I have a file1 that has some PHP code in it. I need to find the following: action="blahblah" and replace it with action="error.php". Problem is, I don't know how many characters are between the quotes in the original.
Here's what I have that doesn't work:
sed 's:action="^[^"]*":action="error.php":' <file1> file2
How can I do this?
Why have you got the ^ start-of-line marker before the character class? Try it with:
sed 's:action="[^"]*":action="error.php":' <file1 > file2
Here's a transcript showing your version alongside that correction:
pax$ echo 'blah action="something" blah' | sed '
...$ s:action="^[^"]*":action="error.php":'
blah action="something" blah
pax$ echo 'blah action="something" blah' | sed '
...$ s:action="[^"]*":action="error.php":'
blah action="error.php" blah
Related
I have a string that goes like this abcd'efgh\ ijkl\ mnop
I want to make the string into abcd\'efgh\ ijkl\ mnop
sed 's/\'/\\\'/g didn't work
You can use
sed 's/'"'"'/\\&/g'
See the online demo:
#!/bin/bash
s="abcd'"'efgh\ ijkl\ mnop'
sed 's/'"'"'/\\&/g' <<< "$s"
# => abcd\'efgh\ ijkl\ mnop
To delete/comment 3 lines befor a pattern (including the line with the pattern):
how can i achive it through sed command
Ref:
sed or awk: delete n lines following a pattern
the above ref blog help to achive the this with after a pattern match but i need to know before match
define host{
use xxx;
host_name pattern;
alias yyy;
address zzz;
}
the below sed command will comment the '#' after the pattern match for example
sed -e '/pattern/,+3 s/^/#/' file.cfg
define host{
use xxx;
#host_name pattern;
#alias yyy;
#address zzz;
#}
like this how can i do this for the before pattern?
can any one help me to resolve this
If tac is allowed :
tac|sed -e '/pattern/,+3 s/^/#/'|tac
If tac isn't allowed :
sed -e '1!G;h;$!d'|sed -e '/pattern/,+3 s/^/#/'|sed -e '1!G;h;$!d'
(source : http://sed.sourceforge.net/sed1line.txt)
Reverse the file, comment the 3 lines after, then re-reverse the file.
tac file | sed '/pattern/ {s/^/#/; N; N; s/\n/&#/g;}' | tac
#define host{
#use xxx;
#host_name pattern;
alias yyy;
address zzz;
}
Although I think awk is a little easier to read:
tac file | awk '/pattern/ {c=3} c-- > 0 {$0 = "#" $0} 1' | tac
This might work for you (GNU sed):
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/^/#/mg;P;D' file
Gather up 4 lines in the pattern space and if the last line contains pattern insert # at the beginning of each line in the pattern space.
To delete those 4 lines, use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/d;P;D' file
To delete the 3 lines before pattern but not the line containing pattern use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/.*\n//;P;D'
I have lines that start like this: 2141058222 11/22/2017 and I want to append a ; at the end of the ten digit number like this: 2141058222; 11/22/2017.
I've tried sed with sed -i 's/^[0-9]\{10\}\\$/;&/g' which does nothing.
What am I missing?
Try this:
echo "2141058222 11/22/2017" | sed -r 's/^([0-9]{10})/&;/'
echo "2141058222 11/22/2017" | sed 's/ /; /'
Output:
2141058222; 11/22/2017
If the input is always in the format specified, GNU cut works, and might even be more efficient than sed:
cut -c -10,11- --output-delimiter ';' <<< "2141058222 11/22/2017"
Output:
2141058222; 11/22/2017
For an input file that'd be:
cut -c -10,11- --output-delimiter ';' file
I have been struggling with this all day. Trying to make variables in sections of a line only contained within braces.
Lines look like this:
blah blah [ae b c] blah [zv y] blah
I need to make this:
blah blah [$ae $b $c] blah [$zv $y] blah
There must be an easy way to do this. However, whenever I try
$ echo "blah blah [ae b c] blah [zv y] blah" | sed 's/\[\(\b.*\b\)\]/$\1/g'
I get greedy matching and just one variable:
blah blah $ae b c] blah [zv y blah
Is there something better?
Thanks,
$ echo "blah blah [ae b c] blah [zv y] blah" | sed -r ':b; s/([[][^]$]* )([[:alnum:]]+)/\1$\2/g; t b; s/[[]([[:alnum:]])/[$\1/g'
blah blah [$ae $b $c] blah [$zv $y] blah
How it works
-r
This turns on extended regex.
:b
This creates a label b.
s/([[][^]$]* )([[:alnum:]]+)/\1$\2/g
This looks for [, followed by anything except ] or $, followed by a space, followed by any alphanumeric characters. It puts a $ in front of the alphanumeric characters.
Note that awk convention that makes [[] match [ while [^]$] matches anything except ] and $. This is more portable than attempting to escape these characters with backslashes.
t b
If the command above resulted in a substitution, this branches back to label b so that the substitution is attempted again.
s/[[]([[:alnum:]])/[$\1/g
The last step is to look for [ followed by an alphanumeric character and put a $ between them.
