The awk function split(s,a,sep) splits a string s into an awk array a using the delimiter sep. set time = 12:34:56 set hr = `echo $time | awk '{split($0,a,":" ); print a[1]}'` # = 12 set sec = `echo $time | awk '{split($0,a,":" ); print a[3]}'` # = 56 # = 12 34 56 set hms = `echo $time | awk '{split($0,a,":" ); print a[1], a[2], a[3]}'` —————————————————————————————————————————— Q: name="76868&5676&435&43526&334&12312312&12321" awk 'BEGIN {print split("$name", filearray, "&")}' 为什么是1 而 awk 'BEGIN {print split("76868&5676&435&43526&334&12312312&12321", filearray, "&")}' 则返回正确的结果，应该是7，有没有人解答一下？ A: 变量引用错误，这样做试试 awk 'BEGIN {print split('"\"$name\""', filearray, "&")}' awk规定引用系统变量必须使用单引号加双引号，即'"$sysvar"'这样的格式，但是split函数也需要双引号来定界，但这个双引号又不能让sh解释，而应留给awk来解释，所以使用了\"和\"组成的双引号 [b]split函数的用法[/b] he awk function split(s,a,sep) splits a string s into an awk array a using the delimiter sep. set time = 12:34:56 set hr = `echo $time | awk '{split($0,a,":" ); print a[1]}'` # = 12 set sec = `echo $time | awk '{split($0,a,":" ); print a[3]}'` # = 56 # = 12 34 56 set hms = `echo $time | awk '{split($0,a,":" ); print a[1], a[2], a[3]}'` set hms = `echo $time | awk '{split($0,a,":" ); for (i=1; i<=3; i++) print a[i]}'` [b]实例一：[/b] cat a a:b:c:d:e:f:g:h:i 使用awk将该字符串冒号两边的段输出 cat a |awk -F':' '{split($0,arr,":")}END{for(i=1;i<=NF;i++)printf("%s\n",arr[i])}' 输出结果如下 a c d e f g h i