Updated description to be clearer.
Say I have a file and it has these lines in it.
one
two
three
five
How do I add a line that says "four" after the line that says "three" so my file now looks like this?
one
two
three
four
five
Assuming you want to do this with the FileEdit class.
Chef::Util::FileEdit.new('/path/to/file').insert_line_after_match(/three/, 'four')
Here is the example ruby block for inserting 2 new line after match:
ruby_block "insert_lines" do
block do
file = Chef::Util::FileEdit.new("/path/of/file")
file.insert_line_after_match("three", "four")
file.insert_line_after_match("four", "five")
file.write_file
end
end
insert_line_after_match searches for the regex/string and it will insert the value in after the match.
The following Ruby script should do what you want quite nicely:
# insert_line.rb
# run with command "ruby insert_line.rb myinputfile.txt", where you
# replace "myinputfile.txt" with the actual name of your input file
$-i = ".orig"
ARGF.each do |line|
puts line
puts "four" if line =~ /^three$/
end
The $-i = ".orig" line makes the script appear to edit the named input file in-place and make a backup copy with ".orig" appended to the name. In reality it reads from the specified file and writes output to a temp file, and on success renames both the original input file (to have the specified suffix) and the temp file (to have the original name).
This particular implementation writes "four" after finding the "three" line, but it would be trivial to alter the pattern being matched, make it count-based, or have it write before some identified line rather than after.
This is an in memory solution. It looks for complete lines rather than doing a string regex search...
def add_after_line_in_memory path, findline, newline
lines = File.readlines(path)
if i = lines.index(findline.to_s+$/)
lines.insert(i+1, newline.to_s+$/)
File.open(path, 'wb') { |file| file.write(lines.join) }
end
end
add_after_line_in_memory 'onetwothreefive.txt', 'three', 'four'
An AWK Solution
While you could do this in Ruby, it's actually trivial to do this in AWK. For example:
# Use the line number to choose the insertion point.
$ awk 'NR == 4 {print "four"}; {print}' lines
one
two
three
four
five
# Use a regex to prepend your string to the matched line.
$ awk '/five/ {print "four"}; {print}' lines
one
two
three
four
five
Related
Suppose I have text as:
This is a sample text.
I have 2 sentences.
text is present there.
I need to replace whole text between two 'text' words. The required solution should be
This is a sample text.
I have new sentences.
text is present there.
I tried using the below command but its not working:
sed -i 's/text.*?text/text\
\nI have new sentence/g' file.txt
With your shown samples please try following. sed doesn't support lazy matching in regex. With awk's RS you could do the substitution with your shown samples only. You need to create variable val which has new value in it. Then in awk performing simple substitution operation will so the rest to get your expected output.
awk -v val="your_new_line_Value" -v RS="" '
{
sub(/text\.\n*[^\n]*\n*text/,"text.\n"val"\ntext")
}
1
' Input_file
Above code will print output on terminal, once you are Happy with results of above and want to save output into Input_file itself then try following code.
awk -v val="your_new_line_Value" -v RS="" '
{
sub(/text\.\n*[^\n]*\n*text/,"text.\n"val"\ntext")
}
1
' Input_file > temp && mv temp Input_file
You have already solved your problem using awk, but in case anyone else will be looking for a sed solution in the future, here's a sed script that does what you needed. Granted, the script is using some advanced sed features, but that's the fun part of it :)
replace.sed
#!/usr/bin/env sed -nEf
# This pattern determines the start marker for the range of lines where we
# want to perform the substitution. In our case the pattern is any line that
# ends with "text." — the `$` symbol meaning end-of-line.
/text\.$/ {
# [p]rint the start-marker line.
p
# Next, we'll read lines (using `n`) in a loop, so mark this point in
# the script as the beginning of the loop using a label called `loop`.
:loop
# Read the next line.
n
# If the last read line doesn't match the pattern for the end marker,
# just continue looping by [b]ranching to the `:loop` label.
/^text/! {
b loop
}
# If the last read line matches the end marker pattern, then just insert
# the text we want and print the last read line. The net effect is that
# all the previous read lines will be replaced by the inserted text.
/^text/ {
# Insert the replacement text
i\
I have a new sentence.