Because [[:alnum:]] is used, this code is unicode-safe.
Mac OSX (BSD) Version
On BSD sed (OSX) limits the ability to combine statements with semicolons. Try this instead:
sed -E -e ':b' -e 's/([[][^]$]* )([[:alnum:]]+)/\1$\2/g' -e 't b' -e 's/[[]([[:alnum:]])/[$\1/g'
To disable it being greedy, instead of matching any character, match any character except closing bracket:
sed 's/\[\(\b[^]]*\b\)\]/$\1/g'
The task you want to do cannot be done with sed because context-sensitive matching cannot be described with regular grammar.
It's difficult to solve it using sed. As alternative, you can use perl with the help of the Text::Balanced module, that extracts text between balanced delimiters, like square brackets. Each call returns an array with the content between delimiters, the text before them and the text after them, so you can apply the regex that insert $ sign to the significative part of the string.
perl -MText::Balanced=extract_bracketed -lne '
BEGIN { $result = q||; }
do {
#result = extract_bracketed($_, q{[]}, q{[^[]*});
if (! defined $result[0]) {
$result .= $result[1];
last;
}
$result[0] =~ s/(\[|\s+)/$1\$/g;
$result .= $result[2] . $result[0];
$_ = $result[1];
} while (1);
END { printf qq|%s\n|, $result; }
' infile
It yields:
blah blah [$ae $b $c] blah [$zv $y] blah
sed 's/\[\([^]]*\)\]/[ \1]/g
:loop
s/\(\(\[[^]$]*\)\([[:blank:]]\)\)\([^][:blank:]$][^]]*\]\)/\1\$\4/g
t loop
s/\[ \([^]]*\)\]/[\1]/g' YourFile
posix version
assuming there is no bracket inside bracket like [a b[c] d ]
algo:
add a space char after opening bracket (needed to use blank as starting word separator an often no space for first one)
label anchor for a loop
add a $ in front of last word between bracket that does not have one (not starting by $). Do it for each bracket group in line, but 1 add per group only
if occuring, retry another time going to label loop
remove the first space added in first operation
This might work for you (GNU sed):
sed -r 'h;s/\</$/g;T;G;s/^/\n/;:a;s/\n[^[]*(\[[^]]*\])(.*\n)([^[]*)[^]]*\]/\3\1\n\2/;ta;s/\n(.*)\n(.*)/\2/' file
Make a copy of the current line. Insert $ infront of all start-of-word boundaries. If nothing is substituted print the current line and bale out. Otherwise append the copy of the unadulterated line and insert a newline at the start of the adulterated current line. Using substitution and pattern matching replace the parts of the line between [...] with the original matching parts using the newline to move the match forwards through the line. When all matches have been made replace the end of the original line and remove the newlines.
I have a file with a certain line, let's say...
AAA BBB CCC
I need to replace that entire line, after finding it, so I did:
q1=`grep -Hnm 1 "AAA" FILE | cut -d : -f 2`
That outputs me the line number of the first occurrence (in q1), because it has more than one occurrence, now, here comes my problem... In a previous step I was using this sed to replace a certain line in the file:
sed -e '3s/.*/WHATEVER/' FILE
To replace (in the example, line 3) the full line with WHATEVER, but now if I try to use $q1 instead of the "3" indicating the line number it doesn't work:
sed -e '$q1s/.*/WHATEVER/' FILE
It's probably a stupid syntax mistake, any help is welcome; thanks in advance
Try:
sed -e "${q1}s/.*/WHATEVER/" FILE
I'd use awk for this:
awk '/AAA/ && !r {print "WHATEVER"; r=1; next} {print}' <<END
a
b
AAA BBB CCC
d
e
AAA foo bar
f
END
a
b
WHATEVER
d
e
AAA foo bar
f
If you want to replace the first occurrence of a string in a file, you could use this awk script:
awk '/occurrence/ && !f++ {print "replacement"; next}1' file
The replacement will only be printed the first time, as !f++ will only evaluate to true once (on subsequent evaluations, f will be greater than zero so !f will be false. The 1 at the end is always true, so for each line other than the matched one, awk does the default action and prints the line.
Testing it out:
$ cat file
blah
blah
occurrence 1 and some other stuff
blah
blah
some more stuff and occurrence 2
blah
$ awk '/occurrence/ && !f++ {print "replacement"; next}1' file
blah
blah
replacement
blah
blah
some more stuff and occurrence 2
blah
The "replacement" string could easily be set to the value of a shell variable in the following way:
awk -v rep="$r" '/occurrence/ && !f++ {print rep; next}1' file
where $r is a shell variable.
Using the same file as above and the example variable in your comment:
$ q2="x=\"Second\""
$ awk -v rep="$q2" '/occurrence/ && !f++ {print rep; next}1' file
blah
blah
x="Second"
stuff
blah
blah
some more stuff and occurrence 2
blah
sed "${q1} c\
WHATEVER" YourFile
but you can directly use
sed '/YourPatternToFound/ {s/.*/WHATEVER/
:a
N;$!ba
}' YourFile