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Dynamic substitution of variables in bash

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Honestly I don't even know how to ask this, or what are the proper terms lol

So what I'm trying to do, is to work on variables with variable names. I think I'd best describe it by actually showing it:
So what I had before was this

if [ -z ${file_var} ]; then

        echo "empty argument"

fi

if [ -z ${pw_var} ]; then

        echo "empty argument"

fi

So what I was trying to do was not have two if statements, but have just one and a for loop, something like this:

for i in file_var, pw_var;do

        if [ -z ${${i}} ]; then

                echo "empty argument"

        fi

done

Now this obviously doesn't work, so I'm wondering how is it properly done.

asked 4 hours ago

user323587

562

add a comment |

Honestly I don't even know how to ask this, or what are the proper terms lol

So what I'm trying to do, is to work on variables with variable names. I think I'd best describe it by actually showing it:
So what I had before was this

if [ -z ${file_var} ]; then

        echo "empty argument"

fi

if [ -z ${pw_var} ]; then

        echo "empty argument"

fi

So what I was trying to do was not have two if statements, but have just one and a for loop, something like this:

for i in file_var, pw_var;do

        if [ -z ${${i}} ]; then

                echo "empty argument"

        fi

done

Now this obviously doesn't work, so I'm wondering how is it properly done.

asked 4 hours ago

user323587

562

add a comment |

Honestly I don't even know how to ask this, or what are the proper terms lol

So what I'm trying to do, is to work on variables with variable names. I think I'd best describe it by actually showing it:
So what I had before was this

if [ -z ${file_var} ]; then

        echo "empty argument"

fi

if [ -z ${pw_var} ]; then

        echo "empty argument"

fi

So what I was trying to do was not have two if statements, but have just one and a for loop, something like this:

for i in file_var, pw_var;do

        if [ -z ${${i}} ]; then

                echo "empty argument"

        fi

done

Now this obviously doesn't work, so I'm wondering how is it properly done.

asked 4 hours ago

user323587

562

Honestly I don't even know how to ask this, or what are the proper terms lol

So what I'm trying to do, is to work on variables with variable names. I think I'd best describe it by actually showing it:
So what I had before was this

if [ -z ${file_var} ]; then

        echo "empty argument"

fi

if [ -z ${pw_var} ]; then

        echo "empty argument"

fi

So what I was trying to do was not have two if statements, but have just one and a for loop, something like this:

for i in file_var, pw_var;do

        if [ -z ${${i}} ]; then

                echo "empty argument"

        fi

done

Now this obviously doesn't work, so I'm wondering how is it properly done.

bash

asked 4 hours ago

user323587

562

asked 4 hours ago

user323587

562

asked 4 hours ago

user323587

562

asked 4 hours ago

user323587

562

asked 4 hours ago

user323587

562

add a comment |

2 Answers
2

active

oldest

votes

What you're trying to do is indirect expansion, which Bash supports using !:

if [ -z "${!i}" ] ; then

If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion.

In this case, "${!i}" expands to whatever the value of the variable whose name is stored in i, and -z will test that indirectly-accessed value.

Alternatively, a nameref is a special type of variable that has this behaviour automatically. You make a nameref variable using declare -n name, and thereafter

All references, assignments, and attribute modifications to name, except for those using or changing the -n attribute itself, are performed on the variable referenced by name’s value.

So if you add

declare -n i

before your loop, it will mean that just

for i in file_var pw_var

do

    if [ -z "$i" ]

    then

    ...

    fi

done

will work as you wanted. Do note though that assignments to i (as opposed to its being set in a loop like this) will modify the target variable.

Note also that there is no comma between items you're looping over in a Bash for loop like how you've written it in the question.

edited 4 hours ago

answered 4 hours ago

Michael Homer

50.8k8141177

Thank you! That's exactly what I needed! Also thanks for the for loop tip :D

– user323587
4 hours ago

add a comment |

This code works, if file_var or pw_var are empty it will print that the specific variable is empty. You can tweak it to fit your usage.

file_var=""

pw_var=""



for i in 'file_var' 'pw_var';do

  if [ -z ${!i} ]; then

    echo "$i is empty"

  fi

done

answered 4 hours ago

Bob Dole

661

New contributor

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

What you're trying to do is indirect expansion, which Bash supports using !:

if [ -z "${!i}" ] ; then

If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion.

In this case, "${!i}" expands to whatever the value of the variable whose name is stored in i, and -z will test that indirectly-accessed value.

Alternatively, a nameref is a special type of variable that has this behaviour automatically. You make a nameref variable using declare -n name, and thereafter

All references, assignments, and attribute modifications to name, except for those using or changing the -n attribute itself, are performed on the variable referenced by name’s value.

So if you add

declare -n i

before your loop, it will mean that just

for i in file_var pw_var

do

    if [ -z "$i" ]

    then

    ...

    fi

done

will work as you wanted. Do note though that assignments to i (as opposed to its being set in a loop like this) will modify the target variable.

