循环遍历文件以计算字段编号 [英] Loop through file to count field number

问题描述

我有一个脚本bash可以从.txt文件中添加用户. 真的很简单:

I have a script bash to add users from a .txt file. It is really simple:

name firstname uid gid

以空格分隔的值

我想用awk检查每一行是否包含4个字段.如果是，我想返回1，如果不是，则返回0.

I want to check with awk if each row contains 4 fields. If yes I want to return 1, if not return 0.

file=/my/file.txt
awk=$(awk -F' ' '{(NF != 4) ? res = 0 : res = 1; print res)}' $file)
echo $awk

现在，awk为每一行返回1，但是我希望它在结尾处返回1或0，而不是文件中的每一行.

Right now, awk returns 1 for each row, but I want it to return 1 or 0 at the end, not for each line in the file.

推荐答案

在UNIX上，如果成功则返回0，如果出错则返回!=0.对我来说，当所有记录都具有4字段时返回0更有意义，而当并非所有记录都具有4个字段时返回1更有意义.

On UNIX you'll return 0 in case of success and !=0 in case of an error. For me it makes more sense to return 0 when all records have 4 fields and 1 when not all records have 4 fields.

要实现此目的，请使用 exit :

To achieve that, use exit:

awk 'NF!=4{exit 1}' file

仅供参考:默认情况下，awk将以0退出.

FYI: awk will exit with 0 by default.

如果要在有条件的shell中使用它:

If you want to use it in a shell conditional:

#!/bin/bash
if ! awk 'NF!=4{exit 1}' file ; then
    echo "file is invalid"
fi

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PS:您的示例中的-F' '是多余的，因为' '是默认字段定界符.

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PS: -F' ' in your example is superfluous because ' ' is the default field delimiter.

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