How to check for space in a variable in bash?

I am taking baby steps at learning bash and I am developing a piece of code which takes an input and checks if it contains any spaces. The idea is that the variable should NOT contain any spaces and the code should consequently echo a suitable message.

asked Nov 21 '18 at 15:35

SSen23

111

add a comment |

asked Nov 21 '18 at 15:35

SSen23

111

add a comment |

asked Nov 21 '18 at 15:35

SSen23

111

bash

asked Nov 21 '18 at 15:35

SSen23

111

asked Nov 21 '18 at 15:35

SSen23

111

asked Nov 21 '18 at 15:35

SSen23

111

asked Nov 21 '18 at 15:35

SSen23

111

asked Nov 21 '18 at 15:35

SSen23

111

add a comment |

5 Answers
5

active

oldest

votes

Try this:

#!/bin/bash

if [[ $1 = *[[:space:]]* ]]

then

    echo "space exist"

fi

answered Nov 21 '18 at 15:47

Incrivel Monstro Verde

41015

2

Pedantically, == is the documented pattern matching operator within [[...]] but = also works

– glenn jackman
Nov 21 '18 at 16:10

Thanks, this works well when I want to check an input which contains space in between words. But what if I want to check if a string has space characters at the beginning or at the end of the string?

– SSen23
Dec 8 '18 at 10:05

use if [[ $var =~ ^[[:space:]] ]] (begin string) and if [[ $var =~ [[:space:]]$ ]] test space at the end of the string

– Incrivel Monstro Verde
Dec 10 '18 at 15:28

add a comment |

You can test simple glob patterns in portable shell by using case, without needing any external programs or Bash extensions (that's a good thing, because then your scripts are useful to more people).

#!/bin/sh



case "$1" in

    *' '*)

        printf 'Invalid argument %s (contains space)n' "$1" >&2

        exit 1

        ;;

esac

You might want to include other whitespace characters in your check - in which case, use *[[:space:]]* as the pattern instead of *' '*.

edited Nov 21 '18 at 21:24

answered Nov 21 '18 at 16:19

Toby Speight

17.2k134367

This has always been my favorite method. It's very effective, and not too hard to read.

– Paul Hodges
Nov 21 '18 at 21:15

add a comment |

You can use wc -w command to check if there are any words. If the result of this output is a number greater than 1, then it means that there are more than 1 words in the input. Here's an example:

#!/bin/bash

read var1

var2=`echo $var1 | wc -w`

if [ $var2 -gt 1 ]

then

    echo "Spaces"

else

    echo "No spaces"

fi

Note: there is a | (pipe symbol) which means that the result of echo $var1 will be given as input to wc -w via the pipe.

Here is the link where I tested the above code: https://ideone.com/aKJdyN

answered Nov 21 '18 at 15:48

kiner_shah

1,31721624

add a comment |

You can use grep, like this:

echo " foo" | grep 's' -c

# 1

echo "foo" | grep 's' -c

# 0

Or you may use something like this:

s=' foo'

if [[ $s =~ " " ]]; then

    echo 'contains space'

else

    echo 'ok'

fi

edited Nov 21 '18 at 15:54

answered Nov 21 '18 at 15:46

Vladimir Kovpak

11.2k43949

I wouldn't call out to grep, bash can do it as shown.

– glenn jackman
Nov 21 '18 at 16:11

add a comment |

You could use parameter expansion to remove everything that isn't a space and see if what's left is the empty string or not:

var1='has space'

var2='nospace'



for var in "$var1" "$var2"; do

    if [[ ${var//[^[:space:]]} ]]; then

        echo "'$var' contains a space"

    fi

done

The key is [[ ${var//[^[:space:]]} ]]:

With ${var//[^[:space:]]}, everything that isn't a space is removed from the expansion of $var.

[[ string ]] has a non-zero exit status if string is empty. It's a shorthand for the equivalent [[ -n string ]].

We could also quote the expansion of ${var//[^[:space:]]}, but [[ ... ]] takes care of the quoting for us.

answered Nov 21 '18 at 16:41

Benjamin W.

21.5k135257

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

Try this:

#!/bin/bash

if [[ $1 = *[[:space:]]* ]]

then

    echo "space exist"

fi

answered Nov 21 '18 at 15:47

Incrivel Monstro Verde

41015

2

Pedantically, == is the documented pattern matching operator within [[...]] but = also works

– glenn jackman
Nov 21 '18 at 16:10

Thanks, this works well when I want to check an input which contains space in between words. But what if I want to check if a string has space characters at the beginning or at the end of the string?

