sed is still giving me headaches, so a little help is extremely appreciated.
In a file I have a string like:
SOME_TEXT="variables"
What I want to accomplish is to add a piece of text (variable) to either the end or the begging of the string for that text.
I tried to use variations of:
sed -i '/^SOME_TEXT="/ s/$/ SOME_TEXT="new text'/' filename
but that is failing, so clearly the quota for the string I want to add to is messing up the syntax.
LE:
A variation further is that I have a variable that I want to use as the replace in that syntax, so I have this:
sed -i "s/^SOME_TEXT="/SOME_TEXT=" $variable/" file
This actually produces this output, as it picks up incorrectly the opening/closing quotas:
SOME_TEXT = text_variable" initial text continuation
So how can I properly close the trailing quota so that I can use the variable after it?
I used
sed 's/^SOME_TEXT="/SOME_TEXT="new text/' filename
and it showed:
SOME_TEXT="new textvariables"
Is that what you want?
Escape the '"' characters with a '\' so that they don't terminate your regex string.
sed -i "s/^TEXT=\"/TEXT=\" $variable/"
Related
I'm having issues matching strings even if they start with any number of white spaces. It's been very little time since I started using regular expressions, so I need some help
Here is an example. I have a file (file.txt) that contains two lines
#String1='Test One'
String1='Test Two'
Im trying to change the value for the second line, without affecting line 1 so I used this
sed -i "s|String1=.*$|String1='Test Three'|g"
This changes the values for both lines. How can I make sed change only the value of the second string?
Thank you
With gnu sed, you match spaces using \s, while other sed implementations usually work with the [[:space:]] character class. So, pick one of these:
sed 's/^\s*AWord/AnotherWord/'
sed 's/^[[:space:]]*AWord/AnotherWord/'
Since you're using -i, I assume GNU sed. Either way, you probably shouldn't retype your word, as that introduces the chance of a typo. I'd go with:
sed -i "s/^\(\s*String1=\).*/\1'New Value'/" file
Move the \s* outside of the parens if you don't want to preserve the leading whitespace.
There are a couple of solutions you could use to go about your problem
If you want to ignore lines that begin with a comment character such as '#' you could use something like this:
sed -i "/^\s*#/! s|String1=.*$|String1='Test Three'|g" file.txt
which will only operate on lines that do not match the regular expression /.../! that begins ^ with optional whiltespace\s* followed by an octothorp #
The other option is to include the characters before 'String' as part of the substitution. Doing it this way means you'll need to capture \(...\) the group to include it in the output with \1
sed -i "s|^\(\s*\)String1=.*$|\1String1='Test Four'|g" file.txt
With GNU sed, try:
sed -i "s|^\s*String1=.*$|String1='Test Three'|" file
or
sed -i "/^\s*String1=/s/=.*/='Test Three'/" file
Using awk you could do:
awk '/String1/ && f++ {$2="Test Three"}1' FS=\' OFS=\' file
#String1='Test One'
String1='Test Three'
It will ignore first hits of string1 since f is not true.
This should be extremely simple, but for the life of me I just can't get gnu-sed to do it this afternoon.
The file in question has lines that look like this:
PART NUMBER PART NUMBER QUANTITY WEIGHT -999 -4,999 -9,999
w/ UL APPROVAL
MIN-3
I need to prepend every line like the "MIN-3" line with a ">" character, and the only thing specifically differentiating those lines from the others are two things:
The first character is a space " ".
The lines do not contain a comma.
I've tried mostly things like any of the following:
/^ +[^,]+$/ s/^/>/
/^ +[\w\-]+$/ s/^/>/
/^ +(\w|\-)+$/ s/^/>/
I will admit, I am somewhat new to sed. :)
Edit: Answers that use perl, or awk could also be appreciated, though my initial target is sed.
try this:
sed '/^ [^,]*$/s/^/>/'
the output is, only the line with MIN-3 with leading >
sed default uses basic regex. so the + should be \+ in your script. I think that could be the problem killing your time. You could add -r however, to let sed use extended-regex.
According to your description this should do:
sed 's/^\([ ][^,]*\)$/> \1/' input
which matches the complete line if the line starts with a space and then contains anything but a comma until the end.
Here is a simple answer:
sed 's/^ [^,]*$/>&/'
I am not sure why sed is not working as expected in this particular instance. I have lines of the form:
12:42:46.675 token
where I expect the timestamp to alwas have that format. Unfortunately every now and then there are lines in the file which do not begin with a timestamp and I want to get rid of those. I tried filtering out everything that does not match the above with:
sed -n /^\d{2}:\d{2}:\d{2}.\d{3}/p
but the above filters everything out, even if I give sed the -r option. What is the correct way of doing that with sed? And is there an alternative with grep?
Using grep to only display lines starting with timestamp format:
grep -E '^([0-9]{2}:){2}[0-9]{2}\.[0-9]{3} ' file
Sed doesn't accept \d, use [0-9] instead. And both { and } are not metacharacters, they are literal for sed so you will need to escape them for the special behaviour, it would result like:
sed -n '/^[0-9]\{2\}:[0-9]\{2\}:[0-9]\{2\}.[0-9]\{3\}/p' infile
EDIT: Also surround the expression between quotes (better singles than double) to avoid shell expansion.
This is a simple question, I'm not sure if i'm able to do this with sed/awk
How can I make sed search for these 3 lines and replace with a line with a determined string?
<Blarg>
<Bllarg>
<Blllarg>
replace with
<test>
I tried with sed "s/<Blarg>\n<Bllarg>\n<Blllarg>/<test>/g" But it just don't seem to find these lines. Probably something with my break line character (?) \n. Am I missing something?
Because sed usually handles only one line at a time, your pattern will never match. Try this:
sed '1N;$!N;s/<Blarg>\n<Bllarg>\n<Blllarg>/<test>/;P;D' filename
This might work for you:
sed '/<Blarg>/ {N;N;s/<Blarg>\n<Bllarg>\n<Blllarg>/<test>/}' <filename>
It works as follows:
Search the file till <Blarg> is found
Then append the two following lines to the current pattern space using N;N;
Check if the current pattern space matches <Blarg>\n<Bllarg>\n<Blllarg>
If so, then substitute it with <test>
You can use range addresses with regular expressions an the c command, which does exactly what you are asking for:
sed '/<Blarg>/,/<Blllarg>/c<test>' filename
I use the following sed syntax in order to replace the timezone string with #timezone
actually sed command replace it ,
but in case after timezone string we have one space or more then sed not replaced it
my question what I need to change in the sed syntax in order to replace the timezone string
on the both cases
After timezone string we not have any space
After timezone string we a have space or more
sed -i '/timezone$/s/timezone/#timezone/' file
You can try:
sed -i.bak -r 's/timezone ?$/#timezone/' file
you need to add the ? after a space, that will make the space optional. Also you'll have to add the -r option to make sed understand extended regex.
Also you are not specifying the backup extension after the -i option.
Stating your two cases as one, you want to replace 'timezone' followed by zero or more spaces, when it is at the end of the line. That should be this:
sed -i 's/timezone *$/#timezone/' file
This should be an ideal solution for both your cases.
sed 's/timezone *$\|timezone/#&/' INPUT_FILE