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how to check if the first line of file contain a specific string?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I need to write a shell script that find and print all files in a directory which starts with the string: #include
Now, I know how to check if a string is in the file, by using:
for f in `ls`; do
if grep -q 'MyString' $f; then:
#DO SOMETHING
fi
but how can I apply this to the first line?
I thought maybe create a variable of the first line and check if it starts with #include, but i'm not sure how to do this, I tried the read command but I fail to read into a variable.
Beside I'd like to here other approaches to this problem, maybe awk?
Anyway remember I need to check if the first line starts with #include, not if it contain that string.
That's why I found those questions: How to print file content only if the first line matches a certain pattern?
https://stackoverflow.com/questions/5536018/how-to-print-matched-regex-pattern-using-awk
they are not completely helping.
shell grep read
edited 2 hours ago
Rui F Ribeiro
38k1475123
asked 2 hours ago
Z E Nir
195
add a comment |
up vote
2
down vote
favorite
I need to write a shell script that find and print all files in a directory which starts with the string: #include
Now, I know how to check if a string is in the file, by using:
for f in `ls`; do
if grep -q 'MyString' $f; then:
#DO SOMETHING
fi
but how can I apply this to the first line?
I thought maybe create a variable of the first line and check if it starts with #include, but i'm not sure how to do this, I tried the read command but I fail to read into a variable.
Beside I'd like to here other approaches to this problem, maybe awk?
Anyway remember I need to check if the first line starts with #include, not if it contain that string.
That's why I found those questions: How to print file content only if the first line matches a certain pattern?
https://stackoverflow.com/questions/5536018/how-to-print-matched-regex-pattern-using-awk
they are not completely helping.
shell grep read
edited 2 hours ago
Rui F Ribeiro
38k1475123
asked 2 hours ago
Z E Nir
195
1
Tip: don't use ls in scripts, it invariably leads to problems.
– ctrl-alt-delor
2 hours ago
I think unix.stackexchange.com/a/232655/117549 is reasonably close -- just anchor the pattern.
– Jeff Schaller
50 mins ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need to write a shell script that find and print all files in a directory which starts with the string: #include
Now, I know how to check if a string is in the file, by using:
for f in `ls`; do
if grep -q 'MyString' $f; then:
#DO SOMETHING
fi
but how can I apply this to the first line?
I thought maybe create a variable of the first line and check if it starts with #include, but i'm not sure how to do this, I tried the read command but I fail to read into a variable.
Beside I'd like to here other approaches to this problem, maybe awk?
Anyway remember I need to check if the first line starts with #include, not if it contain that string.
That's why I found those questions: How to print file content only if the first line matches a certain pattern?
https://stackoverflow.com/questions/5536018/how-to-print-matched-regex-pattern-using-awk
they are not completely helping.
shell grep read
edited 2 hours ago
Rui F Ribeiro
38k1475123
asked 2 hours ago
Z E Nir
195
I need to write a shell script that find and print all files in a directory which starts with the string: #include
Now, I know how to check if a string is in the file, by using:
for f in `ls`; do
if grep -q 'MyString' $f; then:
#DO SOMETHING
fi
but how can I apply this to the first line?
I thought maybe create a variable of the first line and check if it starts with #include, but i'm not sure how to do this, I tried the read command but I fail to read into a variable.
Beside I'd like to here other approaches to this problem, maybe awk?
Anyway remember I need to check if the first line starts with #include, not if it contain that string.
That's why I found those questions: How to print file content only if the first line matches a certain pattern?
https://stackoverflow.com/questions/5536018/how-to-print-matched-regex-pattern-using-awk
they are not completely helping.
shell grep read
shell grep read
edited 2 hours ago
Rui F Ribeiro
38k1475123
asked 2 hours ago
Z E Nir
195
edited 2 hours ago
Rui F Ribeiro
38k1475123
asked 2 hours ago
Z E Nir
195
edited 2 hours ago
Rui F Ribeiro
38k1475123
edited 2 hours ago
Rui F Ribeiro
38k1475123
edited 2 hours ago
Rui F Ribeiro
38k1475123
38k1475123
asked 2 hours ago
Z E Nir
195
asked 2 hours ago
Z E Nir
195
asked 2 hours ago
Z E Nir
195
195
1
Tip: don't use ls in scripts, it invariably leads to problems.
