The comma matches the first comma found in the string. The part inside the brackets (.*) is returned from the function. The symbol $ means the end of the string.
A alternative solution without regular expressions:
select str, substring(str from position(',' in str)+1 for length(str)) from
(select '123,456,789'::text as str) as foo;
You could first turn the string to array and return second and third cell:
select array_to_string((regexp_split_to_array('123,456,789', ','))[2:3], ',')
Or you could use substring-function with regular expressions (pattern matching):
SELECT substring('123,456,789' from '[0-9]+,([0-9]+,[0-9]+)')
[0-9]+ means one or more digits
parentheses tell to return that part from the string
Both solutions work on your specific string.
Your The SQL Server example indicates you just want to remove the first 4 characters, which makes the rest of your question seem misleading because it completely ignores what's in the string. Only the positions matters.
Be that as it may, the simple and cheap way to cut off leading characters is with right():
SELECT right('123,456,789', -4);
SQL Fiddle.
I am trying to find if a string exist in a word and extract it. I have uses the instr() function but this works as the LIKE function: if part or the whole word exists it returns it.
Here I want to get the string 'Services' out, it works but if I change 'Services' to 'Service' it still works. I don't want that. If 'Service' is entered it should return null and not 'Services'
Modified:
What I am trying to do here is abbreviate certain parts of the company name.
This is what my database table looks like :
Word | Abb
---------+-----
Company | com
Limited | ltd
Service | serv
Services | servs
Here is the code:
Declare
Cursor Words Is
SELECT word,abb
FROM abbWords
processingWord VARCHAR2(50);
abbreviatedName VARCHAR(120);
fullName = 'A.D Company Services Limited';
BEGIN
FOR eachWord IN Words LOOP
--find the position of the word in name
wordPosition := INSTR(fullName, eachWord.word);
--extracts the word form the full name that matches the database
processingWord := Substr(fullName,instr(fullName,eachWord.word), length(eachWord.word));
--only process words that exist in name
if wordPosition > 0 then
abbreviatedName = replace(fullName, eachWord.word,eachWord.abb);
end if;
END lOOP;
END;
So if the user enters 'Service' I don't want 'Services' to be returned. By this I mean word position should be 0 if the word 'Service' in not found instead of returning the position for the word 'Services'
One way of doing it:
DECODE(INSTR('A.D Company Seervices Limited','Services'),
0,
NULL,
SUBSTR('A.D Company Services Limited',
INSTR('A.D Company Services Limited','Services'),
length('Services')))
INSTR() will return 0 if text is not found. DECODE() will evaluate the first argument, compare to the second, if match, return third argument, if not, return fourth argument. (sqlfiddle link)
Arguably not the most elegant way, but matches your requirement.
I think you're over-complicating this. You can do everything with regular expressions. For instance; given the following table:
create table names ( name varchar2(100));
insert into names values ('A.D Company Services Limited');
insert into names values ('A.D Company Service Limited');
This query will only return the name 'A.D Company Services Limited'.
select *
from names
where regexp_like( name
, '(^|[[:space:]])services($|[[:space:]])'
, 'i' )
This means match the beginning of the string, ^, or a space followed by services followed the end of the string, $, or a space. This is what differentiates regular expressions from using instr etc. You can make your matches easily conditional on other factors.
However, though this seems to be your question I don't think this is what you're trying to do. You're trying to replace the string 'serv' in your wider string without replacing 'services' or 'service'. For this you need to use regexp_replace().
If I add the following row to the table:
insert into names values ('A.D Company Serv Limited');
and run this query:
select regexp_replace( name
, '(^|[[:space:]])serv($|[[:space:]])'
, ' Services '
, 1, 0, 'i' )
from names
The only thing that will change is ' Serv ', which in this newest line, will be replaced with ' Services '. Note the spaces; as you don't want to replace 'Services' with 'ServServices' these are very important.
Here's a little SQL Fiddle to demonstrate.
Another alternative is to use something like:
select replace(name,' serv ', ' Services ')
from names;
This will replace only the word 'Serv' situated between 2 spaces.
Thank you,
Alex.
INSTR returns a number: the index of the first occurrence of the matching string. You should use regexp_substr instead (10g+):
SQL> select regexp_substr('A.D Company Services Limited', 'Services') match,
2 regexp_substr('A.D Company Service Limited', 'Services') unmatch
3 from dual;
MATCH UNMATCH
-------- -------
Services