I have a log file where data entries are as follows: Each entry starts with time:
Now I want to print only the entries after a specific time. For example, after time : 20130309235926, I want to print all records, so in my case, it should print the last 2 records.
Is there a sed command for doing this?
time: 20130309235926
dn:
changetype: modify
-
replace: modifiersname
modifiersname:
dc=
-
replace: modifytimestamp
modifytimestamp: 20130310045926Z
-
time: 20130309235959
dn:
changetype: modify
-
replace: modifiersname
modifiersname:
dc=
-
replace: modifytimestamp
modifytimestamp: 20130310045926Z
-
time: 20130308025010
dn:
changetype: modify
-
replace: modifiersname
modifiersname:
dc=
-
replace: modifytimestamp
modifytimestamp: 20130310045926Z
-
I like perl for doing paragraph-y things:
perl -00 -ne '$t = (/time: (\d+)/)[0]; print if $t gt "20130309235926"'
The -00 flag provides the input in paragraphs (separated by empty lines)
It depends if the time to find matches exactly or not, because doing aritmethic with sed is difficult. For example, for an exact match, this command uses -n switch to disable automatic printing and uses a range to print from the line that matches your time until end of file ($):
sed -n '/time:[ ]*20130309235926/,$ p' infile
EDIT to fix previous command:
sed -n '
## When found a blank line between a line with your time and end of file,
## jump to label "a".
/time:[ ]*20130309235926/,$ {
/^[ ]*$/ ba
};
## Skip all lines until previous condition be true.
b;
## Label "a".
:a;
## Save all content from next entry until end of file.
$! {
N;
ba
};
## Remove extra newline and print.
s/^\n//;
p
' infile
EDIT to add the previous command as a one-line:
sed -n '/time:[ ]*20130309235926/,$ { /^[ ]*$/ ba }; b; :a; $! { N; ba }; s/^\n//; p' infile
Related
I have file with this text:
mirrors:
docker.io:
endpoint:
- "http://registry:5000"
registry:5000:
endpoint:
- "http://registry:5000"
localhost:
endpoint:
- "http://registry:5000"
I need to replace it with this text in POSIX shell script (not bash):
mirrors:
docker.io:
endpoint:
- "http://docker.io"
registry:5000:
endpoint:
- "http://registry:5000"
localhost:
endpoint:
- "http://localhost"
Replace should be done dynamically in all places without hard-coded names. I mean we should take sub-string from a first line ("docker.io", "registry:5000", "localhost") and replace with it sub-string "registry:5000" in a third line.
I've figure out regex, that splits it on 5 groups: (^ )([^ ]*)(:[^"]*"http:\/\/)([^"]*)(")
Then I've tried to use sed to print group 2 instead of 4, but this didn't work: sed -n 's/\(^ \)\([^ ]*\)\(:[^"]*"http:\/\/\)\([^"]*\)\("\)/\1\2\3\2\5/p'
Please help!
This might work for you (GNU sed):
sed -E '1N;N;/\n.*endpoint:.*\n/s#((\S+):.*"http://)[^"]*#\1\2#;P;D' file
Open up a three line window into the file.
If the second line contains endpoint:, replace the last piece of text following http:// with the first piece of text before :
Print/Delete the first line of the window and then replenish the three line window by appending the next line.
Repeat until the end of the file.
Awk would be a better candidate for this, passing in the string to change to as a variable str and the section to change (" docker.io" or " localhost" or " registry:5000") and so:
awk -v findstr=" docker.io" -v str="http://docker.io" '
$0 ~ findstr { dockfound=1 # We have found the section passed in findstr and so we set the dockfound marker
}
/endpoint/ && dockfound==1 { # We encounter endpoint after the dockfound marker is set and so we set the found marker
found=1;
print;
next
}
found==1 && dockfound==1 { # We know from the found and the dockfound markers being set that we need to process this line
match($0,/^[[:space:]]+-[[:space:]]"/); # Match the start of the line to the beginning quote
$0=substr($0,RSTART,RLENGTH)str"\""; # Print the matched section followed by the replacement string (str) and the closing quote
found=0; # Reset the markers
dockfound=0
}1' file
One liner:
awk -v findstr=" docker.io" -v str="http://docker.io" '$0 ~ findstr { dockfound=1 } /endpoint/ && dockfound==1 { found=1;print;next } found==1 && dockfound==1 { match($0,/^[[:space:]]+-[[:space:]]"/);$0=substr($0,RSTART,RLENGTH)str"\"";found=0;dockfound=0 }1' file
I am trying replace a block of code between two patterns with blank lines
Tried using below command
sed '/PATTERN-1/,/PATTERN-2/d' input.pl
But it only removes the lines between the patterns
PATTERN-1 : "=head"
PATTERN-2 : "=cut"
input.pl contains below text
=head
hello
hello world
world
morning
gud
=cut
Required output :
=head
=cut
Can anyone help me on this?
