I have a string
IsNull(VSK1_DVal.RuntimeSUM,0),
I need to remove IsNull part, so the result would be
VSK1_DVal.RuntimeSUM,
I'm absolute new to RegEx, but it wouldn't be a problem, if not one thing :
VSK1 is dynamic part, can be any combination of A-Z,0-9 and any length. How to replace strings with RegEx? I use MSSQL 2k5, i think it uses general set of RegEx rules.
EDIT : I forgot to say, that I'm doing replacement in SSMS Query window's Replace Box (^H) - not building RegEx query
br
marius
here's a regex that should work:
[^(]+\(([^,]+),[^)]\)
Then use $1 capture group to extract the part that you need.
I did a sanity check in ruby:
orig = "IsNull(VSK1_DVal.RuntimeSUM,0),"
regex = /[^(]*\(([^,]+),[^)]\)/
result = orig.sub(regex){$1} # result => VSK1_DVal.RuntimeSUM,
It gets trickier if you have a prefix that you want to retain. Like if you have this:
"somestuff = IsNull(VSK1_DVal.RuntimeSUM,0),"
In this case, you need someway to identify the start of the pattern. Maybe you can use '=' to identify the start of the pattern? If so, this should work:
orig = "somestuff = IsNull(VSK1_DVal.RuntimeSUM,0),"
regex = /=\s*\w+\(([^,]+),[^)]\)/
result = orig.sub(regex){$1} # result => somestuff = VSK1_DVal.RuntimeSUM,
But then the case where you don't have an equals sign will fail. Maybe you can use 'IsNull' to identify the start of the pattern? If so, try this (note the '/i' representing case insensitive matching):
orig = "somestuff = isnull(VSK1_DVal.RuntimeSUM,0),"
regex = /IsNull\(([^,]+),[^)]\)/i
result = orig.sub(regex){$1} # result => somestuff = VSK1_DVal.RuntimeSUM,
/IsNULL\((A-Z0-9+),0\)/
Then pick group match number 1.
Here's a very useful site: http://www.regexlib.com/RETester.aspx
They have a tester and a cheatsheet that are very useful for quick testing of this sort.
I tested the solution by Dave and it works fine except it also removes the trailing comma you wanted retained. Minor thing to fix.
Try this:
IsNULL\((.*,)0\)
You say in your question
I use MSSQL 2k5, i think it uses
general set of RegEx rules.
This is not true unless you enable CLR and compile and install an assembly. You can use its native pattern matching syntax and LIKE for this as below.
WITH T(C) AS
(
SELECT 'IsNull(VSK1_DVal.RuntimeSUM,0),' UNION ALL
SELECT 'IsNull(VSK1_DVal.RuntimeSUM,123465),' UNION ALL
SELECT 'No Match'
)
SELECT SUBSTRING(C,8,1+LEN(C)-8-CHARINDEX(',',REVERSE(C),2))
FROM T
WHERE C LIKE 'IsNull(%,_%),'
Related
I have a variable rawvalue:
let rawvalue = {abc-def-qwe}
I want to use regex to remove the { and }; I can simply do this by truncating the first and last characters. I built the regex:
^.(.*.).$
I want to know how to apply this regex on my variable to get the desired output?
The syntax you're looking for is like this:
let input = "{abc-def-qwe}";
let re = /^.(.*.).$/;
let fixed = re.exec(input)[1]; // Get the first match group "abc-def-qwe"
Maybe, this RegEx might be a better choice, which creates one group and you can simply call it using $1 and replace your string:
^\{(.+)\}$
For implementation, you might use, maybe these posts: 1, 2, 3.
I have a list of records that are character vectors. Here's an example:
'1mil_0,1_1_1_lb200_ks_drivers_sorted.csv'
'1mil_0_1_lb100_ks_drivers_sorted.csv'
'1mil_1_1_lb2_100_100_ks_drivers_sorted.csv'
'1mil_1_1_lb100_ks_drivers_sorted.csv'
From these names I would like to extract whatever's between the two substrings 1mil_ and _ks_drivers_sorted.csv.
So in this case the output would be:
0,1_1_1_lb200
0_1_lb100
1_1_lb2_100_100
1_1_lb100
I'm using MATLAB so I thought to use regexp to do this, but I can't understand what kind of regular expression would be correct.
Or are there some other ways to do this without using regexp?
Let the data be:
x = {'1mil_0,1_1_1_lb200_ks_drivers_sorted.csv'
'1mil_0_1_lb100_ks_drivers_sorted.csv'
'1mil_1_1_lb2_100_100_ks_drivers_sorted.csv'
'1mil_1_1_lb100_ks_drivers_sorted.csv'};
You can use lookbehind and lookahead to find the two limiting substrings, and match everything in between:
result = cellfun(#(c) regexp(c, '(?<=1mil_).*(?=_ks_drivers_sorted\.csv)', 'match'), x);
Or, since the regular expression only produces one match, the following simpler alternative can be used (thanks #excaza for noticing):
result = regexp(x, '(?<=1mil_).*(?=_ks_drivers_sorted\.csv)', 'match', 'once');
In your example, either of the above gives
result =
4×1 cell array
'0,1_1_1_lb200'
'0_1_lb100'
'1_1_lb2_100_100'
'1_1_lb100'
For me the easy way to do this is just use espace or nothing to replace what you don't need in your string, and the rest is what you need.