– SSen23
Dec 8 '18 at 10:05

use if [[ $var =~ ^[[:space:]] ]] (begin string) and if [[ $var =~ [[:space:]]$ ]] test space at the end of the string

– Incrivel Monstro Verde
Dec 10 '18 at 15:28

add a comment |

Try this:

#!/bin/bash

if [[ $1 = *[[:space:]]* ]]

then

    echo "space exist"

fi

answered Nov 21 '18 at 15:47

Incrivel Monstro Verde

41015

2

Pedantically, == is the documented pattern matching operator within [[...]] but = also works

– glenn jackman
Nov 21 '18 at 16:10

Thanks, this works well when I want to check an input which contains space in between words. But what if I want to check if a string has space characters at the beginning or at the end of the string?

– SSen23
Dec 8 '18 at 10:05

use if [[ $var =~ ^[[:space:]] ]] (begin string) and if [[ $var =~ [[:space:]]$ ]] test space at the end of the string

– Incrivel Monstro Verde
Dec 10 '18 at 15:28

add a comment |

Try this:

#!/bin/bash

if [[ $1 = *[[:space:]]* ]]

then

    echo "space exist"

fi

answered Nov 21 '18 at 15:47

Incrivel Monstro Verde

41015

Try this:

#!/bin/bash

if [[ $1 = *[[:space:]]* ]]

then

    echo "space exist"

fi

answered Nov 21 '18 at 15:47

Incrivel Monstro Verde

41015

answered Nov 21 '18 at 15:47

Incrivel Monstro Verde

41015

answered Nov 21 '18 at 15:47

Incrivel Monstro Verde

41015

answered Nov 21 '18 at 15:47

Incrivel Monstro Verde

41015

2

Pedantically, == is the documented pattern matching operator within [[...]] but = also works

– glenn jackman
Nov 21 '18 at 16:10

Thanks, this works well when I want to check an input which contains space in between words. But what if I want to check if a string has space characters at the beginning or at the end of the string?

– SSen23
Dec 8 '18 at 10:05

use if [[ $var =~ ^[[:space:]] ]] (begin string) and if [[ $var =~ [[:space:]]$ ]] test space at the end of the string

– Incrivel Monstro Verde
Dec 10 '18 at 15:28

add a comment |

2

Pedantically, == is the documented pattern matching operator within [[...]] but = also works

– glenn jackman
Nov 21 '18 at 16:10

Thanks, this works well when I want to check an input which contains space in between words. But what if I want to check if a string has space characters at the beginning or at the end of the string?

– SSen23
Dec 8 '18 at 10:05

use if [[ $var =~ ^[[:space:]] ]] (begin string) and if [[ $var =~ [[:space:]]$ ]] test space at the end of the string

– Incrivel Monstro Verde
Dec 10 '18 at 15:28

Pedantically, == is the documented pattern matching operator within [[...]] but = also works

– glenn jackman
Nov 21 '18 at 16:10

Thanks, this works well when I want to check an input which contains space in between words. But what if I want to check if a string has space characters at the beginning or at the end of the string?

– SSen23
Dec 8 '18 at 10:05

use if [[ $var =~ ^[[:space:]] ]] (begin string) and if [[ $var =~ [[:space:]]$ ]] test space at the end of the string

– Incrivel Monstro Verde
Dec 10 '18 at 15:28

add a comment |

#!/bin/sh



case "$1" in

    *' '*)

        printf 'Invalid argument %s (contains space)n' "$1" >&2

        exit 1

        ;;

esac

You might want to include other whitespace characters in your check - in which case, use *[[:space:]]* as the pattern instead of *' '*.

edited Nov 21 '18 at 21:24

answered Nov 21 '18 at 16:19

Toby Speight

17.2k134367

This has always been my favorite method. It's very effective, and not too hard to read.

– Paul Hodges
Nov 21 '18 at 21:15

add a comment |

#!/bin/sh



case "$1" in

    *' '*)

        printf 'Invalid argument %s (contains space)n' "$1" >&2

        exit 1

        ;;

esac

You might want to include other whitespace characters in your check - in which case, use *[[:space:]]* as the pattern instead of *' '*.

edited Nov 21 '18 at 21:24

answered Nov 21 '18 at 16:19

Toby Speight

17.2k134367

This has always been my favorite method. It's very effective, and not too hard to read.

