内嵌忽略与空间AWK [英] Ignoring embeded spaces with AWK

问题描述

我在寻找一个简单的方法来打印特定领域使用awk，同时允许在该领域的嵌入式空间。

I'm looking for a simple way to print a specific field with awk while allowing for embedded spaces in the field.

示例：字段1字段2领域中的三个字段4

我希望能够做等同于的awk'{$打印3}，但得到三场作为一个单独的领域而不是两个。

I want to be able to do the equivalent to awk '{print $3}' but getting "Field Three" as a single field not two.

更新：更具体地说，我需要得到后场不是$ 3，但在第3的空间被什么东西搞乱了。在$ 3引号之间的空格数是可变的。我只是想能够治疗的报价，即使不是所有的字段都引用了一个单场之间有什么。因此，忽略空格作为字段分隔符，如果引号之间。

Update: More specifically, I need to get later fields not $3 but the space in #3 is what's messing things up. The number of spaces between the quotes in $3 is variable. I'm just wanting to be able to treat what's between quotes as a single field even if not all fields are quoted. So, ignoring the spaces as field separators if between quotes.

推荐答案

可以做到这一点，如果双引号始终存在：

You can do this if the double quotes are always there:

awk -F\" '{print $2}'

具体来说，我告诉 AWK 的字段都用双引号，此时你想要的部分是现成的场2。分离

Specifically, I am telling awk that the fields are separated by double quotes, at which point the part you want is readily available as field 2.

如果您需要获得在随后的字段，您可以分割的空间行的剩余部分，​​并得到一个新的数组领域，比如 F [] ，这样

If you need to get at subsequent fields, you can split the remainder of the line on spaces and get a new array, say F[] of fields, like this:

awk -F\" '{split($3,F," ");print $2,F[1],F[2]}' file

Field Three Field4 Field5

假设你的文件看起来是这样的：

assuming your file looks like this:

Field1 Field2 "Field Three" Field4 Field5 Field6

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