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如何在 IF 语句中检查多个条件

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如何在 IF 语句中检查多个条件

我试图在以下语句中测试多个条件if：

If [[[ "$var1" = "$var2" || "$var1" = "$var3" || "$var1" = "$var4" ]]];

但是，当我执行上述语法时，出现错误。谁能帮我在 IF 语句中检查多个条件？

答案1

括号太多，不if应该大写。尝试这样：

if [[ "$var1" == "$var2" || "$var1" == "$var3" || "$var1" == "$var4" ]]; then

答案2

使用标准sh语法可能比使用（拼写错误，应该[[代替[[[）ksh构造更容易：

case $var1 in
  "$var2" | "$var3" | "$var4") echo match;;
  *) echo no match;;
esac

答案3

[[ "$var1" == "$var2" || "$var1" == "$var3" || "$var1" == "$var4" ]]

或者，为了可读性，

[[ "$var1" == "$var2" ]] ||
[[ "$var1" == "$var3" ]] ||
[[ "$var1" == "$var4" ]]

或者与case,正如斯特凡·查泽拉斯建议的那样，或者对变量名称进行过于复杂的循环并使用名称引用（在bash或中ksh93），

for var in var2 var3 var4; do
    typeset -n v="$var"
    if [[ "$var1" == "$v" ]]; then
        # printf '%s has same value as var1\n' "$var"
        found=1
        break
    fi
done

if [[ "$found" -eq 1 ]]; then ...

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