bash awk 在读取行中的行为
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behavior of awk in read line
提问by user2757300
$ cat file
11 asasaw121
12 saasks122
13 sasjaks22
$ cat no
while read line
do
var=$(awk '{print }' $line)
echo $var
done<file
$ cat yes
while read line
do
var=$(echo $line | awk '{print }')
echo $var
done<file
$ sh no
awk: can't open file 11
source line number 1
awk: can't open file 12
source line number 1
awk: can't open file 13
source line number 1
$ sh yes
11
12
13
Why doesn't the first one work? What does awk expect to find in $1 in it? I think understanding this will help me avoid numerous scripting problems.
为什么第一个不起作用？awk 期望在 $1 中找到什么？我认为理解这将帮助我避免许多脚本问题。
回答by jkshah
awkalways expects a file name as input
awk总是期望一个文件名作为输入
In following, $lineis string not a file.
在下面，$line是字符串而不是文件。
var=$(awk '{print }' $line)
You could say (Note double quotes around variable)
你可以说（注意变量周围的双引号）
var=$(awk '{print }' <<<"$line")
回答by anubhava
Why doesn't the first one work?
Why doesn't the first one work?
Because of this line:
因为这一行：
var=$(awk '{print }' $line)
Which assumes $lineis a file.
假设$line是一个文件。
You can make it:
你可以做到：
var=$(echo "$line" | awk '{print }')
OR
或者
var=$(awk '{print }' <<< "$line")
回答by qwwqwwq
awk '{print }' $line
^^ awk expects to see a file path or list of file paths here
what it is getting from you is the actual file line
What you want to do is pipe the line into awkas you do in your second example.
你想要做的是awk像你在第二个例子中所做的那样将这条线插入。
回答by Ed Morton
You got the answers to your specific questions but I'm not sure it's clear that you would never actually do any of the above.
你得到了你的具体问题的答案，但我不确定你是否真的不会做上述任何事情。
To print the first field from a file you'd either do this:
要从文件打印第一个字段，您可以执行以下操作：
while IFS= read -r first rest
do
printf "%s\n" "$first"
done < file
or this:
或这个：
awk '{print }' file
or this:
或这个：
cut -d ' ' -f1 <file
The shell loop would NOT be recommended.
不推荐使用 shell 循环。
