15468 发表于 2018-8-26 10:56:10

shell判断字符串相等脚本

  #!/bin/sh
  echo -n "login:"
  read name
  echo -n "password:"
  read passwd
  if [ "$name" = "aa" -a "$passwd" = "aaa" ];then
  echo "right!"
  else echo "error"
  fi
  一。运行过程中出现过 [: missing `]'的问题,是[ "$name" = "aa" -a "$passwd" = "aaa" ] "$name"前和"aaa"后都必须要有空格。
  二。if里面的-a相当于是与,-o相当于是或。还可以用&&和||表达。
  if [ "$name" = "aa" -a "$passwd" = "aaa" ];then与
  if [ "$name" = "aa" -o "$passwd" = "aaa" ];then或
  if [ "$name" = "aa" ] && [ "$passwd" = "aaa" ];then与
  if [ "$name" = "aa" ] || [ "$passwd" = "aaa" ];then或
&&是shell本身的语法支持  -a是shell的内部指令的用法

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