report if any one record is not of matching size using sed [closed]

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how can I use sed to report any first record that is not of size 21 in a file?

I don't want sed to scan the complete file and get out as soon as the first record which is not of size 21 is found.

edited May 16 at 19:27

SivaPrasath

4,74212445

asked May 16 at 18:36

Ahmad S

closed as unclear what you're asking by DopeGhoti, steve, G-Man, dr01, X Tian May 17 at 15:59

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, itÃ¢Â€Â™s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

i have the sed command i am not sure how to break out of it .
â€“Â Ahmad S
May 16 at 19:00

If you read the manual it explains how to quit sed.
â€“Â don_crissti
May 16 at 19:02

1

Possible duplicate of sed command to find lines which are not of specific size
â€“Â DopeGhoti
May 16 at 19:31

@DopeGhoti - no, that is a different question that asks how to print all lines. The fact that you changed your answer there to answer the question here after a comment made by the OP doesn't change that (though your one liner there is better that the ones here).
â€“Â don_crissti
May 16 at 19:37

In fairness, this is a duplicate of what the other question was apparently meant to be as opposed to what the other question is stated to be.
â€“Â DopeGhoti
May 16 at 19:50

add a commentÂ |Â

up vote
-3
down vote

favorite

how can I use sed to report any first record that is not of size 21 in a file?

I don't want sed to scan the complete file and get out as soon as the first record which is not of size 21 is found.

edited May 16 at 19:27

SivaPrasath

4,74212445

asked May 16 at 18:36

Ahmad S

closed as unclear what you're asking by DopeGhoti, steve, G-Man, dr01, X Tian May 17 at 15:59

i have the sed command i am not sure how to break out of it .
â€“Â Ahmad S
May 16 at 19:00

If you read the manual it explains how to quit sed.
â€“Â don_crissti
May 16 at 19:02

1

Possible duplicate of sed command to find lines which are not of specific size
â€“Â DopeGhoti
May 16 at 19:31

@DopeGhoti - no, that is a different question that asks how to print all lines. The fact that you changed your answer there to answer the question here after a comment made by the OP doesn't change that (though your one liner there is better that the ones here).
â€“Â don_crissti
May 16 at 19:37

In fairness, this is a duplicate of what the other question was apparently meant to be as opposed to what the other question is stated to be.
â€“Â DopeGhoti
May 16 at 19:50

add a commentÂ |Â

up vote
-3
down vote

favorite

how can I use sed to report any first record that is not of size 21 in a file?

I don't want sed to scan the complete file and get out as soon as the first record which is not of size 21 is found.

edited May 16 at 19:27

SivaPrasath

4,74212445

asked May 16 at 18:36

Ahmad S

how can I use sed to report any first record that is not of size 21 in a file?

I don't want sed to scan the complete file and get out as soon as the first record which is not of size 21 is found.

edited May 16 at 19:27

SivaPrasath

4,74212445

asked May 16 at 18:36

Ahmad S

edited May 16 at 19:27

SivaPrasath

4,74212445

edited May 16 at 19:27

SivaPrasath

4,74212445

edited May 16 at 19:27

SivaPrasath

4,74212445

asked May 16 at 18:36

Ahmad S

asked May 16 at 18:36

Ahmad S

asked May 16 at 18:36

Ahmad S

closed as unclear what you're asking by DopeGhoti, steve, G-Man, dr01, X Tian May 17 at 15:59

i have the sed command i am not sure how to break out of it .
â€“Â Ahmad S
May 16 at 19:00

If you read the manual it explains how to quit sed.
â€“Â don_crissti
May 16 at 19:02

1

Possible duplicate of sed command to find lines which are not of specific size
â€“Â DopeGhoti
May 16 at 19:31

@DopeGhoti - no, that is a different question that asks how to print all lines. The fact that you changed your answer there to answer the question here after a comment made by the OP doesn't change that (though your one liner there is better that the ones here).
â€“Â don_crissti
May 16 at 19:37

In fairness, this is a duplicate of what the other question was apparently meant to be as opposed to what the other question is stated to be.
â€“Â DopeGhoti
May 16 at 19:50

add a commentÂ |Â

i have the sed command i am not sure how to break out of it .
â€“Â Ahmad S
May 16 at 19:00

If you read the manual it explains how to quit sed.
â€“Â don_crissti
May 16 at 19:02

1

Possible duplicate of sed command to find lines which are not of specific size
â€“Â DopeGhoti
May 16 at 19:31

