(21楼又有新问题，请大神关注)getline为何取不到最后一行

jason680

回复 1# bulletmarquis

a|b c|d e|f g

1. 1st time, awk read first data and assign to $0
  awk get "a" to assign $0
A: getline
   get next data "b" to assign to $0
B: getline v
   get next data "b" to assign to v

2. 2nd time, awk read next data and assign to $0
  awk get "c" to assign $0
A: getline
   get next data "d" to assign to $0
B: getline v
   get next data "d" to assign to v

3. 3rd time, awk read next data and assign to $0
  awk get "e" to assign $0
A: getline
   get next data "f" to assign to $0
B: getline v
   get next data "f" to assign to v

3. 4th time, awk read next data and assign to $0
  awk get "g" to assign $0
A: getline
   There is NO data to get and assign to $0
   Note: $0 keep the original data "g"
B: getline v
   There is NO data to get and assign to v
   Note: v keep the original data "f"

Don't forget one thing
程序是
依照写出来(的code),跑出结果
不是依照你想出来的,得到结果

   

ly5066113

回复 10# bulletmarquis


暂时未找到这个问题的原因，RS 设置为正则表达式时，使用 getline 注意一下就好，不用 getline ，其他的是正常的：

1. echo "a+b*c+d*e+f*g+h"|awk 'BEGIN{RS="[*+]"}1'

复制代码

bulletmarquis

回复 11# jason680


    恩，我能理解getline和getline v两种方式打印的最后一个字符不一样

   就是不明白为啥

There is NO data to get

reyleon

回复 11# jason680


   

3. 4th time, awk read next data and assign to $0
  awk get "g" to assign $0
A: getline
   There is NO data to get and assign to $0
   Note: $0 keep the original data "g"
B: getline v
   There is NO data to get and assign to v
   Note: v keep the original data "f"


---
  为什么 就 NO data 了 ? 最后那个 h 呢?

blackold

回复 9# ly5066113


    Tim 老师说得有理。

    另外，好像用[]才有问题。

jason680

回复 13# bulletmarquis

Sorry, I got the some mistake.

It seems have some bug in awk with getline function
   

ly5066113

回复 15# blackold


和 [ ] 无关，这样设置 RS=' |\\|' ，也一样有问题。

jason680

@ly5066113 @ly5066113 @reyleon

It seems have some bug in awk

$ echo "a|b c|d e|f g|h" | awk -vRS='[ |]' '{print NF,NR,$0;getline;print}'
1 1 a
b
1 3 c
d
1 5 e
f
1 7 g
g

$ echo "a/b c/d e/f g/h" | awk -vRS='[ /]' '{print NF,NR,$0;getline;print}'
1 1 a
b
1 3 c
d
1 5 e
f
1 7 g
h

blackold

回复 17# ly5066113


    嗯，刚才又试了，' |\\|' 确实也有问题。

    看起来是在[]中使用元字符或者有|都有问题。

使用命名类正常:

1. $ echo "a|b c|d e|f g|h" | awk -vRS='[[:punct:][:space:]]' '{printf $0"  ";getline;print}'
   
2. a  b
   
3. c  d
   
4. e  f
   
5. g  h

复制代码

blackold

应该是awk正则引擎出了问题，先列为悬案吧。