I have a string variable x=tmp/variable/custom-sqr-sample/test/example
in the script, what I want to do is to replace all the “-” with the /,
after that,I should get the following string
x=tmp/variable/custom/sqr/sample/test/example
Can anyone help me?
I tried the following syntax
it didnot work
exa=tmp/variable/custom-sqr-sample/test/example
exa=$(echo $exa|sed 's/-///g')
sed basically supports any delimiter, which comes in handy when one tries to match a /, most common are |, # and #, pick one that's not in the string you need to work on.
$ echo $x
tmp/variable/custom-sqr-sample/test/example
$ sed 's#-#/#g' <<< $x
tmp/variable/custom/sqr/sample/test/example
In the commend you tried above, all you need is to escape the slash, i.e.
echo $exa | sed 's/-/\//g'
but choosing a different delimiter is nicer.
The tr tool may be a better choice than sed in this case:
x=tmp/variable/custom-sqr-sample/test/example
echo "$x" | tr -- - /
(The -- isn't strictly necessary, but keeps tr (and humans) from mistaking - for an option.)
In bash, you can use parameter substitution:
$ exa=tmp/variable/custom-sqr-sample/test/example
$ exa=${exa//-/\/}
$ echo $exa
tmp/variable/custom/sqr/sample/test/example
Related
I am using the below code for replacing a string
inside a shell script.
echo $LINE | sed -e 's/12345678/"$replace"/g'
but it's getting replaced with $replace instead of the value of that variable.
Could anybody tell what went wrong?
If you want to interpret $replace, you should not use single quotes since they prevent variable substitution.
Try:
echo $LINE | sed -e "s/12345678/${replace}/g"
Transcript:
pax> export replace=987654321
pax> echo X123456789X | sed "s/123456789/${replace}/"
X987654321X
pax> _
Just be careful to ensure that ${replace} doesn't have any characters of significance to sed (like / for instance) since it will cause confusion unless escaped. But if, as you say, you're replacing one number with another, that shouldn't be a problem.
you can use the shell (bash/ksh).
$ var="12345678abc"
$ replace="test"
$ echo ${var//12345678/$replace}
testabc
Not specific to the question, but for folks who need the same kind of functionality expanded for clarity from previous answers:
# create some variables
str="someFileName.foo"
find=".foo"
replace=".bar"
# notice the the str isn't prefixed with $
# this is just how this feature works :/
result=${str//$find/$replace}
echo $result
# result is: someFileName.bar
str="someFileName.sally"
find=".foo"
replace=".bar"
result=${str//$find/$replace}
echo $result
# result is: someFileName.sally because ".foo" was not found
Found a graceful solution.
echo ${LINE//12345678/$replace}
Single quotes are very strong. Once inside, there's nothing you can do to invoke variable substitution, until you leave. Use double quotes instead:
echo $LINE | sed -e "s/12345678/$replace/g"
Let me give you two examples.
Using sed:
#!/bin/bash
LINE="12345678HI"
replace="Hello"
echo $LINE | sed -e "s/12345678/$replace/g"
Without Using sed:
LINE="12345678HI"
str_to_replace="12345678"
replace_str="Hello"
result=${str//$str_to_replace/$replace_str}
echo $result
Hope you will find it helpful!
echo $LINE | sed -e 's/12345678/'$replace'/g'
you can still use single quotes, but you have to "open" them when you want the variable expanded at the right place. otherwise the string is taken "literally" (as #paxdiablo correctly stated, his answer is correct as well)
To let your shell expand the variable, you need to use double-quotes like
sed -i "s#12345678#$replace#g" file.txt
This will break if $replace contain special sed characters (#, \). But you can preprocess $replace to quote them:
replace_quoted=$(printf '%s' "$replace" | sed 's/[#\]/\\\0/g')
sed -i "s#12345678#$replace_quoted#g" file.txt
I had a similar requirement to this but my replace var contained an ampersand. Escaping the ampersand like this solved my problem:
replace="salt & pepper"
echo "pass the salt" | sed "s/salt/${replace/&/\&}/g"
use # if you want to replace things like /. $ etc.
result=$(echo $str | sed "s#$oldstr#$newstr#g")
the above code will replace all occurrences of the specified replacement term
if you want, remove the ending g which means that the only first occurrence will be replaced.