$ awk '/=cut/{f=0} {print (f ? "" : $0)} /=head/{f=1}' file
=head
=cut
To modify the given sed command, try
$ sed '/=head/,/=cut/{//! s/.*//}' ip.txt
=head
=cut
//! to match other than start/end ranges, might depend on sed implementation whether it dynamically matches both the ranges or statically only one of them. Works on GNU sed
s/.*// to clear these lines
awk '/=cut/{found=0}found{print "";next}/=head/{found=1}1' infile
# OR
# ^ to take care of line starts with regexp
awk '/^=cut/{found=0}found{print "";next}/^=head/{found=1}1' infile
Explanation:
awk '/=cut/{ # if line contains regexp
found=0 # set variable found = 0
}
found{ # if variable found is nonzero value
print ""; # print ""
next # go to next line
}
/=head/{ # if line contains regexp
found=1 # set variable found = 1
}1 # 1 at the end does default operation
# print current line/row/record
' infile
Test Results:
$ cat infile
=head
hello
hello world
world
morning
gud
=cut
$ awk '/=cut/{found=0}found{print "";next}/=head/{found=1}1' infile
=head
=cut
This might work for you (GNU sed):
sed '/=head/,/=cut/{//!z}' file
Zap the lines between =head and =cut.
I need to append an asterisk to a line, but only if said line is preceded and followed by empty lines (FYI, said empty lines will NOT have any white space in them).
Suppose I have the following file:
foo
foo
foo
foo
foo
I want the output to look like this:
foo
foo
foo
foo*
foo
I tried modifying the following awk command (found here):
awk 'NR==1 {l=$0; next}
/^$/ {gsub(/test/,"xxx", l)}
{print l; l=$0}
END {print l}' file
to suit my uses, but got all tied up in knots.
Sed or Perl solutions are, of course, welcome also!
UPDATE:
It turned out that the question I asked was not quite correct. What I really needed was code that would append text to non-empty lines that do not start with whitespace AND are followed, two lines down, by non-empty lines that also do not start with whitespace.
For this revised problem, suppose I have the following file:
foo
third line foo
fifth line foo
this line starts with a space foo
this line starts with a space foo
ninth line foo
eleventh line foo
this line starts with a space foo
last line foo
I want the output to look like this:
foobar
third line foobar
fifth line foo
this line starts with a space foo
this line starts with a space foo
ninth line foobar
eleventh line foo
this line starts with a space foo
last line foo
For that, this sed one-liner does the trick:
sed '1N;N;/^[^[:space:]]/s/^\([^[:space:]].*\o\)\(\n\n[^[:space:]].*\)$/\1bar\2/;P;D' infile
Thanks to Benjamin W.'s clear and informative answer below, I was able to cobble this one-liner together!
A sed solution:
$ sed '1N;N;s/^\(\n.*\)\(\n\)$/\1*\2/;P;D' infile
foo
foo
foo
foo*
foo
N;P;D is the idiomatic way to look at two lines at the same time by appending the next one to the pattern space, then printing and deleting the first line.
1N;N;P;D extends that to always having three lines in the pattern space, which is what we want here.
The substitution matches if the first and last line are empty (^\n and \n$) and appends one * to the line between the empty lines.
Notice that this matches and appends a * also for the second line of three empty lines, which might not be what you want. To make sure this doesn't happen, the first capture group has to have at least one non-whitespace character:
sed '1N;N;s/^\(\n[^[:space:]].*\)\(\n\)$/\1*\2/;P;D' infile
Question from comment
Can we not append the * if the line two above begins with abc?
Example input file:
foo
foo
abc
foo
foo
foo
foo
There are three foo between empty lines, but the first one should not get the * appended because the line two above starts with abc. This can be done as follows:
$ sed '1{N;N};N;/^abc/!s/^\(.*\n\n[^[:space:]].*\)\(\n\)$/\1*\2/;P;D' infile
foo
foo
abc
foo
foo*
foo*
foo
This keeps four lines at a time in the pattern space and only makes the substitution if the pattern space does not start with abc:
1 { # On the first line
N # Append next line to pattern space
N # ... again, so there are three lines in pattern space
}
N # Append fourth line
/^abc/! # If the pattern space does not start with abc...
s/^\(.*\n\n[^[:space:]].*\)\(\n\)$/\1*\2/ # Append '*' to 3rd line in pattern space
P # Print first line of pattern space
D # Delete first line of pattern space, start next cycle
Two remarks:
BSD sed requires an extra semicolon: 1{N;N;} instead of 1{N;N}.