@DopeGhoti - no, that is a different question that asks how to print all lines. The fact that you changed your answer there to answer the question here after a comment made by the OP doesn't change that (though your one liner there is better that the ones here).
â€“Â don_crissti
May 16 at 19:37

In fairness, this is a duplicate of what the other question was apparently meant to be as opposed to what the other question is stated to be.
â€“Â DopeGhoti
May 16 at 19:50

i have the sed command i am not sure how to break out of it .
â€“Â Ahmad S
May 16 at 19:00

If you read the manual it explains how to quit sed.
â€“Â don_crissti
May 16 at 19:02

Possible duplicate of sed command to find lines which are not of specific size
â€“Â DopeGhoti
May 16 at 19:31

@DopeGhoti - no, that is a different question that asks how to print all lines. The fact that you changed your answer there to answer the question here after a comment made by the OP doesn't change that (though your one liner there is better that the ones here).
â€“Â don_crissti
May 16 at 19:37

In fairness, this is a duplicate of what the other question was apparently meant to be as opposed to what the other question is stated to be.
â€“Â DopeGhoti
May 16 at 19:50

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

Based on this answer to your previous question

sed -n '/^.21$/! p;q;' file

answered May 16 at 19:08

steeldriver

31.2k34978

add a commentÂ |Â

up vote
1
down vote

Using awk (this would be easiest):

awk 'length != 21 printf("Line of length %d foundn", length); exit ' file

Or, as part of a shell script,

if ! awk 'length != 21 exit 1 ' file; then
 echo 'Line of length != 21 found (or awk failed to execute properly)'
else
 echo 'All lines are 21 characters (or the file is empty)'
fi

Using sed:

sed -nE '/^.21$/!p;q;' file

With GNU sed, you would be able to do

if ! sed -nE '/.21$/!q 1' file; then
 echo 'Line with != 21 characters found (or sed failed to run properly)'
else
 echo 'All lines are 21 characters (or file is empty)'
fi

edited May 16 at 19:12

answered May 16 at 18:42

Kusalananda

102k13199315

using sed please
â€“Â Ahmad S
May 16 at 19:00

@AhmadS See updated answer.
â€“Â Kusalananda
May 16 at 19:12

add a commentÂ |Â

up vote
1
down vote

With GNU grep:

if line=$(grep -Exnvm1 '.21' < file); then
 printf >&2 'Found "%s" which is not 21 characters longn' "$line"
fi

(-n above includes the line number)

answered May 16 at 21:36

StÃ©phane Chazelas

279k53513845

add a commentÂ |Â

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

Based on this answer to your previous question

sed -n '/^.21$/! p;q;' file

answered May 16 at 19:08

steeldriver

31.2k34978

add a commentÂ |Â

up vote
1
down vote

accepted

Based on this answer to your previous question

sed -n '/^.21$/! p;q;' file

answered May 16 at 19:08

steeldriver

31.2k34978

add a commentÂ |Â

up vote
1
down vote

accepted

Based on this answer to your previous question

sed -n '/^.21$/! p;q;' file

answered May 16 at 19:08

steeldriver

31.2k34978

Based on this answer to your previous question

sed -n '/^.21$/! p;q;' file

answered May 16 at 19:08

steeldriver

31.2k34978

answered May 16 at 19:08

steeldriver

31.2k34978

answered May 16 at 19:08

steeldriver

31.2k34978

answered May 16 at 19:08

steeldriver

31.2k34978

add a commentÂ |Â

up vote
1
down vote

Using awk (this would be easiest):

awk 'length != 21 printf("Line of length %d foundn", length); exit ' file

Or, as part of a shell script,

if ! awk 'length != 21 exit 1 ' file; then
 echo 'Line of length != 21 found (or awk failed to execute properly)'
else
 echo 'All lines are 21 characters (or the file is empty)'
fi

Using sed:

sed -nE '/^.21$/!p;q;' file

With GNU sed, you would be able to do

if ! sed -nE '/.21$/!q 1' file; then
 echo 'Line with != 21 characters found (or sed failed to run properly)'
else
 echo 'All lines are 21 characters (or file is empty)'
fi

edited May 16 at 19:12

answered May 16 at 18:42

Kusalananda

102k13199315

using sed please
â€“Â Ahmad S
May 16 at 19:00

@AhmadS See updated answer.
â€“Â Kusalananda
May 16 at 19:12

add a commentÂ |Â

up vote
1
down vote

Using awk (this would be easiest):

awk 'length != 21 printf("Line of length %d foundn", length); exit ' file

Or, as part of a shell script,

if ! awk 'length != 21 exit 1 ' file; then
 echo 'Line of length != 21 found (or awk failed to execute properly)'
else
 echo 'All lines are 21 characters (or the file is empty)'
fi

