How can I check if the number of arguments passed into the shell script is exactly 2?

I have some understanding that to check the number of arguments being passed into the shell script all I have to do inside the shell script is this: Example 1

if [ "$#" -ne 2 ]; then
 do something;
fi

However, I am a little confused and would like to know whether the name of the shell script when being run is considered part of the arguments as well as this is the case for python script when it comes to sys.argv.

An example in the terminal would be like:
Example 2

./script.sh $1 $2

In this case for Example 2 how many arguments would actually be considered to be passed in to Example 1 if else loop.

edited Jan 10 at 18:56

Rui F Ribeiro

39.6k1479132

asked Jan 10 at 10:32

user3187113

What result did you get when you tested it?

– ctrl-alt-delor
Jan 10 at 11:25

add a comment |

I have some understanding that to check the number of arguments being passed into the shell script all I have to do inside the shell script is this: Example 1

if [ "$#" -ne 2 ]; then
 do something;
fi

An example in the terminal would be like:
Example 2

./script.sh $1 $2

In this case for Example 2 how many arguments would actually be considered to be passed in to Example 1 if else loop.

edited Jan 10 at 18:56

Rui F Ribeiro

39.6k1479132

asked Jan 10 at 10:32

user3187113

What result did you get when you tested it?

– ctrl-alt-delor
Jan 10 at 11:25

add a comment |

I have some understanding that to check the number of arguments being passed into the shell script all I have to do inside the shell script is this: Example 1

if [ "$#" -ne 2 ]; then
 do something;
fi

An example in the terminal would be like:
Example 2

./script.sh $1 $2

In this case for Example 2 how many arguments would actually be considered to be passed in to Example 1 if else loop.

edited Jan 10 at 18:56

Rui F Ribeiro

39.6k1479132

asked Jan 10 at 10:32

user3187113

I have some understanding that to check the number of arguments being passed into the shell script all I have to do inside the shell script is this: Example 1

if [ "$#" -ne 2 ]; then
 do something;
fi

An example in the terminal would be like:
Example 2

./script.sh $1 $2

In this case for Example 2 how many arguments would actually be considered to be passed in to Example 1 if else loop.

bash shell-script

edited Jan 10 at 18:56

Rui F Ribeiro

39.6k1479132

asked Jan 10 at 10:32

user3187113

edited Jan 10 at 18:56

Rui F Ribeiro

39.6k1479132

asked Jan 10 at 10:32

user3187113

edited Jan 10 at 18:56

Rui F Ribeiro

39.6k1479132

edited Jan 10 at 18:56

Rui F Ribeiro

39.6k1479132

edited Jan 10 at 18:56

Rui F Ribeiro

39.6k1479132

asked Jan 10 at 10:32

user3187113

asked Jan 10 at 10:32

user3187113

asked Jan 10 at 10:32

user3187113

What result did you get when you tested it?

– ctrl-alt-delor
Jan 10 at 11:25

add a comment |

What result did you get when you tested it?

– ctrl-alt-delor
Jan 10 at 11:25

What result did you get when you tested it?

– ctrl-alt-delor
Jan 10 at 11:25

add a comment |

1 Answer
1

active

oldest

votes

The name of the script is not considered as part of the positional parameters. This means that

somescript arg1 arg2

will set $1 to arg1 and $2 to arg2 and that $# will be 2.

The name of the script will be available in $0, but $0 is special in that it's not included in the array $@, and $# is the length (number of elements) of $@.

answered Jan 10 at 10:39

Kusalananda

126k16239393

1

thank you! I understand now.

– user3187113
Jan 10 at 16:25

add a comment |

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1 Answer
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active

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1 Answer
1

active

oldest

votes

The name of the script is not considered as part of the positional parameters. This means that

somescript arg1 arg2

will set $1 to arg1 and $2 to arg2 and that $# will be 2.

The name of the script will be available in $0, but $0 is special in that it's not included in the array $@, and $# is the length (number of elements) of $@.

answered Jan 10 at 10:39

Kusalananda

126k16239393

1

thank you! I understand now.

– user3187113
Jan 10 at 16:25

add a comment |

The name of the script is not considered as part of the positional parameters. This means that

somescript arg1 arg2

will set $1 to arg1 and $2 to arg2 and that $# will be 2.

The name of the script will be available in $0, but $0 is special in that it's not included in the array $@, and $# is the length (number of elements) of $@.

answered Jan 10 at 10:39

Kusalananda

126k16239393

1

thank you! I understand now.

– user3187113
Jan 10 at 16:25

add a comment |

The name of the script is not considered as part of the positional parameters. This means that

somescript arg1 arg2

will set $1 to arg1 and $2 to arg2 and that $# will be 2.

The name of the script will be available in $0, but $0 is special in that it's not included in the array $@, and $# is the length (number of elements) of $@.

answered Jan 10 at 10:39

Kusalananda

126k16239393

The name of the script is not considered as part of the positional parameters. This means that

somescript arg1 arg2

will set $1 to arg1 and $2 to arg2 and that $# will be 2.

The name of the script will be available in $0, but $0 is special in that it's not included in the array $@, and $# is the length (number of elements) of $@.

answered Jan 10 at 10:39

Kusalananda

126k16239393

answered Jan 10 at 10:39

Kusalananda

126k16239393

answered Jan 10 at 10:39

Kusalananda

126k16239393

answered Jan 10 at 10:39

Kusalananda

126k16239393

1

thank you! I understand now.

– user3187113
Jan 10 at 16:25

add a comment |

1

thank you! I understand now.

– user3187113
Jan 10 at 16:25

thank you! I understand now.

– user3187113
Jan 10 at 16:25

add a comment |

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