If the first and third line of the file are empty, the second line does not get an asterisk appended because we only start checking once there are four lines in the pattern space. This could be solved by adding an extra substitution into the 1{} block:
1{N;N;s/^\(\n[^[:space:]].*\)\(\n\)$/\1*\2/}
(remember the extra ; for BSD sed), but trying to cover all edge cases makes sed even less readable, especially in one-liners:
sed '1{N;N;s/^\(\n[^[:space:]].*\)\(\n\)$/\1*\2/};N;/^abc/!s/^\(.*\n\n[^[:space:]].*\)\(\n\)$/\1*\2/;P;D' infile
One way to think about these problems is as a state machine.
start: state = 0
0: /* looking for a blank line */
if (blank line) state = 1
1: /* leading blank line(s)
if (not blank line) {
nonblank = line
state = 2
}
2: /* saw non-blank line */
if (blank line) {
output noblank*
state = 0
} else {
state = 1
}
And we can translate this pretty directly to an awk program:
BEGIN {
state = 0; # start in state 0
}
state == 0 { # looking for a (leading) blank line
print;
if (length($0) == 0) { # found one
state = 1;
next;
}
}
state == 1 { # have a leading blank line
if (length($0) > 0) { # found a non-blank line
saved = $0; # save it
state = 2;
next;
} else {
print; # multiple leading blank lines (ok)
}
}
state == 2 { # saw the non-blank line
if (length($0) == 0) { # followed by a blank line
print saved "*"; # BINGO!
state = 1; # to the saw a blank-line state
} else { # nope, consecutive non-blank lines
print saved; # as-is
state = 0; # to the looking for a blank line state
}
print;
next;
}
END { # cleanup, might have something saved to show
if (state == 2) print saved;
}
This is not the shortest way, nor likely the fastest, but it's probably the most straightforward and easy to understand.
EDIT
Here is a comparison of Ed's way and mine (see the comments under his answer for context). I replicated the OP's input a million-fold and then timed the runnings:
# ls -l
total 22472
-rw-r--r--. 1 root root 111 Mar 13 18:16 ed.awk
-rw-r--r--. 1 root root 23000000 Mar 13 18:14 huge.in
-rw-r--r--. 1 root root 357 Mar 13 18:16 john.awk
# time awk -f john.awk < huge.in > /dev/null
2.934u 0.001s 0:02.95 99.3% 0+0k 112+0io 1pf+0w
# time awk -f ed.awk huge.in huge.in > /dev/null
14.217u 0.426s 0:14.65 99.8% 0+0k 272+0io 2pf+0w
His version took about 5 times as long, did twice as much I/O, and (not shown in this output) took 1400 times as much memory.
EDIT from Ed Morton:
For those of us unfamiliar with the output of whatever time command John used above, here's the 3rd-invocation results from the normal UNIX time program on cygwin/bash using GNU awk 4.1.3:
$ wc -l huge.in
1000000 huge.in
$ time awk -f john.awk huge.in > /dev/null
real 0m1.264s
user 0m1.232s
sys 0m0.030s
$ time awk -f ed.awk huge.in huge.in > /dev/null
real 0m1.638s
user 0m1.575s
sys 0m0.030s
so if you'd rather write 37 lines than 3 lines to save a third of a second on processing a million line file then John's answer is the right one for you.
EDIT#3
It's the standard "time" built-in from tcsh/csh. And even if you didn't recognize it, the output should be intuitively obvious. And yes, boys and girls, my solution can also be written as a short incomprehensible mess:
s == 0 { print; if (length($0) == 0) { s = 1; next; } }
s == 1 { if (length($0) > 0) { p = $0; s = 2; next; } else { print; } }
s == 2 { if (length($0) == 0) { print p "*"; s = 1; } else { print p; s = 0; } print; next; }
END { if (s == 2) print p; }
Here's a perl filter version, for the sake of illustration — hopefully it's clear to see how it works. It would be possible to write a version that has a lower input-output delay (2 lines instead of 3) but I don't think that's important.
my #lines;
while (<>) {
# Keep three lines in the buffer, print them as they fall out
push #lines, $_;
print shift #lines if #lines > 3;
# If a non-empty line occurs between two empty lines...
if (#lines == 3 && $lines[0] =~ /^$/ && $lines[2] =~ /^$/ && $lines[1] !~ /^$/) {
# place an asterisk at the end
$lines[1] =~ s/$/*/;
}
}
# Flush the buffer at EOF
print #lines;
A perl one-liner
perl -0777 -lne's/(?<=\n\n)(.*?)(\n\n)/$1\*$2/g; print' ol.txt
The -0777 "slurps" in the whole file, assigned to $_, on which the (global) substitution is run and which is then printed.