– ctrl-alt-delor
2 hours ago
I think unix.stackexchange.com/a/232655/117549 is reasonably close -- just anchor the pattern.
– Jeff Schaller
50 mins ago
add a comment |
1
Tip: don't use ls in scripts, it invariably leads to problems.
– ctrl-alt-delor
2 hours ago
I think unix.stackexchange.com/a/232655/117549 is reasonably close -- just anchor the pattern.
– Jeff Schaller
50 mins ago
1
1
Tip: don't use ls in scripts, it invariably leads to problems.
– ctrl-alt-delor
2 hours ago
Tip: don't use ls in scripts, it invariably leads to problems.
– ctrl-alt-delor
2 hours ago
I think unix.stackexchange.com/a/232655/117549 is reasonably close -- just anchor the pattern.
– Jeff Schaller
50 mins ago
I think unix.stackexchange.com/a/232655/117549 is reasonably close -- just anchor the pattern.
– Jeff Schaller
50 mins ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
It is easy to check if the first line starts with #include in sed:
sed -n '1/^#include/p;q' file
That will have an output only if the file contains #include in the first line. That only needs to read the first line to make the check, so it will be very fast.
So, a shell loop for all files (with sed) should be like this:
for file in *
do
[ "$(sed -n '1/^#include/p;q' "$file")" ] && printf '%sn' "$file"
done
If there are only files (not directories) in the pwd.
answered 2 hours ago
Isaac
9,17911342
it prints the name of the file/s, not its content... I just replaced the 'printf' with "cat $file" and it worked! Thanks!
– Z E Nir
2 hours ago
@ZENir good idea, have you tried it?
– ctrl-alt-delor
2 hours ago
@ctrl-alt-delor Yes, it seems that he tried it: He have just said: …and it worked!.
– Isaac
1 hour ago
Yeah i tried this, thank you both!
– Z E Nir
1 hour ago
sorry my bad, I miss-read it as a question.
– ctrl-alt-delor
1 hour ago
add a comment |
up vote
2
down vote
for file in *; do
[ -f "$file" ] || continue
IFS= read -r line < "$file" || [ -n "$line" ] || continue
case $line in
("#include"*) printf '%sn' "$file"
esac
done
To print the content of the file instead of its name, replace the printf command with cat < "$file".
If your awk supports the nextfile extension, and you don't care about the potential side effects of opening non-regular files:
awk '/^#include/print substr(FILENAME, 3); nextfile' ./*
With zsh, you can replace ./* with ./*(-.) to only pass regular files (or symlinks to regular files like for the [ -f ... ] approach above) to awk.
Or to print the file contents instead of name:
awk 'FNR == 1 found = /^#include/; found' ./*
(that one is portable).
answered 2 hours ago
Stéphane Chazelas
292k54544884
I tried this script, for some reason it does nothing, I have a file starts with '#include' and still nothing printes, If I press 'Enter' nine times i'm returning to the bash
– Z E Nir
2 hours ago
@ZENir, in my initial version of the script, I had forgotten the< "$file". Try reloading the page.
– Stéphane Chazelas
1 hour ago
add a comment |
up vote
1
down vote
for file in *
do
[ -f "$file" ] && head -n 1 < "$file" | grep -q '^#include' && cat < "$file"
done
Beware of the fact that, with -q option enabled, grep will exit with a zero status even if an error occurred.
edited 13 mins ago
Stéphane Chazelas
292k54544884
answered 2 hours ago
francescop21
18319
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It is easy to check if the first line starts with #include in sed:
sed -n '1/^#include/p;q' file
That will have an output only if the file contains #include in the first line. That only needs to read the first line to make the check, so it will be very fast.
So, a shell loop for all files (with sed) should be like this:
for file in *
do
[ "$(sed -n '1/^#include/p;q' "$file")" ] && printf '%sn' "$file"
done
If there are only files (not directories) in the pwd.
answered 2 hours ago
Isaac
9,17911342
it prints the name of the file/s, not its content... I just replaced the 'printf' with "cat $file" and it worked! Thanks!
