保持原始格式POST通过AWK | 保持原始格式POST通过AWK

本文介绍了保持原始格式POST通过AWK的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个问题用AWK简单地从流删除字段，如下图所示：

I have an issue with using AWK to simply remove a field from a stream, illustrated below:

  1 int blah (void)
  2 {
  3         if (foo) {
  4                 printf ("blah\n");
  5         }
  6         return 0;
  7 }

我用下面的code，除去第一场：

I use the following code to remove the first field:

$的awk'{$ 1 =;打印}'example.out

$ awk '{ $1=""; print }' example.out

 int blah (void)
 {
 if (foo) {
 printf ("blah\n");
 }
 return 0;
 }

为什么会出现这种情况？这是因为AWK删除所有空白 - ？可这是prevented

Why is this the case? Is this because AWK removes all whitespace - can this be prevented?

亲切的问候提前

推荐答案

包含一个描述如何做到这一点。网站还链接了的含有3解决问题的方案。据我所知，如果分配给字段，则输出字段分隔符是用来连接各个领域在一起。因此，+ suddendly被折叠到一个空格。用一粒盐拿去吧，我敢不awk的专家。例如，尝试分配：来变量 OFS 和冒号代替空格，将导致输出字段之间

Contains a description how to do it. It also links to http://student.northpark.edu/pemente/awk/awktail.txt which contains 3 solutions to the problem. As far as i know, if you assign to a field, then the output field separator is used to concatenate all fields together. So " "+ suddendly is collapsed to one space. Take it with a grain of salt though, i'm no awk expert. For example, try assigning : to the variable OFS, and colons instead of spaces will result in between fields in the output:

echo a b c | awk 'BEGIN{ OFS = ":" } { $1=""; print }'
$ :b:c

如果您使用摹 awk中，那么你可以使用它的 gensub 扩展，我发现pretty直截了当使用方法：

If you use gawk, then you can use its gensub extension which i find pretty straight forward to use:

print gensub($1 "[\t ]*(.*)", "\\1", 1);

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