Можно ли в bash / ksh вызвать переменную переменной? - PullRequest
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Можно ли в bash / ksh вызвать переменную переменной?

3 голосов
/ 11 ноября 2011

Можно ли в bash / ksh вызвать переменную, например: 

set -A MY_ARRAY ${${var}_something}

BR Kolesar

Ответы [ 2 ]

6 голосов
/ 11 ноября 2011

В bash: 

VAR1=aap
VAR2=noot

USEVARNAME=VAR2
echo ${!USEVARNAME}

USEVARNAME=VAR1
echo ${!USEVARNAME}

отпечатки 

noot
aap
2 голосов
/ 11 ноября 2011

Вы, вероятно, ссылаетесь на косвенные ссылки: http://tldp.org/LDP/abs/html/ivr.html

# Indirect reference.
eval a=\$$a

Весь фрагмент кода с этого сайта: 

#!/bin/bash
# ind-ref.sh: Indirect variable referencing.
# Accessing the contents of the contents of a variable.

# First, let's fool around a little.

var=23

echo "\$var   = $var"           # $var   = 23
# So far, everything as expected. But ...

echo "\$\$var  = $$var"         # $$var  = 4570var
#  Not useful ...
#  \$\$ expanded to PID of the script
#  -- refer to the entry on the $$ variable --
#+ and "var" is echoed as plain text.
#  (Thank you, Jakob Bohm, for pointing this out.)

echo "\\\$\$var = \$$var"       # \$$var = $23
#  As expected. The first $ is escaped and pasted on to
#+ the value of var ($var = 23 ).
#  Meaningful, but still not useful. 

# Now, let's start over and do it the right way.

# ============================================== #


a=letter_of_alphabet   # Variable "a" holds the name of another variable.
letter_of_alphabet=z

echo

# Direct reference.
echo "a = $a"          # a = letter_of_alphabet

# Indirect reference.
  eval a=\$$a
# ^^^        Forcing an eval(uation), and ...
#        ^   Escaping the first $ ...
# ------------------------------------------------------------------------
# The 'eval' forces an update of $a, sets it to the updated value of \$$a.
# So, we see why 'eval' so often shows up in indirect reference notation.
# ------------------------------------------------------------------------
  echo "Now a = $a"    # Now a = z

echo

# Now, let's try changing the second-order reference.

t=table_cell_3
table_cell_3=24
echo "\"table_cell_3\" = $table_cell_3"            # "table_cell_3" = 24
echo -n "dereferenced \"t\" = "; eval echo \$$t    # dereferenced "t" = 24
# In this simple case, the following also works (why?).
#         eval t=\$$t; echo "\"t\" = $t"

echo

t=table_cell_3
NEW_VAL=387
table_cell_3=$NEW_VAL
echo "Changing value of \"table_cell_3\" to $NEW_VAL."
echo "\"table_cell_3\" now $table_cell_3"
echo -n "dereferenced \"t\" now "; eval echo \$$t
# "eval" takes the two arguments "echo" and "\$$t" (set equal to $table_cell_3)
...