– Z E Nir
2 hours ago
@ZENir good idea, have you tried it?
– ctrl-alt-delor
2 hours ago
@ctrl-alt-delor Yes, it seems that he tried it: He have just said: …and it worked!.
– Isaac
1 hour ago
Yeah i tried this, thank you both!
– Z E Nir
1 hour ago
sorry my bad, I miss-read it as a question.
– ctrl-alt-delor
1 hour ago
add a comment |
up vote
3
down vote
accepted
It is easy to check if the first line starts with #include in sed:
sed -n '1/^#include/p;q' file
That will have an output only if the file contains #include in the first line. That only needs to read the first line to make the check, so it will be very fast.
So, a shell loop for all files (with sed) should be like this:
for file in *
do
[ "$(sed -n '1/^#include/p;q' "$file")" ] && printf '%sn' "$file"
done
If there are only files (not directories) in the pwd.
answered 2 hours ago
Isaac
9,17911342
it prints the name of the file/s, not its content... I just replaced the 'printf' with "cat $file" and it worked! Thanks!
– Z E Nir
2 hours ago
@ZENir good idea, have you tried it?
– ctrl-alt-delor
2 hours ago
@ctrl-alt-delor Yes, it seems that he tried it: He have just said: …and it worked!.
– Isaac
1 hour ago
Yeah i tried this, thank you both!
– Z E Nir
1 hour ago
sorry my bad, I miss-read it as a question.
– ctrl-alt-delor
1 hour ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It is easy to check if the first line starts with #include in sed:
sed -n '1/^#include/p;q' file
That will have an output only if the file contains #include in the first line. That only needs to read the first line to make the check, so it will be very fast.
So, a shell loop for all files (with sed) should be like this:
for file in *
do
[ "$(sed -n '1/^#include/p;q' "$file")" ] && printf '%sn' "$file"
done
If there are only files (not directories) in the pwd.
answered 2 hours ago
Isaac
9,17911342
It is easy to check if the first line starts with #include in sed:
sed -n '1/^#include/p;q' file
That will have an output only if the file contains #include in the first line. That only needs to read the first line to make the check, so it will be very fast.
So, a shell loop for all files (with sed) should be like this:
for file in *
do
[ "$(sed -n '1/^#include/p;q' "$file")" ] && printf '%sn' "$file"
done
If there are only files (not directories) in the pwd.
answered 2 hours ago
Isaac
9,17911342
edited 1 hour ago
answered 2 hours ago
Isaac
9,17911342
answered 2 hours ago
Isaac
9,17911342
answered 2 hours ago
Isaac
9,17911342
9,17911342
it prints the name of the file/s, not its content... I just replaced the 'printf' with "cat $file" and it worked! Thanks!
– Z E Nir
2 hours ago
@ZENir good idea, have you tried it?
– ctrl-alt-delor
2 hours ago
@ctrl-alt-delor Yes, it seems that he tried it: He have just said: …and it worked!.
– Isaac
1 hour ago
Yeah i tried this, thank you both!
– Z E Nir
1 hour ago
sorry my bad, I miss-read it as a question.
– ctrl-alt-delor
1 hour ago
add a comment |
it prints the name of the file/s, not its content... I just replaced the 'printf' with "cat $file" and it worked! Thanks!
– Z E Nir
2 hours ago
@ZENir good idea, have you tried it?
– ctrl-alt-delor
2 hours ago
@ctrl-alt-delor Yes, it seems that he tried it: He have just said: …and it worked!.
– Isaac
1 hour ago
Yeah i tried this, thank you both!
– Z E Nir
1 hour ago
sorry my bad, I miss-read it as a question.
– ctrl-alt-delor
1 hour ago
it prints the name of the file/s, not its content... I just replaced the 'printf' with "cat $file" and it worked! Thanks!
– Z E Nir
2 hours ago
it prints the name of the file/s, not its content... I just replaced the 'printf' with "cat $file" and it worked! Thanks!
– Z E Nir
2 hours ago
@ZENir good idea, have you tried it?
