I wan't to replace a string like Europe12 with Europe12_yesturday in a file. Without changing the Europe12-36 strings that also exists in the file.
I tried:
$basename=Europe12
sed -i 's/\b$basename\b/${basename}_yesterday/g' file.txt
but this also changed the Europe12-36 strings.
Require a space or end of line character:
sed 's/Europe12\([ ]|$\)/Europe12_yesturday\1/g' input
Manually construct the delimiter list you want instead of using \b, \W or \<. - is not part of the word characters (alphanumericals), so that's why this also matches your other string. So try something like this, expanding the list as needed: [-a-zA-Z0-9].
You can do it in 2 times:
sed -e 's/Europe12/Europe12_yesturday/g' -e 's/Europe12_yesturday-36/Europe12-36/g' file.txt
sed 's/\(Europe12[[:blank:]]\)/\1_yesturday/g;s/Europe12$/&_yesturday/' YourFile
[[:blank:]] could be completeted with any boundary you accept also like .,;:/]) etc (be carrefull of regex meaning of this char in this case)
It is little late to reply..
It can be achieved easily by "word boundary" notation (\<..\>)
sed -i 's/\<$basename\>/${basename}_yesterday/g' file.txt
Related
How to replace double consonants with only one letter using sed Linux command. Example: WILLIAM -> WILIAM. grep -E '(.)\1+' commands finds the words that follow two same consonants in a row pattern, but how do I replace them with only one occurrence of the letter?
I tried
cat test.txt | head | tr -s '[^AEUIO\n]' '?'
tr is all or nothing; it will replace all occurrences of the selected characters, regardless of context. For regex replacement, look at sed - you even included this in your question's tags, but you don't seem to have explored how it might be useful?
sed 's/\(.\)\1/\1/g' test.txt
The dot matches any character; to restrict to only consonants, change it to [b-df-hj-np-tv-xz] or whatever makes sense (maybe extend to include upper case; perhaps include accented characters?)
The regex dialect understood by sed is more like the one understood by grep without -E (hence all the backslashes); though some sed implementations also support this option to select the POSIX extended regular expression dialect.
Neither sed not tr need cat to read standard input for them (though tr obscurely does not accept a file name argument). See tangentially also Useless use of cat?
Match one consonant, remember it in \( \), then match is again with \1 and substitute it for itself.
sed 's/\([bcdfghjklmnpqrstvxzBCDFGHJKLMNPQRSTVXZ]\)\1/\1/'
I'm trying to replace a every reference of \' with ' in a file
I've used variations of: sed -e s/\'/"\'"/g file.txt
But they always replace every.single.(single).quote
Any help would be greatly appreciated.
Not sure it's the best solution,I could do it like this:
sed "s/[\]'/\"\'\"/g" file.txt
(putting the backslash character in a character range so it doesn't interfere with the following quote, and protect with double quotes)
Or just extending your syntax, without quotes but using almost the same trick:
sed -e s/[\\]\'/"\'"/g file.txt
An approach trying to conserve as much of the "single-quotedness" of the sed command as possible:
sed 's/\\'"'"'/\'/g'
Just escaping \ with \\ and getting a single quote into the command with '"'"': the first single quote ends the command so far, then we have a double-quoted single quote ("'"), and finally an opening single quote for the rest of the command.
Alternatively, double quoting the whole command and escaping both the backslash and single quote:
sed "s/\\\'/\'/g"
The correct syntax is:
$ echo "foo'bar" | sed 's/'\''/\'/'
foo'bar
Every script (sed, awk, whatever) should always be enclosed in single quotes and you just us other single quotes to stop/restart the script delimiters break out to shell for the minimal portion of the script that's absolutely necessary, in this case long enough to use \'. You need to break out to shell to specify that ' because per shell rules no script enclosed in 's can contain a ', not even if you try to escape it.
echo "foo'bar" | gawk '{gsub(/\47/,"\\'")}1'
foo'bar
The tricky part here is to replace a single quote with ampersand.
First in order to make the single quote manageable use its octal
code here \47 and then escaping ampersand by two back slash. And all of sudden
it becomes feasible :)
I would like to use sed to remove all occurances of this line if and only if it is this
<ab></ab>
If this line, I would not want to delete it
<ab>keyword</ab>
My attempt that's not working:
sed '/<ab></ab>/d'
Thanks for any insight. I'm not sure what's wrong as I should not have to escape anything?
I'm using a shell script named temp to execute this. My command is this:
cat foobar.html | ./temp
This is my temp shell script:
#!/bin/sh
sed -e '/td/!d' | sed '/<ab></ab>/d'
It looks like we have a couple of problems here. The first is with the / in the close-tag. sed uses this to delimit different parts of the command. Fortunately, all we have to do is escape it with \. Try:
sed '/<ab><\/ab>/d'
Here's an example on my machine:
$ cat test
<ab></ab>
<ab></ab>
<ab>test</ab>
$ sed '/<ab><\/ab>/d' test
<ab>test</ab>
$
The other problem is that I'm not sure what the purpose of sed -e '/td/!d' is. In it's default operating mode, you don't need to tell it not to delete something; just tell it exactly what you want to delete.
