#!/bin/bash# patt-matching.sh# 使用# ## % %%來進行參數(shù)替換操作的模式匹配. parameter substitution operators.var1=abcd12345abc6789pattern1=a*c  # *(通配符)匹配a - c之間的任意字符.echoecho "var1 = $var1"           # abcd12345abc6789echo "var1 = ${var1}"         # abcd12345abc6789                              # (另一種形式)echo "Number of characters in ${var1} = ${#var1}"echoecho "pattern1 = $pattern1"   # a*c  (匹配'a'到'c'之間的任意字符)echo "--------------"echo '${var1#$pattern1}  =' "${var1#$pattern1}"    #         d12345abc6789# 最短的可能匹配, 去掉abcd12345abc6789的前3個字符.#                     |-|               ^^^^^echo '${var1##$pattern1} =' "${var1##$pattern1}"   #                  6789# 最長的可能匹配, 去掉abcd12345abc6789的前12個字符#                     |----------|      ^^^^^^echo; echo; echopattern2=b*9            # 匹配'b'到'9'之間的任意字符echo "var1 = $var1"     # 還是abcd12345abc6789echoecho "pattern2 = $pattern2"echo "--------------"echo '${var1%pattern2}  =' "${var1%$pattern2}"     #     abcd12345a# 最短的可能匹配, 去掉abcd12345abc6789的最后6個字符#                               |----|  ^^^^^^^echo '${var1%%pattern2} =' "${var1%%$pattern2}"    #     a# 最長的可能匹配, 去掉abcd12345abc6789的最后12個字符#                      |-------------|  ^^^^^^^^# 牢記, #和##是從字符串左邊開始, 并且去掉左邊的字符串, #       %和%%從字符串的右邊開始, 并且去掉右邊的字符串. # (譯者注: 有個好記的方法, 那就是察看鍵盤順序, 記住#在%的左邊. ^_^)echoexit 0