Can anyone help me to get substring using sed program?
I have a file with this line:
....
define("BASE", "empty"); # there can be any string (not only "empty").
....
And I need to get "empty" as string variable to my bash script.
At this moment I have:
sed -n '/define(\"BASE\"/p' path/to/file.ext
# returns this line:
# define("BASE", "empty");
# but I need 'empty'
UPD: Thanks to #Jaypal
For now I have bash script:
DBNAME=`sed -n '/define(\"BASE\"/p' path/to/file.ext`
echo $DBNAME | sed -r 's/.*"([a-zA-Z]+)".*/\1/'
It work OK, but if there any way to make the same manipulation with one line of code?
You should use is
sed -n 's/.*\".*\", \"\(.*\)\".*/\1/p' yourFile.txt
which means something (.*) followed by something in quotes (\".*\"), then a comma and a blank space (,), and then again something within quotes (\"\(.*\)\").
The brackets define the part that you later can reuse, i.e. the string within the second quotes. used it with \1.
I put -n front in order to answer the updated question, to get online the line that was manipulated.
This should help -
sed -r 's/.*"([a-zA-Z]+)"\);/\1/' path/to/file.ext
If you are ok with using awk then you can try the following -
awk -F\" '/define\(/{print $(NF-1)}' path/to/file.ext
Update:
DBNAME=$(sed -r '/define\(\"BASE\"/s/.*"([a-zA-Z]+)"\);/\1/' path/to/file.ext)
sed -nr '/^define.*"(.*)".*$/{s//\1/;p}' path/to/file.ext
if your file doesn't change over time (i.e. the line numbers will always be the same) you can take the line, and use delimiters to take your part out:
`sed -n 'Xp' your.file | cut -d ' ' -f 2 |cut -d "\"" -f 2`
assuming X is the line number of your required line
Related
I have to write a function taking as argument a csv file named players.csv and a number giving the line to print.
Indded, i have to print the nth line 2column and 3rd column with a "is" between. For example Mike is John. column delimeter is ";".
I have the following code which is working :
sed -n "$2p" players.csv | cut -d ";" -f 2,3 --output-delimiter=' is '
However, I have to do the same without using cut. I can only use sed and wc. Do you have any idea what sed command I can use to have the same behavior as with cut.
Thank you for your attention and your help.
You were almost there:
sed -En "$2"'s/[^;]*;([^;]*);([^;]*).*/\1 is \2/p' players.csv
I have the following text file with the following lines:
<test="123">
<test="456">
<test="789">
My aim is to have the above text file to be appended with a keyword "HELLO" after the above numbers, as following:
<test="123.HELLO">
<test="456.HELLO">
<test="789.HELLO">
with the grep command and cut, I manage to get the value between the quotation mark.
grep -o "test=".* test.txt | cut -d \" -f2
I tried to use sed on top of it, with this line
grep -o "test=".* test.txt | cut -d \" -f2 | sed -i -- 's/$/.HELLO/' test.txt
however the closest I manage to get is instead a ".HELLO" which directly appended on the end of the line (and not after the numbers in between the quotes)
<test="123">.HELLO
<test="456">.HELLO
<test="789">.HELLO
How can I fix my sed statement to provide me with the requested line?
You can do it with groups in sed. To create new output, you can do this:
sed 's/\(test="[^"]*\)"/\1.HELLO"/g' test.txt
To modify it in-place, you can use the -i switch:
sed -i 's/\(test="[^"]*\)"/\1.HELLO"/g' test.txt
Explanation:
() is a group. You can refer to it with \1. In sed we have to escape the parentheses: \(\)
[^"]* matches everything that's not a quote. So the match will stop before the quote
In the replacement, you have to add the quote manually, since it's outside of the group. So you can put stuff before the quote.
Try this:
This is how your file looks like.
bash > cat a.txt
<test="123">
<test="456">
<test="789">
Your text piped to SED
bash > cat a.txt |sed 's/">/.HELLO">/g'
<test="123.HELLO">
<test="456.HELLO">
<test="789.HELLO">
bash >
Let me know if this worked out for you.
awk 'sub("[0-9]+","&.HELLO")' file
You can accomplish this with sed directly. Cut should not be necessary:
grep "test=" test.txt | sed 's/"\(.*\)"/"\1.HELLO"/'
So I'm trying to print the first word in each line of a txt file. The words are separated by one blank.
cut -c 1 txt file
Thats the code I have so far but it only prints the first character of each line.
