`integer expression expected` [duplicate]

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How to use variables inside single quotes

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I encountered integer expression expected error in the following code:

 #! /bin/bash
 # test integer: evaluate the value of integer
 int=-5
 if [ -z "$int" ]; then
 echo 'int is empty.' >&2
 fi

 if [ "$int" -eq 0 ]; then
 echo "int is zero"
 else
 if [ '$int' -lt 0 ]; then
 echo "int is negative."
 else
 echo 'int is positive.'
 fi
 if [ $((int % 2)) -eq 0 ]; then
 echo 'int is even.'
 else
 echo 'int is odd.'
 fi
 fi

Run it and get error report

 $ bash test_integer.sh
 test_integer.sh: line 14: [: $int: integer expression expected
 int is positive.
 int is odd.

I checked multiple times but fail to locate the bug,

which seem to line-to-line comform to the book

What's the problem with my code?

asked Apr 2 at 15:09

JawSaw

29410

marked as duplicate by Kusalananda, Isaac, Jeff Schaller, Communityâ™¦ Apr 2 at 23:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

FAQ: mywiki.wooledge.org/BashFAQ | Guide: mywiki.wooledge.org/BashGuide | Ref: gnu.org/s/bash/manual | wiki.bash-hackers.org | mywiki.wooledge.org/Quotes | Check your script: shellcheck.net And avoid people recommendations saying to learn with tldp.org web site, the tldp bash guide is outdated, and in some cases just plain wrong.
â€“Â Gilles Quenot
Apr 2 at 15:15

add a commentÂ |Â

up vote
0
down vote

favorite

This question already has an answer here:

How to use variables inside single quotes

2 answers

I encountered integer expression expected error in the following code:

 #! /bin/bash
 # test integer: evaluate the value of integer
 int=-5
 if [ -z "$int" ]; then
 echo 'int is empty.' >&2
 fi

 if [ "$int" -eq 0 ]; then
 echo "int is zero"
 else
 if [ '$int' -lt 0 ]; then
 echo "int is negative."
 else
 echo 'int is positive.'
 fi
 if [ $((int % 2)) -eq 0 ]; then
 echo 'int is even.'
 else
 echo 'int is odd.'
 fi
 fi

Run it and get error report

 $ bash test_integer.sh
 test_integer.sh: line 14: [: $int: integer expression expected
 int is positive.
 int is odd.

I checked multiple times but fail to locate the bug,

which seem to line-to-line comform to the book

What's the problem with my code?

asked Apr 2 at 15:09

JawSaw

29410

marked as duplicate by Kusalananda, Isaac, Jeff Schaller, Communityâ™¦ Apr 2 at 23:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

FAQ: mywiki.wooledge.org/BashFAQ | Guide: mywiki.wooledge.org/BashGuide | Ref: gnu.org/s/bash/manual | wiki.bash-hackers.org | mywiki.wooledge.org/Quotes | Check your script: shellcheck.net And avoid people recommendations saying to learn with tldp.org web site, the tldp bash guide is outdated, and in some cases just plain wrong.
â€“Â Gilles Quenot
Apr 2 at 15:15

add a commentÂ |Â

up vote
0
down vote

favorite

This question already has an answer here:

How to use variables inside single quotes

2 answers

I encountered integer expression expected error in the following code:

 #! /bin/bash
 # test integer: evaluate the value of integer
 int=-5
 if [ -z "$int" ]; then
 echo 'int is empty.' >&2
 fi

 if [ "$int" -eq 0 ]; then
 echo "int is zero"
 else
 if [ '$int' -lt 0 ]; then
 echo "int is negative."
 else
 echo 'int is positive.'
 fi
 if [ $((int % 2)) -eq 0 ]; then
 echo 'int is even.'
 else
 echo 'int is odd.'
 fi
 fi

Run it and get error report

 $ bash test_integer.sh
 test_integer.sh: line 14: [: $int: integer expression expected
 int is positive.
 int is odd.

