I have a field that contains the "#" in a diffrent times. I want to count how many times the "#" occures int this fields.
How do I do that?
Try, untested:
=Split(Fields!Value.Value, "#").Length - 1
Related
I would like to understand why :
re.findall(r"(\d[A-Za-z]+)", "My user name is 3e4r 5fg")
returns
['3e', '4r', '5fg']
while :
re.findall(r"(\d[A-Za-z]+)+", "My user name is 3e4r 5fg")
returns
['4r', '5fg']
I tested some combinations with spaces between groups of "digit-letter" and 2 points clearly are involved in :
spaces between those groups
last "+".
I don't really understand why adding "+" after the group changes the result. Can someone explain me the steps of the process which leads to those different answers? Thank you very much.
When you put + after parenthesis you are searching for a pattern that contains one or more sub pattern with 1 digit and (one or more) letters'
so this phrase: "(\d[A-Za-z]+)+" return 2 matches:
3e4r
5fg
When you put a sub-pattern in parenthesis it means that all matches this sub-pattern will enter in a group, the groups is:
3e
5fg
The function re.findall returns only the groups (Unless there are no groups then it returns the matches ).
I have a field in my database, that contains 10 characters:
Fx: 1234567891
I want to look for the rows where the field has eg. the numbers 8 and 9 in places 5 and 6
So for example,
if the rows are
a) 1234567891
b) 1234897891
c) 1234877891
I only want b) returned in my select.
The type of the field is string/character varying.
I have tried using:
where field like '%89%'
but that won't work, because I need it to be 89 at a specific place in the string.
The fastest solution would be
WHERE substr(field, 8, 2) = '89'
If the positions are not adjacent, you end up with two conditions joined with AND.
You should be able to evaluate the single character using the underscore(_) character. So you should be able to use it as follows.
where field like '____89%'
Can nay one help to add multiple DB field values in one field.
Say i have 3 DB fields:
Name
Address
Age
I want to display all 3 fields in the same field:
John Peter 28.
I tried doing 3 fields next to each other and it did work but when i wrap text. It looks really bad:
Name
Jo.pe.28
hn te
r
My requirement is show data in one text field, for example: John.Peter.26
If you want to put them in one line (which i guess is the case), its straight forward.
Put this as a text box $F{Name} + "." + $F{Address} + "." + $F{Age}.toString()
Or you can use string concatenation (I dont personally like the syntax, take more effort to understand) $F{Name}.concat(".").concat($F{Address}).concat(".").concat($F{Age})
The SQL Method
Why not concatenate all the 3 fields you need in the query you use itself like (Assuming you are with Postgres.),
select (name || address|| to_char(age)) as data from my_table
In Ireport
As suggested,
$F{Name} + "." + $F{Address} + "." + $F{Age}.toString()
too works if needed to make it work from the report.
Make sure that all your fields are of same data type.
I am making program which will go through all possible choices.
Range is from 00000 to 99999.
For example:
00001,00002...01000,01001,01002...99999.
The problem is that i can make string as '00000' but as i convert it to int in order to add extra 1 to keep cycle going only one 0 appears. In that case i will get 0+1 = 1 and i need 00001.
Not completely sure how should i do it with lists because i might need it in the future for certain operations (to get one element from a current number 00450, 01004, 94571...)
Any advice/help would be greatly appreciated! :)
You can use zfill(num) on strings to add leading zeros
def convert_int(number,decimals) :
return str(number).zfill(decimals)
print convert_int(1,6) #prints 000001
I don't know if this is exactly what you want, but you can use string formatting. For example, this will turn int('00000') + 1 into '00001':
new_i = '%05d'%(int('00000')+1)
where %05d adds as many trailing zeros as necessary to whatever comes after % so the total length of the final string formatted number is 5.
I need a regex to match any number between 0 to 100 including decimal numbers example:
my expression should match 1,2,2.3 ,40,40.12 ,100,100.00 like this ..thanks in advance?
Assuming you have to allow for a leading sign, you are best off writing
if ( /(?<![-+.\d])([-+]?\d+(?:\.\d*)?(?![-+.\d])/ and $1 >= 0 and $1 <= 100 ) { .. }
But if you are forced into using a regex, then you need
if ( /(?<![-+.\d])(([-+]?(?:100|\d\d)(?:\.\d*)?(?![-+.\d])/ ) { .. }
These pattern may well be more complex than necessary because they allow for the number appearing anywhere in the string. If you are simply checking an entire string to see if it matches the criteria then it could be much shorter
This would work:
(100(\.0+))|([0-9]{1,2}(\.[0-9]+)?)
match either "100" (with optional dot plus one or more zeroes) or one or two digits, optionally followed by a dot and at least one digit.
EDITED!!!
This problem was much more difficult than I initially realized. With some amount of effort, I have produced a new regex that is without error. Enjoy.
/(?<!\d)(?<!\.)(100(?:(?!\.)|(?:\.0*+|\.))(?=\D)|[0-9]?[0-9](?:\.|\.[0-9]*+)?(?=[\D]))/
This pattern will capture in $1