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

split函数的用法

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]}'`

实例一：

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