为什么(g)awk反转这些输出线？_系统运维

$echo "a,b,c,d,e,f,g" | cut -d,-f-4a,d$echo "a,-f6-f,gecho "a,g" | awk '{ print

$echo "a,g" | awk '{ print $echo "a,g" | awk 'BEGIN {cmd1 = "cut -d,-f-4"; cmd2 = "cut -d,-f6-"} { print [+++] | cmd1; close(cmd1); print [+++] | cmd2; close(cmd2)}'a,df,g
 | "echo hello"; print [+++] | "echo goodbye"; }'hellogoodbye

| "cut -d,-f-4"; print [+++] | "cut -d,-f6-"; }'f,ga,d

$echo "a,b,c,d,e,f,g" | cut -d,-f-4a,d$echo "a,-f6-f,gecho "a,g" | awk '{ print

$echo "a,g" | awk '{ print $echo "a,g" | awk 'BEGIN {cmd1 = "cut -d,-f-4"; cmd2 = "cut -d,-f6-"} { print  | cmd1; close(cmd1); print [+++] | cmd2; close(cmd2)}'a,df,g
 | "echo hello"; print [+++] | "echo goodbye"; }'hellogoodbye

| "cut -d,-f-4"; print [+++] | "cut -d,-f6-"; }'f,ga,d

$echo "a,b,c,d,e,f,g" | cut -d,-f-4a,d$echo "a,-f6-f,gecho "a,g" | awk '{ print

$echo "a,g" | awk '{ print $echo "a,g" | awk 'BEGIN {cmd1 = "cut -d,-f-4"; cmd2 = "cut -d,-f6-"} { print  | cmd1; close(cmd1); print  | cmd2; close(cmd2)}'a,df,g
 | "echo hello"; print [+++] | "echo goodbye"; }'hellogoodbye

| "cut -d,-f-4"; print [+++] | "cut -d,-f6-"; }'f,ga,d

$echo "a,b,c,d,e,f,g" | cut -d,-f-4a,d$echo "a,-f6-f,gecho "a,g" | awk '{ print

$echo "a,g" | awk '{ print $echo "a,g" | awk 'BEGIN {cmd1 = "cut -d,-f-4"; cmd2 = "cut -d,-f6-"} { print  | cmd1; close(cmd1); print  | cmd2; close(cmd2)}'a,df,g
 | "echo hello"; print  | "echo goodbye"; }'hellogoodbye

| "cut -d,-f-4"; print [+++] | "cut -d,-f6-"; }'f,ga,d

$echo "a,b,c,d,e,f,g" | cut -d,-f-4a,d$echo "a,-f6-f,gecho "a,g" | awk '{ print

$echo "a,g" | awk '{ print $echo "a,g" | awk 'BEGIN {cmd1 = "cut -d,-f-4"; cmd2 = "cut -d,-f6-"} { print  | cmd1; close(cmd1); print  | cmd2; close(cmd2)}'a,df,g
 | "echo hello"; print  | "echo goodbye"; }'hellogoodbye

| "cut -d,-f-4"; print | "cut -d,-f6-"; }'f,ga,d

$echo "a,b,c,d,e,f,g" | cut -d,-f-4a,d$echo "a,-f6-f,gecho "a,g" | awk '{ print

$echo "a,g" | awk '{ print $echo "a,g" | awk 'BEGIN {cmd1 = "cut -d,-f-4"; cmd2 = "cut -d,-f6-"} { print  | cmd1; close(cmd1); print  | cmd2; close(cmd2)}'a,df,g
 | "echo hello"; print  | "echo goodbye"; }'hellogoodbye

| "cut -d,-f-4"; print | "cut -d,-f6-"; }'f,ga,d

概述所以,我看到这个输出,我有点惊讶： $echo "a,b,c,d,e,f,g" | cut -d, -f-4a,b,c,d$echo "a,b,c,d,e,f,g" | cut -d, -f6-f,gecho "a,b,c,d,e,f,g" | awk '{ print $0 | "cut -d, -f-4"; print $0 | "cut -d, -f6-"; }'f,ga,b,c 所以,我看到这个输出,我有点惊讶：

$echo "a,b,c,d,e,f,g" | cut -d,-f-4a,d$echo "a,-f6-f,gecho "a,g" | awk '{ print $echo "a,g" | awk '{ print $echo "a,g" | awk 'BEGIN {cmd1 = "cut -d,-f-4"; cmd2 = "cut -d,-f6-"} { print  | cmd1; close(cmd1); print  | cmd2; close(cmd2)}'a,df,g
 | "echo hello"; print  | "echo goodbye"; }'hellogoodbye
 | "cut -d,-f-4"; print  | "cut -d,-f6-"; }'f,ga,d

(作为旁注,我意识到在awk中这是一件非常愚蠢的事情,但这是我见过它的唯一命令！).

据我了解,这应该将记录分为两个命令 – 按顺序.但由于某种原因,输出看起来相反.如果我这样做

然后一切都按照我的预期顺序进行.我认为这肯定是某种竞争条件,但我很惊讶awk不等待管道中的子命令完成.这是一个使用awk或者某些东西来使用gawk的已知问题吗？有没有办法避免这样的陷阱？

编辑：

我也尝试使用mawk ……相同(反转)的结果,似乎两者都是一致的.

解决方法为了确保完成外部命令,您必须执行 close命令.

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