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In a oneliner:
perl -ne 's/(\s*|\n)//g;print $_' inputfile
Inside a multiline script:
perl -e 's/(\s*|\n)//g && print $_ foreach (<>);' < inputfile

The following statement is incomplete:
backup incremental level 1 tag Daily_Diff_Backup
format '/u01/app/oracle/backup/Daily_%T_L1_%db_df_%U_%t';Backup what :)? Database?
(in the example below...

Use raw string notation for regex:
l = re.search(r"\b[A-Z]\S+", word)And just out of curiosity, any particular reason why you want to go for the regex method, as against continuing with python's...

The $1,$2 is a piece of awk code.
You cannot run a string (as assigned to an awk variable) as awk code.
(Unless there were an eval built into awk.)
This following awk code interprets a given...

This works.
ps | awk "{print $var}"

Single quote does not expand variable.

A here document is something that is just read (and expanded) and presented to stdin, and not executed per se. So the echo won't do anything. In the expansion phase, command substitution takes place,...

If you want to disable substitution in a heredoc then you need to quote the tag string:
<<'EOF'
echo "Sum is: `expr $1 + $2`"
EOF

.cshrc is automatically sourced
csh -c 'exec bash'
You need eval

eval $(
sed -rn 's/^\s*setenv\s+(\S+)\s+/export \1=/p' ~/.cshrc
)

Calling a member function "const" is to guarantee to the compiler "this function modifies absolutely nothing in this class". You are calling a non-const member function(strlen) from a const member...

The best option for such requirement is Tie::File

In anyway, you need to seek to the beginning of the file before reading it again.


#! /usr/bin/perl -w

open FH, "+< testfile" or die "$@";...

I suspect you'll want to do both, actually, since there will be a \n on the end in both places.

This is a common enough need I usually make it a function:

void chomp(char *str)
{
int...

fgets(passwd, sizeof(passwd), stdin);
passwd[strlen(passwd) - 1] = '\0';

This looks to me like you're trying to add a null-terminator to the string.

strlen() only works when the string...

You're trying to make a template specialization, which means you need to use a concrete type instead of the generic T. So you could do e.g.:


template<> int* AddFun<int*>(int* i, int* j) {
...

compare :

set prompt="[%/ $] "
# and
set prompt = "-- %T %n %~ -- \n$ "

a Dollar sign followed by ANYTHING is unserstood by the shell as a variable...
Im sure you put:
set prompt="[%/ $ ] "...

Here's the original one-liner:


perl -lne 'print "$1\n$2" if /(HOST.*)\)\((.*?)\)/' input


and here's the only line in the input file on which matches the pattern:


Attempting to...

The options first:
-n --> provides you a while loop by reading records (lines (terminated by new-lines) by default)
from the filename arguments without default printing (similar to sed -n)....

@royalibrahim: perl -n -e '#code' file is equivalent to
#! /usr/bin/perl
open F, "< file";
while (<F>) {
#code
}
close F;So, if you really want to do it by explicitly on command line by...

Of course, yes. You did it, right?

Why would perl complain when what you were doing was syntactically correct but logically wrong?

Never use a for (or foreach) loop to read a file. In this...

$ perl -lane 'if($F[1]!~/test/){$F[1]="test" . "\u$F[1]";}print join " ",@F' input.txt
def testGroupGetVersion(self):
def testSetCurrentGroup(self, listID, listName, path, clientID):

Hi


sed 'n;n;n;n;s/.*/QUERY\n&/;' file

Guru.

Some sed do not know \n in the replacement part:

sed 'n;n;n;n;s/^/QUERY\
/' infile

sed 'n;n;n;n;i\
query\
' infile



--
GNU sed:
sed '0~5s/^/QUERY\n/' infile

It doesn't.

I think you didn't make your int static; static locals act like globals, they only get initialized once and keep their values between different function calls. Without static, the...

with this double quoting, look at PS1 after assignment:


$ PS1="[$PWD $(git branch | grep '*') $] "
[/home/mute $] declare -p PS1
declare -- PS1="[/home/mute \$] "


Your variables are...

echo '1|23||45|||6|' | sed 's/||*/|/g'

When you assign one class to the other, all you're copying is the pointer, not the memory itself. So they both end up holding pointers to the same memory.

When they go out of scope, they both try...