Linux里awk中split函数的用法小结

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]}"`

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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 ",arr[i]）}"
输出结果如下
a

c
d
e
f
g
h
i