Please see my textfile data below
roydwk27:teenaibuchytilibu5762sumonkhan:IJQRiq&76:8801627574057
deonnarsi15:latashajcclaypoolejcv5946sumonkhan:JKVWjv&20:8801627573929
ernaalo68:lindaohschletteoha1797sumonkhan:OPYZoy&84:8801628302709
dorathyshi56:fredrickaslperkinsonsle8932sumonkhan:STJKsj&30:8801621846709
londassg15:nataliaunmcredmondung5478sumonkhan:UVDEud&61:8801624792536
xiaoexu39:miriamfyboatwrightfyr3810sumonkhan:IJZAiz&47:8801626854856
I am want delete first word until :
like
roydwk27:
deonnarsi15:
ernaalo68:
dorathyshi56:
actually I am want if sumonkhan starting line then no problem but if sumonkhan line area 1st position available : with something then need remove this.
below actually data show in my .txt file
nataliaunmcredmondung5478sumonkhan:UVDEud&61:8801624792536
miriamfyboatwrightfyr3810sumonkhan:IJZAiz&47:8801626854856
all line available sumonkhan so if sumon khan starting position like this then good else delete this : full word not full line.
I hope this regex would help you. This regex deletes everything until first colon(:).
If you are reading a file then, read it line by line and run following regex on each line.
$str = 'roydwk27:teenaibuchytilibu5762sumonkhan:IJQRiq&76:8801627574057';
$str =~ s/^(?:.*?):(.*)/$1/g;
This code is in perl, you can re-write equivalent code in any other language.
See this demo at regex101.com.
^[\w\d]+:(.*)
^ // match the beginning of a line
[\w\d]+ // match any letter and any number
: // match ":" literally
( // start of the capturing group
.* // match any characters
) // end of capturing group
Now in all your matches in the first group you have the text you want matched. Note the g (global) and m (multiline) modifiers.
Related
So I have a massive list of numbers where all lines contain the same format.
#976B4B|B|0|0
#970000|B|0|1
#974B00|B|0|2
#979700|B|0|3
#4B9700|B|0|4
#009700|B|0|5
#00974B|B|0|6
#009797|B|0|7
#004B97|B|0|8
#000097|B|0|9
#4B0097|B|0|10
#970097|B|0|11
#97004B|B|0|12
#970000|B|0|13
#974B00|B|0|14
#979700|B|0|15
#4B9700|B|0|16
#009700|B|0|17
#00974B|B|0|18
#009797|B|0|19
#004B97|B|0|20
#000097|B|0|21
#4B0097|B|0|22
#970097|B|0|23
#97004B|B|0|24
#2C2C2C|B|0|25
#979797|B|0|26
#676767|B|0|27
#97694A|B|0|28
#020202|B|0|29
#6894B4|B|0|30
#976B4B|B|0|31
#808080|B|1|0
#800000|B|1|1
#803F00|B|1|2
#808000|B|1|3
What I am trying to do is remove all duplicate lines that contain the same hex codes, regardless of the text after it.
Example, in the first line #976B4B|B|0|0 the hex #976B4B shows up in line 32 as #976B4B|B|0|31. I want all lines EXCEPT the first occurrence to be removed.
I have been attempting to use regex to solve this, and found ^(.*)(\r?\n\1)+$ $1 can remove duplicate lines but obviously not what I need. Looking for some guidance and maybe a possibility to learn from this.
You can use the following regex replacement, make sure you click Replace All as many times as necessary, until no match is found:
Find What: ^((#[[:xdigit:]]+)\|.*(?:\R.+)*?)\R\2\|.*
Replace With: $1
See the regex demo and the demo screenshot:
Details:
^ - start of a line
((#[[:xdigit:]]+)\|.*(?:\R.+)*?) - Group 1 ($1, it will be kept):
(#[[:xdigit:]]+) - Group 2: # and one or more hex chars
\| - a | char
.* - the rest of the line
(?:\R.+)*? - any zero or more non-empty lines (if they can be empty, replace .+ with .*)
\R\2\|.* - a line break, Group 2 value, | and the rest of the line.
so i have (matlab) code .. and of the lines doesnt have (;) after the line
i want to find that line
for a starter :
sad= sdfsdf ; %this is comment
sad = awaww ;
n= sdfdsfd ;
m = (asd + adsf(asd,asd)) %this is comment
lets say i want to find the 4th line because it doesnt have (;) at the end of line ..
so far im stuck at this :
/(^[-a-zA-Z0-9]+\s*=[-a-zA-Z0-9#:%,_\+.()~#?&//= ]+)(?!;)$/gim
so this will work fine.. it will find the fourth line only
but what if i wanted (;) in middle of the line but not at end or before the comment .. ?
w=sss (;)aaa **;** % i dont want this line to be selected
w=sss (;)aaa %i want this line to be selected
http://regexr.com/3cfor
Well, let's find all lines which end with a semicolon:
^.+?;
optionally followed by horizontal whitespace:
^.+?;[ \t]*
and an optional comment:
^.+?;[ \t]*(?:%.*)?
This expression easily matches all the lines you don't want. So, inverse it:
^(?!.+?;[ \t]*(?:%.*)?$).+
Unfortunately, that's too easy. It fails to match lines which contain a semicolon in a comment. We could replace .+? with [^%\r\n]+? but this would fail on lines containing a % in a string.
If you need a more robust pattern, you'll have to account for all of this.