Use this instead
echo $LINE | sed -e 's/12345678/$replace/g'
this works for me just simply remove the quotes
I prefer to use double quotes , as single quptes are very powerful as we used them if dont able to change anything inside it or can invoke the variable substituion .
so use double quotes instaed.
echo $LINE | sed -e "s/12345678/$replace/g"
I have something like this in a file named file.txt
AA.201610.pancake.Paul
AA.201610.hello.Robert
A.201610.hello.Mark
Now, i ONLY get the first three fields in 3 variables like:
field1="A"
field2="201610"
field3='hello'.
I'd like to remove a line, if it contains exactly the first 3 fields, like , in the case described above, i want only the third line to be removed from the file.txt . Is there a way to do that? And is there a way to do that in the same file?
I tried with:
sed -i /$field1"."$field2"."$field3"."/Id file.txt
but of course this removes both the second and the third line
I suggest using awk for this as sed can only do regex search and that requires escaping all special meta-chars and anchors, word boundaries etc to avoid false matches.
Suggested awk with non-regex matching:
awk -F '[.]' -v f1="$field1" -v f2="$field2" -v f3="$field3" '
!($1==f1 && $2==f2 && $3==f3)' file
AA.201610.pancake.Paul
AA.201610.hello.Robert
Use ^ to anchor the pattern at the beginning of the line. Also note that . in a regex means "any character" and not a literal peridio. You have to escape it: either \. (be careful with shell escaping and the difference between single and double quotes) or [.]
Sed cannot do string matches, only regexp matches which becomes horrendously complicated to work around when you simply want to match a literal string (see Is it possible to escape regex metacharacters reliably with sed). Just use awk:
$ awk -v str="${field1}.${field2}.${field3}." 'index($0,str)!=1' file
AA.201610.pancake.Paul
AA.201610.hello.Robert
The question was about bash so in bash:
#!/usr/bin/env bash
field1="A"
field2="201610"
field3='hello'
IFS=
while read -r i
do
case "$i" in
"${field1}.${field2}.${field3}."*) ;;
*) echo -E "$i"
esac
done < file.txt
I trying to find and replace items using bash. I was able to use sed to grab out some of the characters, but I think I might be using it in the wrong matter.
I am basically trying to remove the characters after ";" and before "," including removing ","
sed -e 's/\(;\).*\(,\)/\1\2/'
That is what I used to replace it with nothing. However, it ends up replacing everything in the middle so my output came out like this:
cmd2="BMC,./socflash_x64 if=B600G3_BMC_V0207.ima;,reboot -f"
This is the original text of what I need to replace
cmd2="BMC,./socflash_x64 if=B600G3_BMC_V0207.ima;X,sleep 120;after_BMC,./run-after-bmc-update.sh;hba_fw,./hba_fw.sh;X,sleep 5;DB,2;X,reboot -f"
Is there any way to make it look like this output?
./socflash_x64 if=B600G3_BMC_V0207.ima;sleep 120;./run-after-bmc-update.sh;./hba_fw.sh;sleep 5;reboot -f
Ff there is any way to make this happen other than bash I am fine with any type of language.
Non-greedy search can (mostly) be simulated in programs that don't support it by replacing match-any (dot .) with a negated character class.
Your original command is
sed -e 's/\(;\).*\(,\)/\1\2/'
You want to match everything in between the semi-colon and the comma, but not another comma (non-greedy). Replace .* with [^,]*
sed -e 's/\(;\)[^,]*\(,\)/\1\2/'
You may also want to exclude semi-colons themselves, making the expression
sed -e 's/\(;\)[^,;]*\(,\)/\1\2/'
Note this would treat a string like "asdf;zxcv;1234,qwer" differently, since one would match ;zxcv;1234, and the other would match only ;1234,
In perl:
perl -pe 's/;.*?,/;/g;' -pe 's/^[^,]*,//' foo.txt
will output:
./socflash_x64 if=B600G3_BMC_V0207.ima;sleep 120;./run-after-bmc-update.sh;./hba_fw.sh;sleep 5;2;reboot -f
The .*? is non greedy matching before the comma. The second command is to remove from the beginning to the comma.