– ctrl-alt-delor
2 hours ago
@ZENir good idea, have you tried it?
– ctrl-alt-delor
2 hours ago
@ctrl-alt-delor Yes, it seems that he tried it: He have just said: …and it worked!.
– Isaac
1 hour ago
@ctrl-alt-delor Yes, it seems that he tried it: He have just said: …and it worked!.
– Isaac
1 hour ago
Yeah i tried this, thank you both!
– Z E Nir
1 hour ago
Yeah i tried this, thank you both!
– Z E Nir
1 hour ago
sorry my bad, I miss-read it as a question.
– ctrl-alt-delor
1 hour ago
sorry my bad, I miss-read it as a question.
– ctrl-alt-delor
1 hour ago
add a comment |
up vote
2
down vote
for file in *; do
[ -f "$file" ] || continue
IFS= read -r line < "$file" || [ -n "$line" ] || continue
case $line in
("#include"*) printf '%sn' "$file"
esac
done
To print the content of the file instead of its name, replace the printf command with cat < "$file".
If your awk supports the nextfile extension, and you don't care about the potential side effects of opening non-regular files:
awk '/^#include/print substr(FILENAME, 3); nextfile' ./*
With zsh, you can replace ./* with ./*(-.) to only pass regular files (or symlinks to regular files like for the [ -f ... ] approach above) to awk.
Or to print the file contents instead of name:
awk 'FNR == 1 found = /^#include/; found' ./*
(that one is portable).
answered 2 hours ago
Stéphane Chazelas
292k54544884
I tried this script, for some reason it does nothing, I have a file starts with '#include' and still nothing printes, If I press 'Enter' nine times i'm returning to the bash
– Z E Nir
2 hours ago
@ZENir, in my initial version of the script, I had forgotten the< "$file". Try reloading the page.
– Stéphane Chazelas
1 hour ago
add a comment |
up vote
2
down vote
for file in *; do
[ -f "$file" ] || continue
IFS= read -r line < "$file" || [ -n "$line" ] || continue
case $line in
("#include"*) printf '%sn' "$file"
esac
done
To print the content of the file instead of its name, replace the printf command with cat < "$file".
If your awk supports the nextfile extension, and you don't care about the potential side effects of opening non-regular files:
awk '/^#include/print substr(FILENAME, 3); nextfile' ./*
With zsh, you can replace ./* with ./*(-.) to only pass regular files (or symlinks to regular files like for the [ -f ... ] approach above) to awk.
Or to print the file contents instead of name:
awk 'FNR == 1 found = /^#include/; found' ./*
(that one is portable).
answered 2 hours ago
Stéphane Chazelas
292k54544884
I tried this script, for some reason it does nothing, I have a file starts with '#include' and still nothing printes, If I press 'Enter' nine times i'm returning to the bash
– Z E Nir
2 hours ago
@ZENir, in my initial version of the script, I had forgotten the< "$file". Try reloading the page.
– Stéphane Chazelas
1 hour ago
add a comment |
up vote
2
down vote
up vote
2
down vote
for file in *; do
[ -f "$file" ] || continue
IFS= read -r line < "$file" || [ -n "$line" ] || continue
case $line in
("#include"*) printf '%sn' "$file"
esac
done
To print the content of the file instead of its name, replace the printf command with cat < "$file".
If your awk supports the nextfile extension, and you don't care about the potential side effects of opening non-regular files:
awk '/^#include/print substr(FILENAME, 3); nextfile' ./*
With zsh, you can replace ./* with ./*(-.) to only pass regular files (or symlinks to regular files like for the [ -f ... ] approach above) to awk.
Or to print the file contents instead of name:
awk 'FNR == 1 found = /^#include/; found' ./*
(that one is portable).
answered 2 hours ago
Stéphane Chazelas
292k54544884
for file in *; do
[ -f "$file" ] || continue
IFS= read -r line < "$file" || [ -n "$line" ] || continue
case $line in
("#include"*) printf '%sn' "$file"
esac
done
To print the content of the file instead of its name, replace the printf command with cat < "$file".