The lookbehind (?<=text) is needed for repeating patterns, [empty][line][empty][line][empty]. It is a "zero-width assertion" that only checks that the pattern is there without consuming it. That way the pattern stays available for next matches.
Such consecutive repeating patterns trip up the /(\n\n)(.*?)(\n\n)/$1$2\*$3/, posted initially, since the trailing \n\n are not considered for the start of the very next pattern, having been just matched.
Update: My solution also fails after two consecutive matches as described above and needs the same lookback: s/(?<=\n\n)(\w+)\n\n/\1\2*\n\n/mg;
The easiest way is to use multi-line match:
local $/; ## slurp mode
$file = <DATA>;
$file =~ s/\n\n(\w+)\n\n/\n\n\1*\n\n/mg;
printf $file;
__DATA__
foo
foo
foo
foo
foo
It's simplest and clearest to do this in 2 passes:
$ cat tst.awk
NR==FNR { nf[NR]=NF; nr=NR; next }
FNR>1 && FNR<nr && NF && !nf[FNR-1] && !nf[FNR+1] { $0 = $0 "*" }
{ print }
$ awk -f tst.awk file file
foo
foo
foo
foo*
foo
The above takes one pass to record the number of fields on each line (NF is zero for an empty line) and then the second pass just checks your requirements - the current line is not the first or last in the file, it is not empty and the lines before and after are empty.
alternative awk solution (single pass)
$ awk 'NR>2 && !pp && !NF {p=p"*"}
NR>1{print p}
{pp=length(p);p=$0}
END{print p}' foo
foo
foo
foo
foo*
foo
Explanation: defer printing to next line for decision making, so need to keep previous line in p and state of the second previous line in pp (length zero assumed to be empty). Do the bookkeeping assignments and at the end print the last line.
Sed editing is always a new challenge to me when it comes to multiple line editing. In this case I have the following pattern:
RECORD 4,4 ,5,48 ,7,310 ,10,214608 ,12,199.2 ,13,-19.2 ,15,-83 ,17,35 \
,18,0.8 ,21,35 ,22,31.7 ,23,150 ,24,0.8 ,25,150 ,26,0.8 ,28,25 ,29,6 \
,30,1200 ,31,1 ,32,0.2 ,33,15 ,36,0.4 ,37,1 ,39,1.1 ,41,4 ,80,2 \
,82,1000 ,84,1 ,85,1
which I want to convert into:
#RECORD 4,4 ,5,48 ,7,310 ,10,214608 ,12,199.2 ,13,-19.2 ,15,-83 ,17,35 \
# ,18,0.8 ,21,35 ,22,31.7 ,23,150 ,24,0.8 ,25,150 ,26,0.8 ,28,25 ,29,6\
# ,30,1200 ,31,1 ,32,0.2 ,33,15 ,36,0.4 ,37,1 ,39,1.1 ,41,4 ,80,2 \
# ,82,1000 ,84,1 ,85,1
Besides this I would like to preserve the entirety of these 4 lines (which may be more or less than 4 (unpredictable as the appear in the input) into one (long) line without the backslashes or line wraps.
Two tasks in one so to say.
sed is mandatory.
It's not terribly clear how you recognize the blocks you want to comment out, so I'll use blocks from a line that starts with RECORD and process as long as there are backslashes at the end (if your requirements differ, the patterns used will need to be amended accordingly).
For that, you could use
sed '/^RECORD/ { :a /\\$/ { N; ba }; s/[[:space:]]*\\\n[[:space:]]*/ /g; s/^/#/ }' filename
This works as follows:
/^RECORD/ { # if you find a line that starts with
# RECORD:
:a # jump label for looping
/\\$/ { # while there's a backslash at the end
# of the pattern space
N # fetch the next line
ba # loop.
}
# After you got the whole block:
s/[[:space:]]*\\\n[[:space:]]*/ /g # remove backslashes, newlines, spaces
# at the end, beginning of lines
s/^/#/ # and put a comment sign at the
# beginning.
}
Addendum: To keep the line structure intact, instead use
sed '/^RECORD/ { :a /\\$/ { N; ba }; s/\(^\|\n\)/&#/g }' filename
This works pretty much the same way, except the newline-removal is removed, and the comment signs are inserted after every line break (and once at the beginning).
