从awk输出中读取两个变量

I have code of something like the following.

我有类似以下的代码。

awk '{ print $1, $2}' file | 
while IFS= read A B
do 
echo $A
if [ "$B" -eq "16" ];
then 
  grep -A 1 $A $1 | python unreverse.py  
else
  grep -A 1 $A
fi
done    

where the awk command outputs a string and a number for each line in file. I don't quite understand how to get the while loop to work, however. It is not splitting the line properly because B is empty and I get "integer expression expected".

其中awk命令为文件中的每一行输出一个字符串和一个数字。但是，我不太明白如何让while循环工作。它没有正确分割线，因为B是空的，我得到“预期的整数表达式”。

1 个解决方案

#1

You are essentially undefining IFS with IFS= read A B, so word splitting does not occur and all the input is captured into $A. Simply remove IFS=, or use IFS=" ", and you should be satisfied

您基本上是在IFS =读取A B时取消定义IFS，因此不会发生分词，并且所有输入都被捕获到$ A中。只需删除IFS =，或使用IFS =“”，您应该感到满意