find: current filename evaluates to empty in nested command

up vote
0
down vote

favorite

I am trying to rename a bunch of images in a directory according to their dimensions. To that end, I am building up my command step by step, using of Imagemagik's identify command. I currently am using echo and plan to switch that to mv once echo outputs correctly.

find * -type f -exec echo "$(identify -format '%w-%h' )" ;

This outputs a bunch of empty lines. However, when I just run it without the echo directly,

find * -type f -exec identify -format '%w-%h' ;

It outputs the dimensions as expected. Why isnt the inner command executing properly?

asked Sep 25 '17 at 17:33

Jeff

1033

Why would you want to do a command substitution and then just echo?
â€“Â Kusalananda
Sep 25 '17 at 20:48

1

Once I get it to echo, Im going to swap that out with a mv in order to rename the file.
â€“Â Jeff
Oct 11 '17 at 14:57

add a commentÂ |Â

up vote
0
down vote

favorite

find * -type f -exec echo "$(identify -format '%w-%h' )" ;

This outputs a bunch of empty lines. However, when I just run it without the echo directly,

find * -type f -exec identify -format '%w-%h' ;

It outputs the dimensions as expected. Why isnt the inner command executing properly?

asked Sep 25 '17 at 17:33

Jeff

1033

Why would you want to do a command substitution and then just echo?
â€“Â Kusalananda
Sep 25 '17 at 20:48

1

Once I get it to echo, Im going to swap that out with a mv in order to rename the file.
â€“Â Jeff
Oct 11 '17 at 14:57

add a commentÂ |Â

up vote
0
down vote

favorite

find * -type f -exec echo "$(identify -format '%w-%h' )" ;

This outputs a bunch of empty lines. However, when I just run it without the echo directly,

find * -type f -exec identify -format '%w-%h' ;

It outputs the dimensions as expected. Why isnt the inner command executing properly?

asked Sep 25 '17 at 17:33

Jeff

1033

find * -type f -exec echo "$(identify -format '%w-%h' )" ;

This outputs a bunch of empty lines. However, when I just run it without the echo directly,

find * -type f -exec identify -format '%w-%h' ;

It outputs the dimensions as expected. Why isnt the inner command executing properly?

find exec

asked Sep 25 '17 at 17:33

Jeff

1033

asked Sep 25 '17 at 17:33

Jeff

1033

asked Sep 25 '17 at 17:33

Jeff

1033

asked Sep 25 '17 at 17:33

Jeff

1033

asked Sep 25 '17 at 17:33

Jeff

1033

Why would you want to do a command substitution and then just echo?
â€“Â Kusalananda
Sep 25 '17 at 20:48

1

Once I get it to echo, Im going to swap that out with a mv in order to rename the file.
â€“Â Jeff
Oct 11 '17 at 14:57

add a commentÂ |Â

Why would you want to do a command substitution and then just echo?
â€“Â Kusalananda
Sep 25 '17 at 20:48

1

Once I get it to echo, Im going to swap that out with a mv in order to rename the file.
â€“Â Jeff
Oct 11 '17 at 14:57

Why would you want to do a command substitution and then just echo?
â€“Â Kusalananda
Sep 25 '17 at 20:48

Once I get it to echo, Im going to swap that out with a mv in order to rename the file.
â€“Â Jeff
Oct 11 '17 at 14:57

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
1
down vote

accepted

Use:

find . -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

Explanation

The problem is with the way bash expands variables

find * -type f -exec echo "$(identify -format '%w-%h' )" ;

First bash expands what is in side $(...):

find * -type f -exec echo 'identify: unable to open image `': No such file or directory...' ;

At which point find will use the same exec statement for all commands with a rather useless argument. What you want to do is suppress the bash expansion for inside the finds exec. You can stop bash expanding things by placing them inside single quotes

find * -type f -exec echo '$(identify -format %w-%h )' ;

But here we lose expansion completely, we can inject it back by running the expression with sh -c

find * -type f -exec sh -c 'echo "$(identify -format %w-%h )"' ;

Having within the shell script can also lead to some unexpected errors when dealing with whitespace or quotes or any character special to the shell in file names. To get around this you should move the as an argument to sh with:

find * -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

See this question/answer for a more detailed explanation of this.

Finally you can replace the * with . to let find find the files in the current directory by itself (and include hidden ones and not fail for filenames starting with -).