# [print] the end-marker line
p
}
# Exit the script, so that we don't hit the [p]rint command below.
b
}
# Print all other lines.
p
Usage
$ cat lines.txt
foo
This is a sample text.
I have many sentences.
I have many sentences.
I have many sentences.
I have many sentences.
text is present there.
bar
$
$ ./replace.sed lines.txt
foo
This is a sample text.
I have a new sentence.
text is present there.
bar
Substitue
sed -i 's/I have 2 sentences./I have new sentences./g'
sed -i 's/[A-Z]\s[a-z].*/I have new sentences./g'
Insert
sed -i -e '2iI have new sentences.' -e '2d'
I need to replace whole text between two 'text' words.
If I understand, first text. (with a dot) is at the end of first line and second text at the beginning of third line. With awk you can get the required solution adding values to var s:
awk -v s='\nI have new sentences.\n' '/text.?$/ {s=$0 s;next} /^text/ {s=s $0;print s;s=""}' file
This is a sample text.
I have new sentences.
text is present there.
I have the config file on local, whom I am appending some variables from different remote machines. The file content is as:
#!/bin/bash
name=bob
department=(Production)
name=alice
department=(R&D)
name=maggie
department=(Production R&D)
The latest values updated in the file are the last one. So the expected output in the config file should be:
#!/bin/bash
name=maggie
department=(Production R&D)
I want to remove the first two data of name and address except for the latest one which is last. But this should happen only if there are multiple same variables.
I referred and try this for my solution but not getting expected output:
https://backreference.org/2011/11/17/remove-duplicates-but-keeping-only-the-last-occurrence/
Would you please try the following:
tac file | awk '{ # print "file" reversing the line order: last line first
line = $0 # backup the line
sub(/#.*/, "") # remove comments (not sure if comment line exists)
if (match($0, /([[:alnum:]_]+)=/)) { # look like an assignment to a variable
varname = substr($0, RSTART, RLENGTH - 1)
# extract the variable name (-1 to remove "=")
if (! seen[varname]++) print line # print the line if the variable is seen irst time
} else { # non-assignment line
print line
}
}' | tac # reverse the lines again
Output:
#!/bin/bash
name=maggie
department=(Production R&D)
Please note the parser to extract variable names is a lousy one. You may need to tweak the code depending on the actual file.
I have a following file:
old_file
new_file
Some string.
end
Text in the middle that is not supposed to go to any of files.
new_file
Another text.
end
How using regex can I create two files with the following content:
file1
new_file
Some string.
end
file2
new_file
Another text.
end
How can I get information which is between keywords 'new_file' and 'end' to write it to the file?
If your files are not that large, you can read them in as a string, (use File.read(file_name)), and then run the following regex:
file_contents.scan(/^new_file$.*?^end$/m).select { |block| WRITE_TO_FILE_CODE_HERE }
See the regex demo
The ^new_file$.*?^end$ regex matches new_file that is a whole line content, then 0+ any characters as few as possible (incl. a newline as /m modifier is used), and then end (a whole line).
Else, you may adapt this answer here as
printing = false
File.open(my_file).each_line do |line|
printing = true if line =~ /^new_file$/
puts line if printing
printing = false if line =~ /^end$/
end
Open the file when the starting line is found, write to it where puts line is in the example above, and close when printing false occurs.
You can also read the file chunk by chunk by changing what constitutes a "line" in ruby:
File.open("file1.txt", "w") do |file1|
File.open("file2.txt", "w") do |file2|
enum = IO.foreach("old_file.txt", sep="\n\n")
file1.puts enum.next.strip
enum.next #discard
file2.puts enum.next.strip
end #automatically closes file2
end #automatically closes file1
By designating the separator as "\n\n" ruby will read all the characters up to and including two consecutive newlines--and return that as a "line".