Thanks
To print a whole word, you want -f 1, not -c 1. And since the default field delimiter is TAB rather than SPACE, you need to use the -d option.
cut -d' ' -f1 filename
To print the last two words not possible with cut, AFAIK, because it can only count from the beginning of the line. Use awk instead:
awk '{print $(NF-1), $NF;}' filename
you can try
awk '{print $1}' your_file
read word _ < file
echo "$word"
What's nice about this solution is it doesn't read beyond the first line of the file. Even awk, which has some very clean, terse syntax, has to be explicitly told to stop reading past the first line. read just reads one line at a time. Plus it's a bash builtin (and a builtin in many shells), so you don't need a new process to run.
If you want to print the first word in each line:
while read word _; do printf '%s\n' "$word"; done < file
But if the file is large then awk or cut will win out for reading every line.
You can use:
cut -d\ -f1 file
Where:
-d is the delimiter (here using \ for a space)
-f is the field selector
Notice that there is a space after the \.
-c is for characters, you want -f for fields, and -d to indicate your separator of space instead of the default tab:
cut -d " " -f 1 file
I want to replace new lines in text with coma or space but do not change the last new line.
I know of this question: How to replace newlines with tab characters? - but it does produce an tab on end instead of new line.
So far I have come with:
awk 'NR>1{printf","} {printf $1} END{printf"\n"}'
Is there an easier way to do this? This is not an assignment, I am just curious want to level up my scripting.
This might work for you:
paste -s -d, file
or
paste -s -d" " file
This should do a job:
echo ${$(tr '\n' ',' < file)%?}
or perhaps using sed:
sed -e ':a' -e 'N' -e '$!ba' -e 's/\n/,/g' file
Using perl is not quite as simple as paste, but it does generalize to more cases:
perl -0pe 's/\n(?!$)/,/g' filename # comma
perl -0pe 's/\n(?!$)/ /g' filename # space
The (?!$) is a lookahead assertion that means "not at the end", so the last newline is left alone.
You can use sed command to remove the end ","
# sed -i 's/,$//g' endmodify.txt
# cat endmodify.txt
root
bin
daemon
adm
lp
sync
games
gopher
Thanks.
I am writing shell script for embedded Linux in a small industrial box. I have a variable containing the text pid: 1234 and I want to strip first X characters from the line, so only 1234 stays. I have more variables I need to "clean", so I need to cut away X first characters and ${string:5} doesn't work for some reason in my system.
The only thing the box seems to have is sed.
I am trying to make the following to work:
result=$(echo "$pid" | sed 's/^.\{4\}//g')
Any ideas?
The following should work:
var="pid: 1234"
var=${var:5}
Are you sure bash is the shell executing your script?
Even the POSIX-compliant
var=${var#?????}
would be preferable to using an external process, although this requires you to hard-code the 5 in the form of a fixed-length pattern.
Here's a concise method to cut the first X characters using cut(1). This example removes the first 4 characters by cutting a substring starting with 5th character.
echo "$pid" | cut -c 5-
Use the -r option ("use extended regular expressions in the script") to sed in order to use the {n} syntax:
$ echo 'pid: 1234'| sed -r 's/^.{5}//'
1234
Cut first two characters from string:
$ string="1234567890"; echo "${string:2}"
34567890
pipe it through awk '{print substr($0,42)}' where 42 is one more than the number of characters to drop. For example:
$ echo abcde| awk '{print substr($0,2)}'
bcde
$
Chances are, you'll have cut as well. If so:
[me#home]$ echo "pid: 1234" | cut -d" " -f2
1234
Well, there have been solutions here with sed, awk, cut and using bash syntax. I just want to throw in another POSIX conform variant:
$ echo "pid: 1234" | tail -c +6
1234
-c tells tail at which byte offset to start, counting from the end of the input data, yet if the the number starts with a + sign, it is from the beginning of the input data to the end.
Another way, using cut instead of sed.
result=`echo $pid | cut -c 5-`
I found the answer in pure sed supplied by this question (admittedly, posted after this question was posted). This does exactly what you asked, solely in sed:
result=\`echo "$pid" | sed '/./ { s/pid:\ //g; }'\``
The dot in sed '/./) is whatever you want to match. Your question is exactly what I was attempting to, except in my case I wanted to match a specific line in a file and then uncomment it. In my case it was:
# Uncomment a line (edit the file in-place):
sed -i '/#\ COMMENTED_LINE_TO_MATCH/ { s/#\ //g; }' /path/to/target/file
The -i after sed is to edit the file in place (remove this switch if you want to test your matching expression prior to editing the file).
(I posted this because I wanted to do this entirely with sed as this question asked and none of the previous answered solved that problem.)
Rather than removing n characters from the start, perhaps you could just extract the digits directly. Like so...
$ echo "pid: 1234" | grep -Po "\d+"
This may be a more robust solution, and seems more intuitive.
This will do the job too:
echo "$pid"|awk '{print $2}'