Question

在我给自己增加一个变数的情况下,我这样做了。

while read line 
do
      let variable=$variable+$someOtherVariable
done
    return $variable

当我重复应计价时,我没有产出......

Is this the correct way to add some value back to the variable itself (i.e. i = i+j) Also, in the context of bash scripting what is the scope in this case..

Answer 1

问题是,变数在范围外看不见(转让在外传播)。

头一种思维方式是把指挥控制在小片中,迫使护堤排放变数:

variable=$(variable=0; while read line; do variable=$((variable+someOtherVariable)); done; echo $variable)

Answer 2

<代码>return 回归“exit”代码,编号,不是你所期望的。页: 1

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