If your awk supports the nextfile extension, and you don't care about the potential side effects of opening non-regular files:
awk '/^#include/print substr(FILENAME, 3); nextfile' ./*
With zsh, you can replace ./* with ./*(-.) to only pass regular files (or symlinks to regular files like for the [ -f ... ] approach above) to awk.
Or to print the file contents instead of name:
awk 'FNR == 1 found = /^#include/; found' ./*
(that one is portable).
answered 2 hours ago
Stéphane Chazelas
292k54544884
edited 1 hour ago
answered 2 hours ago
Stéphane Chazelas
292k54544884
answered 2 hours ago
Stéphane Chazelas
292k54544884
answered 2 hours ago
Stéphane Chazelas
292k54544884
292k54544884
I tried this script, for some reason it does nothing, I have a file starts with '#include' and still nothing printes, If I press 'Enter' nine times i'm returning to the bash
– Z E Nir
2 hours ago
@ZENir, in my initial version of the script, I had forgotten the< "$file". Try reloading the page.
– Stéphane Chazelas
1 hour ago
add a comment |
I tried this script, for some reason it does nothing, I have a file starts with '#include' and still nothing printes, If I press 'Enter' nine times i'm returning to the bash
– Z E Nir
2 hours ago
@ZENir, in my initial version of the script, I had forgotten the< "$file". Try reloading the page.
– Stéphane Chazelas
1 hour ago
I tried this script, for some reason it does nothing, I have a file starts with '#include' and still nothing printes, If I press 'Enter' nine times i'm returning to the bash
– Z E Nir
2 hours ago
I tried this script, for some reason it does nothing, I have a file starts with '#include' and still nothing printes, If I press 'Enter' nine times i'm returning to the bash
– Z E Nir
2 hours ago
@ZENir, in my initial version of the script, I had forgotten the
< "$file". Try reloading the page.– Stéphane Chazelas
1 hour ago
@ZENir, in my initial version of the script, I had forgotten the
< "$file". Try reloading the page.– Stéphane Chazelas
1 hour ago
add a comment |
up vote
1
down vote
for file in *
do
[ -f "$file" ] && head -n 1 < "$file" | grep -q '^#include' && cat < "$file"
done
Beware of the fact that, with -q option enabled, grep will exit with a zero status even if an error occurred.
edited 13 mins ago
Stéphane Chazelas
292k54544884
answered 2 hours ago
francescop21
18319
add a comment |
up vote
1
down vote
for file in *
do
[ -f "$file" ] && head -n 1 < "$file" | grep -q '^#include' && cat < "$file"
done
Beware of the fact that, with -q option enabled, grep will exit with a zero status even if an error occurred.
edited 13 mins ago
Stéphane Chazelas
292k54544884
answered 2 hours ago
francescop21
18319
add a comment |
up vote
1
down vote
up vote
1
down vote
for file in *
do
[ -f "$file" ] && head -n 1 < "$file" | grep -q '^#include' && cat < "$file"
done
Beware of the fact that, with -q option enabled, grep will exit with a zero status even if an error occurred.
edited 13 mins ago
Stéphane Chazelas
292k54544884
answered 2 hours ago
francescop21
18319
for file in *
do
[ -f "$file" ] && head -n 1 < "$file" | grep -q '^#include' && cat < "$file"
done
Beware of the fact that, with -q option enabled, grep will exit with a zero status even if an error occurred.
edited 13 mins ago
Stéphane Chazelas
292k54544884
answered 2 hours ago
francescop21
18319
edited 13 mins ago
Stéphane Chazelas
292k54544884
edited 13 mins ago
Stéphane Chazelas
292k54544884
edited 13 mins ago
Stéphane Chazelas
292k54544884
292k54544884
answered 2 hours ago
francescop21
18319
answered 2 hours ago
francescop21
18319
answered 2 hours ago
francescop21
18319
18319
add a comment |
add a comment |
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1
Tip: don't use ls in scripts, it invariably leads to problems.
– ctrl-alt-delor
2 hours ago
I think unix.stackexchange.com/a/232655/117549 is reasonably close -- just anchor the pattern.
– Jeff Schaller
50 mins ago