So let's start the same way, by defining what a "correct" line should look like. I'll use the PCRE syntax for atomic grouping, so you'll have to use perl = TRUE.
A string is: '(?>[^']+|'')*'
Other code (except string, comments and semicolons) is covered by: [^%';\r\n]+
So "normal" code is:
(?>[^%';\r\n]+|'(?>[^']+|'')*'|;)+?
Then, we add the required semicolon and optional comment:
(?>[^%';\r\n]+|'(?>[^']+|'')*'|;)+?;[ \t]*(?:%.*)?$
Finally, we invert all of this:
^(?!(?>[^%';\r\n]+|'(?>[^']+|'')*'|;)+?;[ \t]*(?:%.*)?$).+
And we have the final pattern. Demo.
You don't need to fully tokenize the input, you only have to recognize the different "lexer modes". I hope handling strings and comments is enough, but I didn't check the Matlab syntax thoroughly.
You could use this with other regex engines that do not support atomic groups by replacing (?> with (?: but you'll expose yourself to the catastrophic backtracking problem.
I have a text file
#sp_id int,
#sp_name varchar(120),
#sp_gender varchar(10),
#sp_date_of_birth varchar(10),
#sp_address varchar(120),
#sp_is_active int,
#sp_role int
Here, I want to get only the first word from each line. How can I do this? The spaces between the words may be space or tab etc.
Here is what I suggest:
Find what: ^([^ \t]+).*
Replace with: $1
Explanation: ^ matches the start of line, ([^ \t]+) matches 1 or more (due to +) characters other than space and tab (due to [^ \t]), and then any number of characters up to the end of the line with .*.
See settings:
In case you might have leading whitespace, you might want to use
^\s*([^ \t]+).*
I did something similar with this:
with open('handles.txt', 'r') as handles:
handlelist = [line.rstrip('\n') for line in handles]
newlist = [str(re.findall("\w+", line)[0]) for line in handlelist]
This gets a list containing all the lines in the document,
then it changes each line to a string and uses regex to extract the first word (ignoring white spaces)
My file (handles.txt) contained info like this:
JoIyke - personal twitter link;
newMan - another twitter handle;
yourlink - yet another one.
The code will return this list:
[JoIyke, newMan, yourlink]
Find What: ^(\S+).*$
Replace by : \1
You can simply use this to get the first word.Here we are capturing the first word in a group and replace the while line by the captured group.
Find the first word of each line with /^\w+/gm.
I try to extract the name1 (first-row), name2 (second-row), name3 (third-row) and the street-name (last-row) with regex:
Company Inc.
JohnDoe
Foobar
Industrieterrein 13
The very last row is the street name and this part is already working (the text is stored in the variable "S2").
REGEXREPLACE(S2, "(.*\n)+(?!(.*\n))", "")
This expression will return me the very last line. I am also able the extract the first row:
REGEXREPLACE(S2, "(\n.*)", "")
My problem is, that I do not know how to extract the second and third row....
Also how do I test if the text contains one, two, three or more rows?
Update:
The regex is used in the context of Scribe (a ETL tool). The problem is I can not execute sourcecode, I only have the following functions:
REGEXMATCH(input, pattern)
REGEXREPLACE(input, pattern, replacement)
If the regex language provides support for lookaheads you may count rows backwards and thus get (assuming . does not match newline)
(.*)$ # matching the last line
(.*)(?=(\n.*){1}$) # matching the second last line (excl. newline)
(.*)(?=(\n.*){2}$) # matching the third last line (excl. newline)
just use this regex:
(.+)+
explain:
.
Wildcard: Matches any single character except \n.
+
Matches the previous element one or more times.
As for a regular expression that will match each of four rows, how about this:
(.*?)\n(.*?)\n(.*?)\n(.*)
The parentheses will match, and the \n will match a new line. Note: you may have to use \r\n instead of just \n depending; try both.
You can try the following:
((.*?)\n){3}
I have this very long cfg file, where I need to find the latest occurrence of a line starting with a specific string. An example of the cfg file:
...
# format: - search.index.[number] = [search field]:element.qualifier
...
search.index.1 = author:dc.contributor.*
...
search.index.12 = language:dc.language.iso
...
jspui.search.index.display.1 = ANY
...
I need to be able to get the last occurrence of the line starting with search.index.[number] , more specific: I need that number. For the above snippet, that number would be 12.
As you can see, there are other lines too containing that pattern, but I do not want to match those.
I'm using Groovy as a programming/scripting language.
Any help is appreciated!
Have you tried:
def m = lines =~ /(?m)^search\.index\.(\d+)/
m[ -1 ][ 1 ]
Try this as your expression :
^search\.index\.(\d+)/
And then with Groovy you can get your result with:
matcher[0][0]
Here is an explanation page.
I don't think you should go for it but...
If you can do a multi-line search (anyway you have to here), the only way would be to read the file backward. So first, eat everything with a .* (om nom nom)(if you can make the dot match all, (?:.|\s)* if you can't). Now match your pattern search\.index\.(\d+). And you want to match this pattern at the beginning of a line: (?:^|\n) (hoping you're not using some crazy format that doesn't use \n as new line character).
So...
(?:.|\s)*(?:^|\n)search\.index\.(\d+)
The number should be in the 1st matching group. (Test in JavaScript)
PS: I don't know groovy, so sorry if it's totally not appropriate.
Edit:
This should also work:
search\.index\.(\d+)(?!(?:.|\s)*?(?:^|\n)search\.index\.\d+)