If that kind of format is fixed, then you may try this (new_file\n.*\nend)
I am trying to print all the lines from a file before the first match. I have the same entries again in the file, but I don't need that lines. Tried
awk "{print} /${pattern}/ {exit}" and sed "/$pattern/q" (my serach is based on a variable). But both these commands are printing all the line before the last match
ex: my file is like
abc
bcd
def
xyz
def
lmno
def
xvd
when my pattern is 'def', i just need abc and bcd . but the above commands are printing, all the lines before the last 'def'. could you please provide some idea
This should work:
awk '!'"/${pattern}/{print} /${pattern}/ {exit}" input_file.txt
A program creates HTML files from a database. There are headings and stuff in between the headings.
There are not a set amount of headings.
After each heading the program places the text:
$WHITE*("5")$
$WHITE*("20")$
$HRULE$
I need every occurrence of these 4 lines to be replaced with:
$WHITE*("20")$
$HRULE$
$WHITE*("10")$
I am not fussed what program is used :)
I have tried:
sed 's:\$WHITE\*(\"5\")\$\n\n\$WHITE\*(\"20\")\$\n\$HRULE\$:\$WHITE\*(\"20\")\$\
\$HRULE$\
\$WHITE*("10")$:g'
and various other permutations
If that'S your input file, and this is the spec, you can do:
sed -n '3,$p;$a$WHITE*("10")$' INPUTFILE
But I assume that's not the case, so you might want to rephrase your question and/or giving some more detailes.
More specific solution with sed:
sed '/^\$WHITE\*("5")\$$/,/^$/d;/\$HRULE\$/ a$WHITE*("10")$' INPUTFILE
(Searches for the $WHITE*("5")$ line and deletes it till (including!) the next empty line. Then searches for the next $HRULE$ line and appends an $WHITE*("10")$ line.
awk solution:
awk '/\$WHITE\*\("5"\)\$/ { getline ; next }
/\$WHITE\*\("20"\)\$/ { print ;
getline ;
if ($0 ~ /\$HRULE\$/) { print ;
print "$WHITE*(\"10\")$" ;
}
else { print }
}
1 ' INPUTFILE
This reads the file and prints every line - that's why the 1 is there, except if it finds the $WHITE*("5") pattern it drops it, reads the next line and drops that too. if it finds the $WHITE*("20") prints it. Reads the next line and if its $HRULE$ then prints that and the appended $WHITE*("10") line. Else just prints the line.
HTH
UPDATE #2
From the sed faq, section 4.23.3
If you need to match a static block of text (which may occur any number of times throughout a file), where the contents of the block are known in advance, then this script is easy to use
UPDATE #1
Python?
$ cat input
first line
second line
3rd line
$WHITE*("5")$
$WHITE*("20")$
$HRULE$
some more lines
yet another
$WHITE*("5")$
$WHITE*("20")$
$HRULE$
THE END
the script:
#!/usr/bin/env python
## Use these 3 lines for python version < 2.5
#fd=open('input')
#text=fd.read()
#fd.close()
## Use these 2 lines for python version >= 2.5
with open('input') as fd:
text=fd.read()
old="""$WHITE*("5")$
$WHITE*("20")$
$HRULE$
"""
new="""$WHITE*("20")$
$HRULE$
$WHITE*("10")$
"""
print text.replace(old,new)
output:
first line
second line
3rd line
$WHITE*("20")$
$HRULE$
$WHITE*("10")$
some more lines
yet another
$WHITE*("20")$
$HRULE$
$WHITE*("10")$
THE END
Try something like
sed -e '${p;};/$WHITE\*("5")\$/,/$HRULE\$/{H;/$HRULE\$/{g;s/$HRULE\$//;s/20/10/;s/5/20/;s/\n/&$HRULE$/2p;s/.*//p;x;d;};d;};' white.txt
Crude, but it should work.
This might work for you:
sed '/^\$WHITE\*(\"5\")\$/{N;N;N;s/.*\n\n\(\(\$WHITE\*(\"\)20\(\")\$\s*\)\n\$HRULE\$\s*$\)/\1\n\210\3/}' file
Explanation:
Match on first string $WHITE*("5")$, read the next 3 lines and match on remainder. Use grouping and back references to formulate output lines.