If is a list, you can use a loop to do this.
Exemple to replace "1mil_" with "" and "_ks_drivers_sorted.csv" with ""
newChr = strrep(chr,'1mil_','')
newChr = strrep(chr,'_ks_drivers_sorted.csv','')
I have a list of label names in a text file I'd like to manipulate using Find and Replace in Notepad++, they are listed as follows:
MyLabel_01
MyLabel_02
MyLabel_03
MyLabel_04
MyLabel_05
MyLabel_06
I want to rename them in Notepad++ to the following:
Label_A_One
Label_A_Two
Label_A_Three
Label_B_One
Label_B_Two
Label_B_Three
The Regex I'm using in the Notepad++'s replace dialog to capture the label name is the following:
((MyLabel_0)((1)|(2)|(3)|(4)|(5)|(6)))
I want to replace each capture group as follows:
\1 = Label_
\2 = A_One
\3 = A_Two
\4 = A_Three
\5 = B_One
\6 = B_Two
\7 = B_Three
My problem is that Notepad++ doesn't register the syntax of the regex above. When I hit Count in the Replace Dialog, it returns with 0 occurrences. Not sure what's misesing in the syntax. And yes I made sure the Regular Expression radio button is selected. Help is appreciated.
UPDATE:
Tried escaping the parenthesis, still didn't work:
\(\(MyLabel_0\)\((1\)|\(2\)|\(3\)|\(4\)|\(5\)|\(6\)\)\)
Ed's response has shown a working pattern since alternation isn't supported in Notepad++, however the rest of your problem can't be handled by regex alone. What you're trying to do isn't possible with a regex find/replace approach. Your desired result involves logical conditions which can't be expressed in regex. All you can do with the replace method is re-arrange items and refer to the captured items, but you can't tell it to use "A" for values 1-3, and "B" for 4-6. Furthermore, you can't assign placeholders like that. They are really capture groups that you are backreferencing.
To reach the results you've shown you would need to write a small program that would allow you to check the captured values and perform the appropriate replacements.
EDIT: here's an example of how to achieve this in C#
var numToWordMap = new Dictionary<int, string>();
numToWordMap[1] = "A_One";
numToWordMap[2] = "A_Two";
numToWordMap[3] = "A_Three";
numToWordMap[4] = "B_One";
numToWordMap[5] = "B_Two";
numToWordMap[6] = "B_Three";
string pattern = #"\bMyLabel_(\d+)\b";
string filePath = #"C:\temp.txt";
string[] contents = File.ReadAllLines(filePath);
for (int i = 0; i < contents.Length; i++)
{
contents[i] = Regex.Replace(contents[i], pattern,
m =>
{
int num = int.Parse(m.Groups[1].Value);
if (numToWordMap.ContainsKey(num))
{
return "Label_" + numToWordMap[num];
}
// key not found, use original value
return m.Value;
});
}
File.WriteAllLines(filePath, contents);
You should be able to use this easily. Perhaps you can download LINQPad or Visual C# Express to do so.
If your files are too large this might be an inefficient approach, in which case you could use a StreamReader and StreamWriter to read from the original file and write it to another, respectively.
Also be aware that my sample code writes back to the original file. For testing purposes you can change that path to another file so it isn't overwritten.
Bar bar bar - Notepad++ thinks you're a barbarian.
(obsolete - see update below.) No vertical bars in Notepad++ regex - sorry. I forget every few months, too!
Use [123456] instead.
Update: Sorry, I didn't read carefully enough; on top of the barhopping problem, #Ahmad's spot-on - you can't do a mapping replacement like that.
Update: Version 6 of Notepad++ changed the regular expression engine to a Perl-compatible one, which supports "|". AFAICT, if you have a version 5., auto-update won't update to 6. - you have to explicitly download it.
A regular expression search and replace for
MyLabel_((01)|(02)|(03)|(04)|(05)|(06))
with
Label_(?2A_One)(?3A_Two)(?4A_Three)(?5B_One)(?6B_Two)(?7B_Three)
works on Notepad 6.3.2
The outermost pair of brackets is for grouping, they limit the scope of the first alternation; not sure whether they could be omitted but including them makes the scope clear. The pattern searches for a fixed string followed by one of the two-digit pairs. (The leading zero could be factored out and placed in the fixed string.) Each digit pair is wrapped in round brackets so it is captured.
In the replacement expression, the clause (?4A_Three) says that if capture group 4 matched something then insert the text A_Three, otherwise insert nothing. Similarly for the other clauses. As the 6 alternatives are mutually exclusive only one will match. Thus only one of the (?...) clauses will have matched and so only one will insert text.
