變數中的變數? - Linux

Dora · 2013-12-13

Table of Contents

i=5
var5=hello

echo ${var$i}
bad substitution

請問要如何更改${}內的東西使得他會印出var5內的值?

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All Comments

Heather2013-12-13

varname=var$i ; echo ${!varname}

Dorothy2013-12-15

eval('$var'.$i)

Hedwig2013-12-20

BASH> i=5;var[5]=hello;echo ${var[$i]}