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Bash Indirect variable referencing Bash  Download (.zip)#!/bin/bash # Indirect variable referencing. a=letter_of_alphabet letter_of_alphabet=z echo # Direct reference. echo "a = $a" # Indirect reference. eval a=\$$a echo "Now a = $a" echo # Now, let's try changing the second order reference. t=table_cell_3 table_cell_3=24 echo "\"table_cell_3\" = $table_cell_3" echo -n "dereferenced \"t\" = "; eval echo \$$t # In this simple case, # eval t=\$$t; echo "\"t\" = $t" # also works (why?). echo t=table_cell_3 NEW_VAL=387 table_cell_3=$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) echo # (Thanks, S.C., for clearing up the above behavior.) # Another method is the ${!t} notation, discussed in "Bash, version 2" section. # See also example "ex78.sh". exit 0
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