The easiest way to do this that I would recommend is to use AWK. If you're on Windows, look for the mingw32 precompiled binaries out there for free download (it'll be called gawk).
BEGIN {
FS = "_0";
a[1]="A_One";
a[2]="A_Two";
a[3]="A_Three";
a[4]="B_One";
a[5]="B_Two";
a[6]="B_Three";
}
{
printf("Label_%s\n", a[$2]);
}
Execute on Windows as follows:
C:\Users\Mydir>gawk -f test.awk awk.in
Label_A_One
Label_A_Two
Label_A_Three
Label_B_One
Label_B_Two
Label_B_Three
I want to use a regular expression that would do the following thing ( i extracted the part where i'm in trouble in order to simplify ):
any character for 1 to 5 first characters, then an "underscore", then some digits, then an "underscore", then some digits or dot.
With a restriction on "underscore" it should give something like that:
^([^_]{1,5})_([\\d]{2,3})_([\\d\\.]*)$
But i want to allow the "_" in the 1-5 first characters in case it still match the end of the regular expression, for example if i had somethink like:
to_to_123_12.56
I think this is linked to an eager problem in the regex engine, nevertheless, i tried to do some lazy stuff like explained here but without sucess.
Any idea ?
I used the following regex and it appeared to work fine for your task. I've simply replaced your initial [^_] with ..
^.{1,5}_\d{2,3}_[\d\.]*$
It's probably best to replace your final * with + too, unless you allow nothing after the final '_'. And note your final part allows multiple '.' (I don't know if that's what you want or not).
For the record, here's a quick Python script I used to verify the regex:
import re
strs = [ "a_12_1",
"abc_12_134",
"abcd_123_1.",
"abcde_12_1",
"a_123_123.456.7890.",
"a_12_1",
"ab_de_12_1",
]
myre = r"^.{1,5}_\d{2,3}_[\d\.]+$"
for str in strs:
m = re.match(myre, str)
if m:
print "Yes:",
if m.group(0) == str:
print "ALL",
else:
print "No:",
print str
Output is:
Yes: ALL a_12_1
Yes: ALL abc_12_134
Yes: ALL abcd_134_1.
Yes: ALL abcde_12_1
Yes: ALL a_123_123.456.7890.
Yes: ALL a_12_1
Yes: ALL ab_de_12_1
^(.{1,5})_(\d{2,3})_([\d.]*)$
works for your example. The result doesn't change whether you use a lazy quantifier or not.
While answering the comment ( writing the lazy expression ), i saw that i did a mistake... if i simply use the folowing classical regex, it works:
^(.{1,5})_([\\d]{2,3})_([\\d\\.]*)$
Thank you.
Can anyone help me with a regex to turn:
filename_author
to
author_filename
I am using MS Word 2003 and am trying to do this with Word's Find-and-Replace. I've tried the use wildcards feature but haven't had any luck.
Am I only going to be able to do it programmatically?
Here is the regex:
([^_]*)_(.*)
And here is a C# example:
using System;
using System.Text.RegularExpressions;
class Program
{
static void Main()
{
String test = "filename_author";
String result = Regex.Replace(test, #"([^_]*)_(.*)", "$2_$1");
}
}
Here is a Python example:
from re import sub
test = "filename_author";
result = sub('([^_]*)_(.*)', r'\2_\1', test)
Edit: In order to do this in Microsoft Word using wildcards use this as a search string:
(<*>)_(<*>)
and replace with this:
\2_\1
Also, please see Add power to Word searches with regular expressions for an explanation of the syntax I have used above:
The asterisk (*) returns all the text in the word.
The less than and greater than symbols (< >) mark the start and end
of each word, respectively. They
ensure that the search returns a
single word.
The parentheses and the space between them divide the words into
distinct groups: (first word) (second
word). The parentheses also indicate
the order in which you want search to
evaluate each expression.
Here you go:
s/^([a-zA-Z]+)_([a-zA-Z]+)$/\2_\1/
Depending on the context, that might be a little greedy.
Search pattern:
([^_]+)_(.+)
Replacement pattern:
$2_$1
In .NET you could use ([^_]+)_([^_]+) as the regex and then $2_$1 as the substitution pattern, for this very specific type of case. If you need more than 2 parts it gets a lot more complicated.
Since you're in MS Word, you might try a non-programming approach. Highlight all of the text, select Table -> Convert -> Text to Table. Set the number of columns at 2. Choose Separate Text At, select the Other radio, and enter an _. That will give you a table. Switch the two columns. Then convert the table back to text using the _ again.
Or you could copy the whole thing to Excel, construct a formula to split and rejoin the text and then copy and paste that back to Word. Either would work.
In C# you could also do something like this.
string[] parts = "filename_author".Split('_');
return parts[1] + "_" + parts[0];
You asked about regex of course, but this might be a good alternative.