I checked multiple times but fail to locate the bug,

which seem to line-to-line comform to the book

What's the problem with my code?

asked Apr 2 at 15:09

JawSaw

29410

This question already has an answer here:

How to use variables inside single quotes

2 answers

I encountered integer expression expected error in the following code:

 #! /bin/bash
 # test integer: evaluate the value of integer
 int=-5
 if [ -z "$int" ]; then
 echo 'int is empty.' >&2
 fi

 if [ "$int" -eq 0 ]; then
 echo "int is zero"
 else
 if [ '$int' -lt 0 ]; then
 echo "int is negative."
 else
 echo 'int is positive.'
 fi
 if [ $((int % 2)) -eq 0 ]; then
 echo 'int is even.'
 else
 echo 'int is odd.'
 fi
 fi

Run it and get error report

 $ bash test_integer.sh
 test_integer.sh: line 14: [: $int: integer expression expected
 int is positive.
 int is odd.

I checked multiple times but fail to locate the bug,

which seem to line-to-line comform to the book

What's the problem with my code?

This question already has an answer here:

How to use variables inside single quotes

2 answers

asked Apr 2 at 15:09

JawSaw

29410

asked Apr 2 at 15:09

JawSaw

29410

asked Apr 2 at 15:09

JawSaw

29410

asked Apr 2 at 15:09

JawSaw

29410

marked as duplicate by Kusalananda, Isaac, Jeff Schaller, Communityâ™¦ Apr 2 at 23:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Kusalananda, Isaac, Jeff Schaller, Communityâ™¦ Apr 2 at 23:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

FAQ: mywiki.wooledge.org/BashFAQ | Guide: mywiki.wooledge.org/BashGuide | Ref: gnu.org/s/bash/manual | wiki.bash-hackers.org | mywiki.wooledge.org/Quotes | Check your script: shellcheck.net And avoid people recommendations saying to learn with tldp.org web site, the tldp bash guide is outdated, and in some cases just plain wrong.
â€“Â Gilles Quenot
Apr 2 at 15:15

add a commentÂ |Â

FAQ: mywiki.wooledge.org/BashFAQ | Guide: mywiki.wooledge.org/BashGuide | Ref: gnu.org/s/bash/manual | wiki.bash-hackers.org | mywiki.wooledge.org/Quotes | Check your script: shellcheck.net And avoid people recommendations saying to learn with tldp.org web site, the tldp bash guide is outdated, and in some cases just plain wrong.
â€“Â Gilles Quenot
Apr 2 at 15:15

FAQ: mywiki.wooledge.org/BashFAQ | Guide: mywiki.wooledge.org/BashGuide | Ref: gnu.org/s/bash/manual | wiki.bash-hackers.org | mywiki.wooledge.org/Quotes | Check your script: shellcheck.net And avoid people recommendations saying to learn with tldp.org web site, the tldp bash guide is outdated, and in some cases just plain wrong.
â€“Â Gilles Quenot
Apr 2 at 15:15

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
4
down vote

accepted

'$int' is a quoting error which you would have easily seen if you had run the script through

bash -vx schript.sh

There is no expansion within single quotes. You need:

if [ "$int" -lt 0 ]

answered Apr 2 at 15:15

Hauke Laging

53.2k1282130

add a commentÂ |Â

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
4
down vote

accepted

'$int' is a quoting error which you would have easily seen if you had run the script through

bash -vx schript.sh

There is no expansion within single quotes. You need:

if [ "$int" -lt 0 ]

answered Apr 2 at 15:15

Hauke Laging

53.2k1282130

add a commentÂ |Â

up vote
4
down vote

accepted

'$int' is a quoting error which you would have easily seen if you had run the script through

bash -vx schript.sh

There is no expansion within single quotes. You need:

if [ "$int" -lt 0 ]

answered Apr 2 at 15:15

Hauke Laging

53.2k1282130

add a commentÂ |Â

up vote
4
down vote

accepted

'$int' is a quoting error which you would have easily seen if you had run the script through

bash -vx schript.sh

There is no expansion within single quotes. You need:

if [ "$int" -lt 0 ]

answered Apr 2 at 15:15

Hauke Laging

53.2k1282130

'$int' is a quoting error which you would have easily seen if you had run the script through

bash -vx schript.sh

There is no expansion within single quotes. You need:

if [ "$int" -lt 0 ]

answered Apr 2 at 15:15

Hauke Laging

53.2k1282130

answered Apr 2 at 15:15

Hauke Laging

53.2k1282130

answered Apr 2 at 15:15

Hauke Laging

53.2k1282130

answered Apr 2 at 15:15

Hauke Laging

53.2k1282130

add a commentÂ |Â

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