Something like:
echo $cmd2 | tr ';' '\n' | cut -d',' -f2- | tr '\n' ';' ; echo
result is:
./socflash_x64 if=B600G3_BMC_V0207.ima;sleep 120;./run-after-bmc-update.sh;./hba_fw.sh;sleep 5;2;reboot -f;
however, I thing your requirements are a few more complex, because 'DB,2' seems a particular case. After "tr" command, insert a "grep" or "grep -v" to include/exclude these cases.
How can I fix this:
abc="a/b/c"; echo porc | sed -r "s/^/$abc/"
sed: -e expression #1, char 7: unknown option to `s'
The substitution of variable $abc is done correctly, but the problem is that $abc contains slashes, which confuse sed. Can I somehow escape these slashes?
Note that sed(1) allows you to use different characters for your s/// delimiters:
$ abc="a/b/c"
$ echo porc | sed -r "s|^|$abc|"
a/b/cporc
$
Of course, if you go this route, you need to make sure that the delimiters you choose aren't used elsewhere in your input.
The GNU manual for sed states that "The / characters may be uniformly replaced by any other single character within any given s command."
Therefore, just use another character instead of /, for example ::
abc="a/b/c"; echo porc | sed -r "s:^:$abc:"
Do not use a character that can be found in your input. We can use : above, since we know that the input (a/b/c/) doesn't contain :.
Be careful of character-escaping.
If using "", Bash will interpret some characters specially, e.g. ` (used for inline execution), ! (used for accessing Bash history), $ (used for accessing variables).
If using '', Bash will take all characters literally, even $.
The two approaches can be combined, depending on whether you need escaping or not, e.g.:
abc="a/b/c"; echo porc | sed 's!^!'"$abc"'!'
You don't have to use / as pattern and replace separator, as others already told you. I'd go with : as it is rather rarely used in paths (it's a separator in PATH environment variable). Stick to one and use shell built-in string replace features to make it bullet-proof, e.g. ${abc//:/\\:} (which means replace all : occurrences with \: in ${abc}) in case of : being the separator.
$ abc="a/b/c"; echo porc | sed -r "s:^:${abc//:/\\:}:"
a/b/cporc
backslash:
abc='a\/b\/c'
space filling....
As for the escaping part of the question I had the same issue and resolved with a double sed that can possibly be optimized.
escaped_abc=$(echo $abc | sed "s/\//\\\AAA\//g" | sed "s/AAA//g")
The triple A is used because otherwise the forward slash following its escaping backslash is never placed in the output, no matter how many backslashes you put in front of it.
If I run these commands from a script:
#my.sh
PWD=bla
sed 's/xxx/'$PWD'/'
...
$ ./my.sh
xxx
bla
it is fine.
But, if I run:
#my.sh
sed 's/xxx/'$PWD'/'
...
$ ./my.sh
$ sed: -e expression #1, char 8: Unknown option to `s'
I read in tutorials that to substitute environment variables from shell you need to stop, and 'out quote' the $varname part so that it is not substituted directly, which is what I did, and which works only if the variable is defined immediately before.
How can I get sed to recognize a $var as an environment variable as it is defined in the shell?
Your two examples look identical, which makes problems hard to diagnose. Potential problems:
You may need double quotes, as in sed 's/xxx/'"$PWD"'/'
$PWD may contain a slash, in which case you need to find a character not contained in $PWD to use as a delimiter.
To nail both issues at once, perhaps
sed 's#xxx#'"$PWD"'#'
In addition to Norman Ramsey's answer, I'd like to add that you can double-quote the entire string (which may make the statement more readable and less error prone).
So if you want to search for 'foo' and replace it with the content of $BAR, you can enclose the sed command in double-quotes.
sed 's/foo/$BAR/g'
sed "s/foo/$BAR/g"
In the first, $BAR will not expand correctly while in the second $BAR will expand correctly.
Another easy alternative:
Since $PWD will usually contain a slash /, use | instead of / for the sed statement:
sed -e "s|xxx|$PWD|"
You can use other characters besides "/" in substitution:
sed "s#$1#$2#g" -i FILE
一. bad way: change delimiter
sed 's/xxx/'"$PWD"'/'
sed 's:xxx:'"$PWD"':'
sed 's#xxx#'"$PWD"'#'
maybe those not the final answer,
you can not known what character will occur in $PWD, / : OR #.
if delimiter char in $PWD, they will break the expression
the good way is replace(escape) the special character in $PWD.