– Paul Hodges
Nov 21 '18 at 21:15

add a comment |

#!/bin/sh



case "$1" in

    *' '*)

        printf 'Invalid argument %s (contains space)n' "$1" >&2

        exit 1

        ;;

esac

You might want to include other whitespace characters in your check - in which case, use *[[:space:]]* as the pattern instead of *' '*.

edited Nov 21 '18 at 21:24

answered Nov 21 '18 at 16:19

Toby Speight

17.2k134367

#!/bin/sh



case "$1" in

    *' '*)

        printf 'Invalid argument %s (contains space)n' "$1" >&2

        exit 1

        ;;

esac

You might want to include other whitespace characters in your check - in which case, use *[[:space:]]* as the pattern instead of *' '*.

edited Nov 21 '18 at 21:24

answered Nov 21 '18 at 16:19

Toby Speight

17.2k134367

edited Nov 21 '18 at 21:24

answered Nov 21 '18 at 16:19

Toby Speight

17.2k134367

answered Nov 21 '18 at 16:19

Toby Speight

17.2k134367

answered Nov 21 '18 at 16:19

Toby Speight

17.2k134367

This has always been my favorite method. It's very effective, and not too hard to read.

– Paul Hodges
Nov 21 '18 at 21:15

add a comment |

This has always been my favorite method. It's very effective, and not too hard to read.

– Paul Hodges
Nov 21 '18 at 21:15

This has always been my favorite method. It's very effective, and not too hard to read.

– Paul Hodges
Nov 21 '18 at 21:15

add a comment |

You can use wc -w command to check if there are any words. If the result of this output is a number greater than 1, then it means that there are more than 1 words in the input. Here's an example:

#!/bin/bash

read var1

var2=`echo $var1 | wc -w`

if [ $var2 -gt 1 ]

then

    echo "Spaces"

else

    echo "No spaces"

fi

Note: there is a | (pipe symbol) which means that the result of echo $var1 will be given as input to wc -w via the pipe.

Here is the link where I tested the above code: https://ideone.com/aKJdyN

answered Nov 21 '18 at 15:48

kiner_shah

1,31721624

add a comment |

You can use wc -w command to check if there are any words. If the result of this output is a number greater than 1, then it means that there are more than 1 words in the input. Here's an example:

#!/bin/bash

read var1

var2=`echo $var1 | wc -w`

if [ $var2 -gt 1 ]

then

    echo "Spaces"

else

    echo "No spaces"

fi

Note: there is a | (pipe symbol) which means that the result of echo $var1 will be given as input to wc -w via the pipe.

Here is the link where I tested the above code: https://ideone.com/aKJdyN

answered Nov 21 '18 at 15:48

kiner_shah

1,31721624

add a comment |

You can use wc -w command to check if there are any words. If the result of this output is a number greater than 1, then it means that there are more than 1 words in the input. Here's an example:

#!/bin/bash

read var1

var2=`echo $var1 | wc -w`

if [ $var2 -gt 1 ]

then

    echo "Spaces"

else

    echo "No spaces"

fi

Note: there is a | (pipe symbol) which means that the result of echo $var1 will be given as input to wc -w via the pipe.

Here is the link where I tested the above code: https://ideone.com/aKJdyN

answered Nov 21 '18 at 15:48

kiner_shah

1,31721624

You can use wc -w command to check if there are any words. If the result of this output is a number greater than 1, then it means that there are more than 1 words in the input. Here's an example:

#!/bin/bash

read var1

var2=`echo $var1 | wc -w`

if [ $var2 -gt 1 ]

then

    echo "Spaces"

else

    echo "No spaces"

fi

Note: there is a | (pipe symbol) which means that the result of echo $var1 will be given as input to wc -w via the pipe.

Here is the link where I tested the above code: https://ideone.com/aKJdyN

answered Nov 21 '18 at 15:48

kiner_shah

1,31721624

answered Nov 21 '18 at 15:48

kiner_shah

1,31721624

answered Nov 21 '18 at 15:48

kiner_shah

1,31721624

answered Nov 21 '18 at 15:48

kiner_shah

1,31721624

add a comment |

You can use grep, like this:

echo " foo" | grep 's' -c

# 1

echo "foo" | grep 's' -c

# 0

Or you may use something like this:

s=' foo'

if [[ $s =~ " " ]]; then

    echo 'contains space'

else

    echo 'ok'

fi

edited Nov 21 '18 at 15:54

answered Nov 21 '18 at 15:46

Vladimir Kovpak

11.2k43949

I wouldn't call out to grep, bash can do it as shown.

– glenn jackman
Nov 21 '18 at 16:11

add a comment |

You can use grep, like this:

echo " foo" | grep 's' -c

# 1

echo "foo" | grep 's' -c

# 0

Or you may use something like this:

s=' foo'

if [[ $s =~ " " ]]; then

    echo 'contains space'

else

    echo 'ok'

fi

edited Nov 21 '18 at 15:54

answered Nov 21 '18 at 15:46

Vladimir Kovpak

11.2k43949

I wouldn't call out to grep, bash can do it as shown.