Addendum 2: To just put RECORD blocks onto a single line:
sed '/^RECORD/ { :a /\\$/ { N; ba }; s/[[:space:]]*\\\n[[:space:]]*/ /g }' filename
This is just the first script with the s/^/#/ bit removed.
Addendum 3: To isolate RECORD blocks while putting them onto a single line at the same time,
sed -n '/^RECORD/ { :a /\\$/ { N; ba }; s/[[:space:]]*\\\n[[:space:]]*/ /g; p }' filename
The -n flag suppresses the normal default printing action, and the p command replaces it for those lines that we want printed.
To write those records out to a file while commenting them out in the normal output at the same time,
sed -e '/^RECORD/ { :a /\\$/ { N; ba }; h; s/[[:space:]]*\\\n[[:space:]]*/ /g; w saved_records.txt' -e 'x; s/\(^\|\n\)/&#/g }' foo.txt
There's actually new stuff in this. Shortly annotated:
#!/bin/sed -f
/^RECORD/ {
:a
/\\$/ {
N
ba
}
# after assembling the lines
h # copy them to the hold buffer
s/[[:space:]]*\\\n[[:space:]]*/ /g # put everything on a line
w saved_records.txt # write that to saved_records.txt
x # swap the original lines back
s/\(^\|\n\)/&#/g # and insert comment signs
}
When specifying this code directly on the command line, it is necessary to split it into several -e options because the w command is not terminated by ;.
This problem does not arise when putting the code into a file of its own (say foo.sed) and running sed -f foo.sed filename instead. Or, for the advanced, putting a #!/bin/sed -f shebang on top of the file, chmod +xing it and just calling ./foo.sed filename.
Lastly, to edit the input file in-place and print the records to stdout, this could be amended as follows:
sed -i -e '/^RECORD/ { :a /\\$/ { N; ba }; h; s/[[:space:]]*\\\n[[:space:]]*/ /g; w /dev/stdout' -e 'x; s/\(^\|\n\)/&#/g }' filename
The new things here are the -i flag for inplace editing of the file, and to have /dev/stdout as target for the w command.
sed '/^RECORD.*\\$/,/[^\\]$/ s/^/#/
s/^RECORD.*/#&/' YourFile
After several remark of #Wintermute and more information from OP
Assuming:
line with RECORD at start are a trigger to modify the next lines
structure is the same (no line with \ with a RECORD line following directly or empty lines)
Explain:
take block of line starting with RECORD and ending with \
add # in front of each line
take line (so after ana eventual modification from earlier block that leave only RECORD line without \ at the end or line without record) and add a # at the start if starting with RECORD
I have two files one with 17k lines and another one with 4k lines. I wanted to compare position 115 to position 125 with each line in the second file and if there is a match, write the entire line from the first file into a new file. I had come up with a solution where i read the file using 'cat $filename | while read LINE'. but it's taking around 8 mins to complete. is there any other way like using 'awk' to reduce this process time.
my code
cat $filename | while read LINE
do
#read 115 to 125 and then remove trailing spaces and leading zeroes
vid=`echo "$LINE" | cut -c 115-125 | sed 's,^ *,,; s, *$,,' | sed 's/^[0]*//'`
exist=0
#match vid with entire line in id.txt
exist=`grep -x "$vid" $file_dir/id.txt | wc -l`
if [[ $exist -gt 0 ]]; then
echo "$LINE" >> $dest_dir/id.txt
fi
done
How is this:
FNR==NR { # FNR == NR is only true in the first file
s = substr($0,115,10) # Store the section of the line interested in
sub(/^\s*/,"",s) # Remove any leading whitespace
sub(/\s*$/,"",s) # Remove any trailing whitespace
lines[s]=$0 # Create array of lines
next # Get next line in first file
}
{ # Now in second file
for(i in lines) # For each line in the array
if (i~$0) { # If matches the current line in second file
print lines[i] # Print the matching line from file1
next # Get next line in second file
}
}
Save it to a script script.awk and run like:
$ awk -f script.awk "$filename" "${file_dir}/id.txt" > "${dest_dir}/id.txt"
This will still be slow because for each line in second file you need to look at ~50% of the unique lines in first (assuming most line do in fact match). This can be significantly improved if you can confirmed that the lines in the second file are full line matches against the substrings.
For full line matches this should be faster:
FNR==NR { # FNR == NR is only true in the first file
s = substr($0,115,10) # Store the section of the line interested in
sub(/^\s*/,"",s) # Remove any leading whitespace
sub(/\s*$/,"",s) # Remove any trailing whitespace
lines[s]=$0 # Create array of lines
next # Get next line in first file
}
($0 in lines) { # Now in second file
print lines[$0] # Print the matching line from file1
}