二. good way: escape delimiter
for example:
try to replace URL as $url (has : / in content)
x.com:80/aa/bb/aa.js
in string $tmp
URL
A. use / as delimiter
escape / as \/ in var (before use in sed expression)
## step 1: try escape
echo ${url//\//\\/}
x.com:80\/aa\/bb\/aa.js #escape fine
echo ${url//\//\/}
x.com:80/aa/bb/aa.js #escape not success
echo "${url//\//\/}"
x.com:80\/aa\/bb\/aa.js #escape fine, notice `"`
## step 2: do sed
echo $tmp | sed "s/URL/${url//\//\\/}/"
URL
echo $tmp | sed "s/URL/${url//\//\/}/"
URL
OR
B. use : as delimiter (more readable than /)
escape : as \: in var (before use in sed expression)
## step 1: try escape
echo ${url//:/\:}
x.com:80/aa/bb/aa.js #escape not success
echo "${url//:/\:}"
x.com\:80/aa/bb/aa.js #escape fine, notice `"`
## step 2: do sed
echo $tmp | sed "s:URL:${url//:/\:}:g"
x.com:80/aa/bb/aa.js
With your question edit, I see your problem. Let's say the current directory is /home/yourname ... in this case, your command below:
sed 's/xxx/'$PWD'/'
will be expanded to
sed `s/xxx//home/yourname//
which is not valid. You need to put a \ character in front of each / in your $PWD if you want to do this.
Actually, the simplest thing (in GNU sed, at least) is to use a different separator for the sed substitution (s) command. So, instead of s/pattern/'$mypath'/ being expanded to s/pattern//my/path/, which will of course confuse the s command, use s!pattern!'$mypath'!, which will be expanded to s!pattern!/my/path!. I’ve used the bang (!) character (or use anything you like) which avoids the usual, but-by-no-means-your-only-choice forward slash as the separator.
Dealing with VARIABLES within sed
[root#gislab00207 ldom]# echo domainname: None > /tmp/1.txt
[root#gislab00207 ldom]# cat /tmp/1.txt
domainname: None
[root#gislab00207 ldom]# echo ${DOMAIN_NAME}
dcsw-79-98vm.us.oracle.com
[root#gislab00207 ldom]# cat /tmp/1.txt | sed -e 's/domainname: None/domainname: ${DOMAIN_NAME}/g'
--- Below is the result -- very funny.
domainname: ${DOMAIN_NAME}
--- You need to single quote your variable like this ...
[root#gislab00207 ldom]# cat /tmp/1.txt | sed -e 's/domainname: None/domainname: '${DOMAIN_NAME}'/g'
--- The right result is below
domainname: dcsw-79-98vm.us.oracle.com
VAR=8675309
echo "abcde:jhdfj$jhbsfiy/.hghi$jh:12345:dgve::" |\
sed 's/:[0-9]*:/:'$VAR':/1'
where VAR contains what you want to replace the field with
I had similar problem, I had a list and I have to build a SQL script based on template (that contained #INPUT# as element to replace):
for i in LIST
do
awk "sub(/\#INPUT\#/,\"${i}\");" template.sql >> output
done
If your replacement string may contain other sed control characters, then a two-step substitution (first escaping the replacement string) may be what you want:
PWD='/a\1&b$_' # these are problematic for sed
PWD_ESC=$(printf '%s\n' "$PWD" | sed -e 's/[\/&]/\\&/g')
echo 'xxx' | sed "s/xxx/$PWD_ESC/" # now this works as expected
for me to replace some text against the value of an environment variable in a file with sed works only with quota as the following:
sed -i 's/original_value/'"$MY_ENVIRNONMENT_VARIABLE"'/g' myfile.txt
BUT when the value of MY_ENVIRONMENT_VARIABLE contains a URL (ie https://andreas.gr) then the above was not working.
THEN use different delimiter:
sed -i "s|original_value|$MY_ENVIRNONMENT_VARIABLE|g" myfile.txt