Note also that there is no comma between items you're looping over in a Bash for loop like how you've written it in the question.

edited 4 hours ago

answered 4 hours ago

Michael Homer

50.8k8141177

Thank you! That's exactly what I needed! Also thanks for the for loop tip :D

– user323587
4 hours ago

add a comment |

What you're trying to do is indirect expansion, which Bash supports using !:

if [ -z "${!i}" ] ; then

If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion.

In this case, "${!i}" expands to whatever the value of the variable whose name is stored in i, and -z will test that indirectly-accessed value.

Alternatively, a nameref is a special type of variable that has this behaviour automatically. You make a nameref variable using declare -n name, and thereafter

All references, assignments, and attribute modifications to name, except for those using or changing the -n attribute itself, are performed on the variable referenced by name’s value.

So if you add

declare -n i

before your loop, it will mean that just

for i in file_var pw_var

do

    if [ -z "$i" ]

    then

    ...

    fi

done

will work as you wanted. Do note though that assignments to i (as opposed to its being set in a loop like this) will modify the target variable.

Note also that there is no comma between items you're looping over in a Bash for loop like how you've written it in the question.

edited 4 hours ago

answered 4 hours ago

Michael Homer

50.8k8141177

Thank you! That's exactly what I needed! Also thanks for the for loop tip :D

– user323587
4 hours ago

add a comment |

What you're trying to do is indirect expansion, which Bash supports using !:

if [ -z "${!i}" ] ; then

If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion.

In this case, "${!i}" expands to whatever the value of the variable whose name is stored in i, and -z will test that indirectly-accessed value.

Alternatively, a nameref is a special type of variable that has this behaviour automatically. You make a nameref variable using declare -n name, and thereafter

All references, assignments, and attribute modifications to name, except for those using or changing the -n attribute itself, are performed on the variable referenced by name’s value.

So if you add

declare -n i

before your loop, it will mean that just

for i in file_var pw_var

do

    if [ -z "$i" ]

    then

    ...

    fi

done

will work as you wanted. Do note though that assignments to i (as opposed to its being set in a loop like this) will modify the target variable.

Note also that there is no comma between items you're looping over in a Bash for loop like how you've written it in the question.

edited 4 hours ago

answered 4 hours ago

Michael Homer

50.8k8141177

What you're trying to do is indirect expansion, which Bash supports using !:

if [ -z "${!i}" ] ; then

If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion.

In this case, "${!i}" expands to whatever the value of the variable whose name is stored in i, and -z will test that indirectly-accessed value.

Alternatively, a nameref is a special type of variable that has this behaviour automatically. You make a nameref variable using declare -n name, and thereafter

All references, assignments, and attribute modifications to name, except for those using or changing the -n attribute itself, are performed on the variable referenced by name’s value.

So if you add

declare -n i

before your loop, it will mean that just

for i in file_var pw_var

do

    if [ -z "$i" ]

    then

    ...

    fi

done

will work as you wanted. Do note though that assignments to i (as opposed to its being set in a loop like this) will modify the target variable.

Note also that there is no comma between items you're looping over in a Bash for loop like how you've written it in the question.

edited 4 hours ago

answered 4 hours ago

Michael Homer

50.8k8141177

edited 4 hours ago

answered 4 hours ago

Michael Homer

50.8k8141177

answered 4 hours ago

Michael Homer

50.8k8141177

answered 4 hours ago

Michael Homer

50.8k8141177

Thank you! That's exactly what I needed! Also thanks for the for loop tip :D

– user323587
4 hours ago

add a comment |

Thank you! That's exactly what I needed! Also thanks for the for loop tip :D

– user323587
4 hours ago

Thank you! That's exactly what I needed! Also thanks for the for loop tip :D

– user323587
4 hours ago

add a comment |

This code works, if file_var or pw_var are empty it will print that the specific variable is empty. You can tweak it to fit your usage.

file_var=""

pw_var=""



for i in 'file_var' 'pw_var';do

  if [ -z ${!i} ]; then

    echo "$i is empty"

  fi

done

answered 4 hours ago

Bob Dole

661

New contributor

add a comment |

This code works, if file_var or pw_var are empty it will print that the specific variable is empty. You can tweak it to fit your usage.

file_var=""

pw_var=""



for i in 'file_var' 'pw_var';do

  if [ -z ${!i} ]; then

    echo "$i is empty"

  fi

done

answered 4 hours ago

Bob Dole

661

New contributor

add a comment |

This code works, if file_var or pw_var are empty it will print that the specific variable is empty. You can tweak it to fit your usage.

file_var=""

pw_var=""



for i in 'file_var' 'pw_var';do

  if [ -z ${!i} ]; then

    echo "$i is empty"

  fi

done

answered 4 hours ago

Bob Dole

661

New contributor

This code works, if file_var or pw_var are empty it will print that the specific variable is empty. You can tweak it to fit your usage.

file_var=""

pw_var=""



for i in 'file_var' 'pw_var';do

  if [ -z ${!i} ]; then

    echo "$i is empty"

  fi

done

answered 4 hours ago

Bob Dole

661

New contributor

answered 4 hours ago

Bob Dole

661

New contributor

answered 4 hours ago

Bob Dole

661

answered 4 hours ago

Bob Dole

661

New contributor

Bob Dole is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment |

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