Additionally, you can add the -x flag to sh to get it to print out what was it executed to help with debugging the command - this can be very noisy for a lot of files.

edited Sep 25 '17 at 19:47

StÃ©phane Chazelas

284k53523860

answered Sep 25 '17 at 17:44

Michael Daffin

2,6101517

1

See unix.stackexchange.com/a/156010/116858
â€“Â Kusalananda
Sep 25 '17 at 18:04

The identify: unable to open image... error will be output by identify on its stderr, so won't be captured by $(...). as there won't be anything on stdout, $(...) will expand to an empty string, so it's like running find ... -exec echo '' ;
â€“Â StÃ©phane Chazelas
Sep 25 '17 at 19:43

add a commentÂ |Â

up vote
2
down vote

That's because echo doesn't know what is inside itself to expand it, only exec understand , in this case you should use in sh -c mode.

find . -type f -exec sh -c 'echo "$(identify -format '%w-%h' "$1")"' sh '' ;

edited Sep 25 '17 at 17:53

Kusalananda

106k14209327

answered Sep 25 '17 at 17:41

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.7k92563

add a commentÂ |Â

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "106"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f394365%2ffind-current-filename-evaluates-to-empty-in-nested-command%23new-answer', 'question_page');

);

Post as a guest

Name

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
1
down vote

accepted

Use:

find . -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

Explanation

The problem is with the way bash expands variables

find * -type f -exec echo "$(identify -format '%w-%h' )" ;

First bash expands what is in side $(...):

find * -type f -exec echo 'identify: unable to open image `': No such file or directory...' ;

find * -type f -exec echo '$(identify -format %w-%h )' ;

But here we lose expansion completely, we can inject it back by running the expression with sh -c

find * -type f -exec sh -c 'echo "$(identify -format %w-%h )"' ;

find * -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

See this question/answer for a more detailed explanation of this.

Finally you can replace the * with . to let find find the files in the current directory by itself (and include hidden ones and not fail for filenames starting with -).

Additionally, you can add the -x flag to sh to get it to print out what was it executed to help with debugging the command - this can be very noisy for a lot of files.

edited Sep 25 '17 at 19:47

StÃ©phane Chazelas

284k53523860

answered Sep 25 '17 at 17:44

Michael Daffin

2,6101517

1

See unix.stackexchange.com/a/156010/116858
â€“Â Kusalananda
Sep 25 '17 at 18:04

The identify: unable to open image... error will be output by identify on its stderr, so won't be captured by $(...). as there won't be anything on stdout, $(...) will expand to an empty string, so it's like running find ... -exec echo '' ;
â€“Â StÃ©phane Chazelas
Sep 25 '17 at 19:43

add a commentÂ |Â

up vote
1
down vote

accepted

Use:

find . -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

Explanation

The problem is with the way bash expands variables

find * -type f -exec echo "$(identify -format '%w-%h' )" ;

First bash expands what is in side $(...):

find * -type f -exec echo 'identify: unable to open image `': No such file or directory...' ;

find * -type f -exec echo '$(identify -format %w-%h )' ;

But here we lose expansion completely, we can inject it back by running the expression with sh -c

find * -type f -exec sh -c 'echo "$(identify -format %w-%h )"' ;

find * -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

See this question/answer for a more detailed explanation of this.

Finally you can replace the * with . to let find find the files in the current directory by itself (and include hidden ones and not fail for filenames starting with -).

Additionally, you can add the -x flag to sh to get it to print out what was it executed to help with debugging the command - this can be very noisy for a lot of files.

edited Sep 25 '17 at 19:47

StÃ©phane Chazelas

284k53523860

answered Sep 25 '17 at 17:44

Michael Daffin

2,6101517

1

See unix.stackexchange.com/a/156010/116858
â€“Â Kusalananda
Sep 25 '17 at 18:04

The identify: unable to open image... error will be output by identify on its stderr, so won't be captured by $(...). as there won't be anything on stdout, $(...) will expand to an empty string, so it's like running find ... -exec echo '' ;
â€“Â StÃ©phane Chazelas
Sep 25 '17 at 19:43

add a commentÂ |Â

up vote
1
down vote

accepted

Use:

find . -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

Explanation

The problem is with the way bash expands variables

find * -type f -exec echo "$(identify -format '%w-%h' )" ;

First bash expands what is in side $(...):

find * -type f -exec echo 'identify: unable to open image `': No such file or directory...' ;

find * -type f -exec echo '$(identify -format %w-%h )' ;

But here we lose expansion completely, we can inject it back by running the expression with sh -c

find * -type f -exec sh -c 'echo "$(identify -format %w-%h )"' ;

find * -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

See this question/answer for a more detailed explanation of this.

Finally you can replace the * with . to let find find the files in the current directory by itself (and include hidden ones and not fail for filenames starting with -).

