In a bash-script, only the case if a regular expression does not match is relevant.herefore I used the exclamation mark !. But where to place it?
These two work fine, but are they equivalent?
Code: if ! [[ $abc =~ $pattern ]]; then or
Code: if [[ ! $abc =~ $pattern ]]; then Where is the ! placed more correct?
In a bash shell script, I want to do something like this:
Code:
if [[ $(ls -ld /some/dir/foo_* | wc -l) -gt 4 ]]; then
rm -rf first_match_of /some/dir/foo_*
[code]....
I'm trying to find out how to extract the string between the 2 <title> tags: <title>this is what i want</title>.I found lots of results but nothing I've tried works.. EG:$page =~ m/<title>($.)</title>/gism;
View 2 Replies View RelatedI'm having a small issue with regex matching in Perl. I'm pretty certain it's a simple fix, but it all looks correct to me...
If I run the following:
Code:
It prints out all the lines containing a 'P', as one would expect. But when the regex is
Code:
I get zero lines printed. It seems to match only single-character patterns.
The file I'm reading is: (It has the same effect whether I leave it with Windows linebreaks or convert them to unix).
Code:
I have written a regular expression (tested in regexpal and regextester alpha something) with which I want to replace something like code...
but it only matches functions which occupy one line only, despite my tests showing multiple line matching in javascript testers online and using the m and s flags (which should make it multi line no?)
I'm writing a script to read user input for a computername.I need a check that a given userinput is valid.Right now I use grep like this (for sure not optimal):Quote:
if echo "$name" | grep -q '[^a-z][^A-Z][^0-9]'; then
echo error
else
[code]....
the following works and BASH doesn't complain, but VIM highlights the closing square bracket is if it sees a syntax error. Is there a better way to express regex in a case statement or is this an issue with VIM?
Code:
#!/bin/bash
case $1 in
[code]...
Code:
g echo ${mm[$j]}
4 BashNotes
[code]...
Are there some good tutorials or reference materials on how do pattern matching and text manipulation in Linux?I have a few simple tasks I'd like taken care of...like formatting numbers in file names, stripping some text from directory names, etc
View 2 Replies View RelatedI'd like to print only few groups from text matching a regex. How can I do that in the Linux command line?
What other tools for text processing are there in Linux?
I have two table files with x (1st column) ,y (2nd column) coordinates and intensity (3rd column). I need to match these two tables and divide the intensities at the consecutive coordinates on the 3rd column. The problem is the size of the tables are not same and I want to ignore the lines if they are not in one of the other file.
Here is Table 1:
Code:
-7.500-30.00013.006
-7.500-22.50037.952
-7.500-15.00060.962
-7.500-7.50040.922
-7.5000.00014.348
code....
I have been messing with diff and grep for 2 days now without result
I am trying to match a file consisting of words to many separate other wordfiles in a specific directory. one by one.
What i want the script to do is to report how many matching words my main file has with every file in the directory, each in turn
setup:
Each of em are plain text files with 1 word per line
Output should be something like:
SCRIPT REPORT:
I have a file with joker character patterns:
./include/*
./src/*
etc.
From the current directory I would like to recursively get the list of files that do not match these patterns.
If I have files named like this:
abc_one.c
egx_two.c
tsf_two.c
[code]...
I need a regex expression to read IP address from lines like these for my perl script:
Code:
2011.05.13-14:54:58, test@xyz.com, 180.211.51.11, pop3
2011.05.13-14:57:26, test@xyz.com, 160.234.47.12, pop3
2011.05.13-14:57:54, www@abcd.com, 166.22.22.224, imap
2011.05.13-14:57:55, www@somedomain.com, 172.58.22.154, imap
2011.05.13-15:03:08, www@example.com, 190.22.120.44, pop3
I'm just starting out with bash scripting (yesterday, really). I want to add a file to each user's home directory, pretty simple really, and send it out via our Apple Remote Desktop system to our Macs. Here is my script: Code: #!/bin/bash
for i in $(ls -d /Users/*)
doif [ -e $i/.tcshrc ]
thenecho "$i/.tcshrc exists!"elseecho "$i/.tcshrc does not exist"
[code]....
