I got the following string: '/transfer/IN/name/test.txt'
Now I'm trying to split this for the string name, cause I need it for further operations.
How can I split this correctly?
I've tried with cut (would sed be better?), but I'm not able to find the right approach.
Thanks for your help in advance.
This should help too:
awk -F'/' '{print $(NF-1)}' <<<"/a/b/c/d"
And it outputs:
c
If you like using sed:
sed 's#/[^/]*$##;s#.*/##' <<<"/a/b/c/d"
Just use the proper tools dirname and basename chained together:
echo $(basename $(dirname /transfer/IN/name/test.txt))
dirname => /transfer/IN/name
basename => name
sed solution looks more complex BTW:
sed -e "s#.*/\([^/]*\)/[^/]*#\1#" -e "s#/.*##" <<< name/test.txt
(2 expressions to handle the full relative case name/test.txt)
You can use Parameter expansion:
#!/bin/bash
path=/transfer/IN/name/test.txt
path1=${path%/*} # Remove everything from the last /.
path1=${path1##*/} # Remove everything up to the last /.
echo "$path1"
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 a file, the content of file has a string like this:
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
I want to check the file has this string. But when I use grep to check, It always return false. I try some ways:
grep "'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'" foo.txt
grep "'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'" foo.txt
str="'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'"
grep "$str" foo.txt
Can you help me? Maybe, another command line.
This is my case:
while read str; do
if [ ! -z "$str" ]; then
if grep -Fxq "$str" "$file_path"; then
do somthing
fi
fi
done < <(cat /usr/local/caotoc/db.dat)
Thank you so much!
First, you need to make sure the string is quoted properly. This is a bit of an art form, since your string contains both single and double quotes.
One thought would be to use read and a here-document to avoid having to escape anything.
Second, you need to use -F to perform exact string matching instead of more general regular-expression matching.
IFS= read -r str <<'EOF'
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
EOF
grep -F "$str" foo.txt
Based on the update, you can use a simple loop to read them one at a time.
while IFS= read -r str; do
grep -F "$str" foo.txt
done < /usr/local/caotoc/db.dat
You may be able to simply use the -f option to grep, which will cause grep to output lines from foo.txt that match any line from db.dat.
grep -f /usr/local/caotoc/db.dat -F foo.txt
Instead of trying to workaround regexes, the simplest way is to turn off regular expressions using -F (or --fixed-strings) option, which makes grep act like a simple string search
-F, --fixed-strings PATTERN is a set of newline-separated strings
like this:
grep -F "'/ad/e','#'.base64_decode(\"ZXZhbA==\").'(\$zad)', 'add'" test
Note: because of the shell, you still need to escape:
double quotes
dollar sign or else $zad is evaluated as an environment variable
I have a file that contains strings in this format:
MD5 (TestImages/IMG_0627.JPG) = 6ed611b3e777c5f7b729fa2f2412d656
I am trying to figure out a way to extract the file path, so that I would get a string like this:
TestImages/IMG_0627.JPG
For a different part of my script, I am using this code to remove everything before and after the brackets, and I could of course do something similar, however I'm sure there is a better way?
shortFile=${line#*MD5 }
shortFile=${shortFile%%)*}
Anyone have any suggestions?
You could use sed but that has the overhead of starting a new process.
echo $line | sed -r 's/MD5 \((.*)\).*/\1/'
Just to throw a non-sed answer onto the pile. (Also slightly cheaper since it avoids the pipeline and sub-shell.)
awk -F '[()]' '{print $2}' <<<"$line"
That said the substring expansion option is a reasonable one if it does what you need. (Though it looks like you missed the ( in the first expansion.)
Another way with cut can be :
echo $line|cut -d "(" -f2|cut -d ")" -f1
sed -e 's/^.*(\([^)]*\)).*$/\1/' < infile.txt
I Want to search variable and replace with its absolute path in file.
setenv ABC /home/xyz
cat file.txt
${ABC}/Test/Folder_1
${ABC}/Test/Folder_2
I want to replace all occurance of the ${ABC} by /home/xyz.
I tried by the below mentioned way, but does not work,
sed -i 's/\$ABC/echo $ABC/g' file.txt
I can do by below mentioned way, but I do not want to do this way.(I have to put so many back slash)
$ echo $ABC | sed -i 's/\$ABC/\/home\/xyz/g' file.txt
Please give me some suggestion for this question.
Thank You.
If you really want to use the value from a variable in your replacement string, you could use
sed "s#\${ABC}#$ABC#g" file.txt
Character after s in sed is the delimiter and it can be any one character of your choice and it works as long as it's not in the string-to-be-matched and string-to-be-replaced.
Example :
sed 's:string-to-be-matched:string-to-be-replaced:g' file-to-be-edited
: is the delimiter
g means global replacement.
In your case, as the string-to-be-replaced contains the / , the same you are using as sed delimiter.
Simple Solution will be :
sed -i 's:${ABC}:'"$ABC"':g' fill.txt
'" is at either end of $ABC in the replacement string. Purpose is to expand shell variable to use with sed
Another way to get the absolute path is readlink -f ${ABC}/Test/Folder_2
Or the perl alternative to your slash hungry command
$ echo $ABC | sed -i 's/\$ABC/\/home\/xyz/g' file.txt
would be
$ echo $ABC | perl -p -i -e 's!\$ABC!/home/xyz!g'
the first character after the 's above will be used as the delimiter in the replacement expression (i.e. 's#foo#bar#g')
I am currently working on a project where I would like to change a picture in multiple places.
old file dir: /images/icons/helpPop.png
new file dir: /public/website_pngs/icons-buttons/button_question_mark.png
I want to try
grep -rl 'images/icons/helpPop' . | xargs sed -i 's/images/icons/helpPop/public/website_pngs/icons-buttons/button_question_mark/g'
But I know that will not work. I am looking into delimiters and would like some extra advice please.
I am looking into delimiters and would like some extra advice please.
some suggestions:
if the old/new dirs are fixed, try to save them into variables
use different delimiter in sed e.g. "s#$old#$new#g" or "s#$old#$new#g" (double quotes!!)
to be safer, you better escape the . (period/dot) in your old dir text. since it indicates any char in regex. it could be a dangerous operation with plain text like that. (fortunately you don't have .* in your path ^_^ )
or you want the working codes?
EDIT for comment:
you don't need to "setup" the delimiter, just use it, like you are using /, take a look this example:
kent$ echo "/////"|sed 's#/#?#g'
?????
kent$ echo "/////"|sed 's^/^?^g'
?????
kent$ echo "/////"|sed 's%/%?%g'
?????
kent$ echo "/////"|sed 's;/;?;g'
?????
for i in grep -rl 'images/icons/helpPop' . ; do cp $i $i.bak; sed -i 's/images\/icons\/helpPop/public\/website_pngs\/icons-buttons\/button_question_mark/g' < $i.bak > $i; done