Additionally, you can add the -x flag to sh to get it to print out what was it executed to help with debugging the command - this can be very noisy for a lot of files.

edited Sep 25 '17 at 19:47

StÃ©phane Chazelas

284k53523860

answered Sep 25 '17 at 17:44

Michael Daffin

2,6101517

Use:

find . -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

Explanation

The problem is with the way bash expands variables

find * -type f -exec echo "$(identify -format '%w-%h' )" ;

First bash expands what is in side $(...):

find * -type f -exec echo 'identify: unable to open image `': No such file or directory...' ;

find * -type f -exec echo '$(identify -format %w-%h )' ;

But here we lose expansion completely, we can inject it back by running the expression with sh -c

find * -type f -exec sh -c 'echo "$(identify -format %w-%h )"' ;

find * -type f -exec sh -c 'echo "$(identify -format %w-%h "$1")"' sh ;

See this question/answer for a more detailed explanation of this.

Finally you can replace the * with . to let find find the files in the current directory by itself (and include hidden ones and not fail for filenames starting with -).

Additionally, you can add the -x flag to sh to get it to print out what was it executed to help with debugging the command - this can be very noisy for a lot of files.

edited Sep 25 '17 at 19:47

StÃ©phane Chazelas

284k53523860

answered Sep 25 '17 at 17:44

Michael Daffin

2,6101517

edited Sep 25 '17 at 19:47

StÃ©phane Chazelas

284k53523860

edited Sep 25 '17 at 19:47

StÃ©phane Chazelas

284k53523860

edited Sep 25 '17 at 19:47

StÃ©phane Chazelas

284k53523860

answered Sep 25 '17 at 17:44

Michael Daffin

2,6101517

answered Sep 25 '17 at 17:44

Michael Daffin

2,6101517

answered Sep 25 '17 at 17:44

Michael Daffin

2,6101517

1

See unix.stackexchange.com/a/156010/116858
â€“Â Kusalananda
Sep 25 '17 at 18:04

The identify: unable to open image... error will be output by identify on its stderr, so won't be captured by $(...). as there won't be anything on stdout, $(...) will expand to an empty string, so it's like running find ... -exec echo '' ;
â€“Â StÃ©phane Chazelas
Sep 25 '17 at 19:43

add a commentÂ |Â

1

See unix.stackexchange.com/a/156010/116858
â€“Â Kusalananda
Sep 25 '17 at 18:04

The identify: unable to open image... error will be output by identify on its stderr, so won't be captured by $(...). as there won't be anything on stdout, $(...) will expand to an empty string, so it's like running find ... -exec echo '' ;
â€“Â StÃ©phane Chazelas
Sep 25 '17 at 19:43

See unix.stackexchange.com/a/156010/116858
â€“Â Kusalananda
Sep 25 '17 at 18:04

The identify: unable to open image... error will be output by identify on its stderr, so won't be captured by $(...). as there won't be anything on stdout, $(...) will expand to an empty string, so it's like running find ... -exec echo '' ;
â€“Â StÃ©phane Chazelas
Sep 25 '17 at 19:43

add a commentÂ |Â

up vote
2
down vote

That's because echo doesn't know what is inside itself to expand it, only exec understand , in this case you should use in sh -c mode.

find . -type f -exec sh -c 'echo "$(identify -format '%w-%h' "$1")"' sh '' ;

edited Sep 25 '17 at 17:53

Kusalananda

106k14209327

answered Sep 25 '17 at 17:41

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.7k92563

add a commentÂ |Â

up vote
2
down vote

That's because echo doesn't know what is inside itself to expand it, only exec understand , in this case you should use in sh -c mode.

find . -type f -exec sh -c 'echo "$(identify -format '%w-%h' "$1")"' sh '' ;

edited Sep 25 '17 at 17:53

Kusalananda

106k14209327

answered Sep 25 '17 at 17:41

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.7k92563

add a commentÂ |Â

up vote
2
down vote

That's because echo doesn't know what is inside itself to expand it, only exec understand , in this case you should use in sh -c mode.

find . -type f -exec sh -c 'echo "$(identify -format '%w-%h' "$1")"' sh '' ;

edited Sep 25 '17 at 17:53

Kusalananda

106k14209327

answered Sep 25 '17 at 17:41

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.7k92563

That's because echo doesn't know what is inside itself to expand it, only exec understand , in this case you should use in sh -c mode.

find . -type f -exec sh -c 'echo "$(identify -format '%w-%h' "$1")"' sh '' ;

edited Sep 25 '17 at 17:53

Kusalananda

106k14209327

answered Sep 25 '17 at 17:41

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.7k92563

edited Sep 25 '17 at 17:53

Kusalananda

106k14209327

edited Sep 25 '17 at 17:53

Kusalananda

106k14209327

edited Sep 25 '17 at 17:53

Kusalananda

106k14209327

answered Sep 25 '17 at 17:41

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.7k92563

answered Sep 25 '17 at 17:41

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.7k92563

answered Sep 25 '17 at 17:41

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.7k92563

add a commentÂ |Â

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

搜尋此網誌

mjhjmtu