So, to do this on a file called input.html:
sed '/<ab><\/ab>/d' input.html
Or, to edit the file in-place, you can just do:
sed -i -e '/<ab><\/ab>/d' input.html
Additionally, sed lets you use any character you want as a delimiter; you don't have to use /. So if you'd prefer not to escape your input, you can do:
sed '\#<ab></ab>#d' input.html
Edit
In the comments, you mentioned wanting to delete lines that only contain </ab> and nothing else. To do that, you need to do what's called anchoring the match. The ^ character represents the beginning of the line for anchoring, and $ represents the end of the line.
sed '/^<\/ab>$/d' input.html
This will only match a line that contains (literally) </ab> and nothing else at all, and delete the line. If you want to match lines that contain whitespace too, but no text other than </ab>:
sed '/^[[:blank:]]*<\/ab>[[:blank:]]*$/d' input.html
[[:blank:]]* matches "0 or more whitespace characters" and is called a "POSIX bracket expression".
I have a simple sed question.
I have data like this:
2600,Sale,"Approved 911973",244.72
2601,Sale,"Approved 04735C",490.51
2602,Sale,"Approved 581068",52.82
2603,Sale,"Approved 009275",88.10
How do I make it like this:
2600,Sale,Approved,244.72
2601,Sale,Approved,490.51
2602,Sale,Approved,52.82
2603,Sale,Approved,88.10
Notice the numbers after approved are gone as well as the quotes. I can remove quotes with:
sed 's/,$//gn' file
but I don't know how to remove the spaces and digits.
Thanks!
sed "s/\"Approved[^,]*/Approved/g"
It finds the quoted "Approved" followed by any non-comma character, up until the first comma encountered, and replaces it with Approved (no quotes)
2600,Sale,Approved,244.72
2601,Sale,Approved,490.51
2602,Sale,Approved,52.82
2603,Sale,Approved,88.10
Using extended regex with sed:
sed -r 's/"([^[:space:]]*)[^"]*"/\1/g' file
The above regex targets for any quoted string. If you want to target the string Approved, then:
sed -r 's/"(Approved)[^"]*"/\1/g' file
With basic regex:
sed 's/"\(Approved\)[^"]*"/\1/g' file
To target any quoted string, change Approved to [^[:space:]]*
One way using awk(only if the other columns does not contain multiple words as in your sample):
awk -F"[ ,]" '{gsub("\"","");$1=$1}1' OFS=, file
awk -F'[," ]' '{OFS=","; print $1,$2,$4,$7}' file
Output:
2600,Sale,Approved,244.72
2601,Sale,Approved,490.51
2602,Sale,Approved,52.82
2603,Sale,Approved,88.10
I suppose there is no other whitespace.
I am trying to find the following text
get_pins {
and replace it with
get_pins -hierarchical {proc_top_*/
I've tried using sed but I'm not sure what I'm doing wrong. I know that you need # in front of curly braces but I still can't get the command to work properly.
The closest I've come is to this:
sed 's/get_pins #{#/get_pins -hierarchical #{#proc_top_*\//g' filename.txt > output
but it doesn't do the replacement I wanted above.
#merlin2011's answer shows you how to do it with alternative delimiters, but as for why your command didn't work:
It's actually perfectly fine, if you just remove all # chars. from your statement:
sed 's/get_pins {/get_pins -hierarchical {proc_top_*\//'g filename.txt > output
There are two distinct escaping requirements involved here:
Escaping literal use of the regex delimiter: this is what you did correctly, by escaping the / as \/.
Escaping characters with special meaning inside a regex in general: this escaping is always done with \-prefixing, but in your case there is NO need for such escaping: since you're NOT using -E or -r to indicate use of extended regexes - and are therefore using a basic regex - { is actually NOT a special character, so you need NOT escape it. If, by contrast, you had used -E (-r), then you should have escaped { as \{.
The problem is not in the curly braces, it's in the /.
This is exactly why sed lets you do alternate delimiters.
The line below uses ! as a delimiter instead, and works correctly for a simple file with get_pins { in it.
sed 's!get_pins {!get_pins -hierarchical {proc_top_*/!g' Input.txt
Output:
get_pins -hierarchical {proc_top_*/
Update: Based mklement0's comment, and testing with the csh shell, the following should work in csh.
sed 's#get_pins {#get_pins -hierarchical {proc_top_*/#g' Input.txt
This awk should do the replace:
awk '{sub(/get_pins {/,"get_pins -hierarchical {proc_top_*/")}1'