Using sed:

sed -nE '/^.21$/!p;q;' file

With GNU sed, you would be able to do

if ! sed -nE '/.21$/!q 1' file; then
 echo 'Line with != 21 characters found (or sed failed to run properly)'
else
 echo 'All lines are 21 characters (or file is empty)'
fi

edited May 16 at 19:12

answered May 16 at 18:42

Kusalananda

102k13199315

using sed please
â€“Â Ahmad S
May 16 at 19:00

@AhmadS See updated answer.
â€“Â Kusalananda
May 16 at 19:12

add a commentÂ |Â

up vote
1
down vote

Using awk (this would be easiest):

awk 'length != 21 printf("Line of length %d foundn", length); exit ' file

Or, as part of a shell script,

if ! awk 'length != 21 exit 1 ' file; then
 echo 'Line of length != 21 found (or awk failed to execute properly)'
else
 echo 'All lines are 21 characters (or the file is empty)'
fi

Using sed:

sed -nE '/^.21$/!p;q;' file

With GNU sed, you would be able to do

if ! sed -nE '/.21$/!q 1' file; then
 echo 'Line with != 21 characters found (or sed failed to run properly)'
else
 echo 'All lines are 21 characters (or file is empty)'
fi

edited May 16 at 19:12

answered May 16 at 18:42

Kusalananda

102k13199315

Using awk (this would be easiest):

awk 'length != 21 printf("Line of length %d foundn", length); exit ' file

Or, as part of a shell script,

if ! awk 'length != 21 exit 1 ' file; then
 echo 'Line of length != 21 found (or awk failed to execute properly)'
else
 echo 'All lines are 21 characters (or the file is empty)'
fi

Using sed:

sed -nE '/^.21$/!p;q;' file

With GNU sed, you would be able to do

if ! sed -nE '/.21$/!q 1' file; then
 echo 'Line with != 21 characters found (or sed failed to run properly)'
else
 echo 'All lines are 21 characters (or file is empty)'
fi

edited May 16 at 19:12

answered May 16 at 18:42

Kusalananda

102k13199315

edited May 16 at 19:12

answered May 16 at 18:42

Kusalananda

102k13199315

answered May 16 at 18:42

Kusalananda

102k13199315

answered May 16 at 18:42

Kusalananda

102k13199315

using sed please
â€“Â Ahmad S
May 16 at 19:00

@AhmadS See updated answer.
â€“Â Kusalananda
May 16 at 19:12

add a commentÂ |Â

using sed please
â€“Â Ahmad S
May 16 at 19:00

@AhmadS See updated answer.
â€“Â Kusalananda
May 16 at 19:12

using sed please
â€“Â Ahmad S
May 16 at 19:00

@AhmadS See updated answer.
â€“Â Kusalananda
May 16 at 19:12

add a commentÂ |Â

up vote
1
down vote

With GNU grep:

if line=$(grep -Exnvm1 '.21' < file); then
 printf >&2 'Found "%s" which is not 21 characters longn' "$line"
fi

(-n above includes the line number)

answered May 16 at 21:36

279k53513845

add a commentÂ |Â

up vote
1
down vote

With GNU grep:

if line=$(grep -Exnvm1 '.21' < file); then
 printf >&2 'Found "%s" which is not 21 characters longn' "$line"
fi

(-n above includes the line number)

answered May 16 at 21:36

279k53513845

add a commentÂ |Â

up vote
1
down vote

With GNU grep:

if line=$(grep -Exnvm1 '.21' < file); then
 printf >&2 'Found "%s" which is not 21 characters longn' "$line"
fi

(-n above includes the line number)

answered May 16 at 21:36

279k53513845

With GNU grep:

if line=$(grep -Exnvm1 '.21' < file); then
 printf >&2 'Found "%s" which is not 21 characters longn' "$line"
fi

(-n above includes the line number)

answered May 16 at 21:36

279k53513845

answered May 16 at 21:36

279k53513845

answered May 16 at 21:36

279k53513845

answered May 16 at 21:36

279k53513845

add a commentÂ |Â

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