– glenn jackman
Nov 21 '18 at 16:11

add a comment |

You can use grep, like this:

echo " foo" | grep 's' -c

# 1

echo "foo" | grep 's' -c

# 0

Or you may use something like this:

s=' foo'

if [[ $s =~ " " ]]; then

    echo 'contains space'

else

    echo 'ok'

fi

edited Nov 21 '18 at 15:54

answered Nov 21 '18 at 15:46

Vladimir Kovpak

11.2k43949

You can use grep, like this:

echo " foo" | grep 's' -c

# 1

echo "foo" | grep 's' -c

# 0

Or you may use something like this:

s=' foo'

if [[ $s =~ " " ]]; then

    echo 'contains space'

else

    echo 'ok'

fi

edited Nov 21 '18 at 15:54

answered Nov 21 '18 at 15:46

Vladimir Kovpak

11.2k43949

edited Nov 21 '18 at 15:54

answered Nov 21 '18 at 15:46

Vladimir Kovpak

11.2k43949

answered Nov 21 '18 at 15:46

Vladimir Kovpak

11.2k43949

answered Nov 21 '18 at 15:46

Vladimir Kovpak

11.2k43949

I wouldn't call out to grep, bash can do it as shown.

– glenn jackman
Nov 21 '18 at 16:11

add a comment |

I wouldn't call out to grep, bash can do it as shown.

– glenn jackman
Nov 21 '18 at 16:11

I wouldn't call out to grep, bash can do it as shown.

– glenn jackman
Nov 21 '18 at 16:11

add a comment |

You could use parameter expansion to remove everything that isn't a space and see if what's left is the empty string or not:

var1='has space'

var2='nospace'



for var in "$var1" "$var2"; do

    if [[ ${var//[^[:space:]]} ]]; then

        echo "'$var' contains a space"

    fi

done

The key is [[ ${var//[^[:space:]]} ]]:

With ${var//[^[:space:]]}, everything that isn't a space is removed from the expansion of $var.

[[ string ]] has a non-zero exit status if string is empty. It's a shorthand for the equivalent [[ -n string ]].

We could also quote the expansion of ${var//[^[:space:]]}, but [[ ... ]] takes care of the quoting for us.

answered Nov 21 '18 at 16:41

Benjamin W.

21.5k135257

add a comment |

You could use parameter expansion to remove everything that isn't a space and see if what's left is the empty string or not:

var1='has space'

var2='nospace'



for var in "$var1" "$var2"; do

    if [[ ${var//[^[:space:]]} ]]; then

        echo "'$var' contains a space"

    fi

done

The key is [[ ${var//[^[:space:]]} ]]:

With ${var//[^[:space:]]}, everything that isn't a space is removed from the expansion of $var.

[[ string ]] has a non-zero exit status if string is empty. It's a shorthand for the equivalent [[ -n string ]].

We could also quote the expansion of ${var//[^[:space:]]}, but [[ ... ]] takes care of the quoting for us.

answered Nov 21 '18 at 16:41

Benjamin W.

21.5k135257

add a comment |

You could use parameter expansion to remove everything that isn't a space and see if what's left is the empty string or not:

var1='has space'

var2='nospace'



for var in "$var1" "$var2"; do

    if [[ ${var//[^[:space:]]} ]]; then

        echo "'$var' contains a space"

    fi

done

The key is [[ ${var//[^[:space:]]} ]]:

With ${var//[^[:space:]]}, everything that isn't a space is removed from the expansion of $var.

[[ string ]] has a non-zero exit status if string is empty. It's a shorthand for the equivalent [[ -n string ]].

We could also quote the expansion of ${var//[^[:space:]]}, but [[ ... ]] takes care of the quoting for us.

answered Nov 21 '18 at 16:41

Benjamin W.

21.5k135257

You could use parameter expansion to remove everything that isn't a space and see if what's left is the empty string or not:

var1='has space'

var2='nospace'



for var in "$var1" "$var2"; do

    if [[ ${var//[^[:space:]]} ]]; then

        echo "'$var' contains a space"

    fi

done

The key is [[ ${var//[^[:space:]]} ]]:

With ${var//[^[:space:]]}, everything that isn't a space is removed from the expansion of $var.

[[ string ]] has a non-zero exit status if string is empty. It's a shorthand for the equivalent [[ -n string ]].

We could also quote the expansion of ${var//[^[:space:]]}, but [[ ... ]] takes care of the quoting for us.

answered Nov 21 '18 at 16:41

Benjamin W.

21.5k135257

answered Nov 21 '18 at 16:41

Benjamin W.

21.5k135257

answered Nov 21 '18 at 16:41

Benjamin W.

21.5k135257

answered Nov 21 '18 at 16:41

Benjamin W.

21.5k135257

add a comment |

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