I'm using bash scripting to find any file that matches a path governed by the following regular expression:
"(monthly|nightly).[0-9]+/home/(user1|user2)/.mailbox/"
to match files like:
monthly.9/home/user1/.mailbox/l23131564
nightly.15/home/user2/.mailbox/cur/6546213
I've tried:
Code:
myRegex="(monthly|nightly).[0-9]+/home/(user1|user2)/.mailbox/"
find ./ -regextype posix-egrep -regex $myRegex
and it just spins and never gives me an answer, even though the file structure isn't that big.
So those of you that know me will agree that when it comes to awk I don't usually ask a lot of questions ... however this one has me stumped. I am guessing I have missed something obvious but for the life of me (and I have tested at great length) I cannot find it So the scenario is this: The following awk code should identify all versions of libgpg-error within the attached file (see below) and only show one for each version:
[Code]...
I'm fairly new to Perl and regular expressions. I have a large collection of files with their file names in the following general format: string - another string with spaces (2004) [year].ext I would like to know how I could create a regex to separate out:
the first string
another string with spaces
year
extension
If you know of a better way of doing it without regular expressions, I would be happy to hear that way too.
I was wondering if you insert a variable into a regex in Perl, will the contents of the variable be expanded by the regex engine?
View 14 Replies View RelatedI have .txt.gz files that store queries made on a browser, d my job is to analyze them.The information is stored in a xml-like style.Quote:
<browser>lwp-trivial/1.41</browser>
<http_code>200</http_code>
<keywords />
[code]...
how do you include a string variable as part of a regex in Perl?
View 5 Replies View RelatedI want to match for this string:
Code:
Content-Transfer-Encoding:[:space:]base64
Content-Disposition:[:space:]attachment;[:space:]filename="%variable%"
Both lines are new lines, so they won't be inline. Other than that, they are all constants, I want this regex match to be an if statement rather than returning match string. so if the $content variable contains some string that matches:
Code:
if `sed "//p"` ;
Im trying to extract the href of a <link> tag from a html page however as some links contain further preferences I seem to be unable to extract them, do you have any idea how I can write this: Link:
[Code]...
am trying to find a proper regex to match the two numbers in the following log entry.
Code:
15:08:16.142 INF Found 64468
15:08:16.142 ERR [Uniform test code=64469]
Basically the pattern I'm looking for will match the two different numbers spanned across two lines.Thought I need to use multi-line mode as follow but this doesn't match on [URL]...
Code:
/^($[0-9]{5}
[0-9]{5})$/m;
How do i use perl regex to extract the hostname from a FQDN?
I have
Quote:
$host=ganymede.a.linux.com
$host=io.a.linux.com
$host=europa.a.linux.com
i just want the characters which are to the left of the first .(dot) in FQDN name. I could get it using substr and split function,but how do i get it through regex.
I have a sed match that matches for certain string of a regex expression:
Code:
tname=$(echo "$contents" | sed -n 'some pattern')
How do I match for multiple strings in the $contents and return them as an array? for example
Code:
contents="this is a text, just to match patterns, here is another text to be matched"
the sed func would be able to recognize both "text"s, but only one is outputted?
Possible to put it in an array? so ${bar[0]} gives one and ${bar[1]} gives another
I need to add some text using sed before and after the matching pattern. Does any one have any clue?e.g.cat /my/file | sed -e "s/first pattern/New Pattern/g" . /my/file.bakNow I need a result like New Pattern
View 7 Replies View RelatedI am trying to check if the 9th character in a file on each line is a v and if so, then print the first word. I've tried a number of variations and am stuck !If it's possible to also check if character position 1 begins with a s in the same awk, that would make it cleaner instead of using egrep.
egrep '^s' file | nawk '{virtual=substr($0,9,1); if ($virtual=="v") {printf "%s", $1}}'
nawk: illegal field $(e)
input record number 1
source line number 1
I am struggling to understand why nawk matches values that are either 1 or 2 digits in length, but not 100 (3 digits).
Am I missing something obvious ? Should I use substr to remove the % ?
I am sure it's something to do with character and string matching behaviour with nawk.
Code: