需要将文本文件的内容分配给bash脚本中的变量

Question

我是制作bash脚本的新手,但我的目标是获取一个.txt文件,并将txt文件中的字符串分配给变量.我试过这个(如果我在正确的轨道上没有线索).

#!/bin/bash
FILE="answer.txt"
file1="cat answer.txt"
print $file1

当我跑这个时,我明白了

Warning: unknown mime-type for "cat" -- using "application/octet-stream"
Error: no such file "cat"
Error: no "print" mailcap rules found for type "text/plain"

我能做些什么来完成这项工作？

编辑**当我将其更改为:

#!/bin/bash
    FILE="answer.txt"
    file1=$(cat answer.txt)
    print $file1

我得到了这个:

Warning: unknown mime-type for "This" -- using "application/octet-stream"
Warning: unknown mime-type for "text" -- using "application/octet-stream"
Warning: unknown mime-type for "string" -- using "application/octet-stream"
Warning: unknown mime-type for "should" -- using "application/octet-stream"
Warning: unknown mime-type for "be" -- using "application/octet-stream"
Warning: unknown mime-type for "a" -- using "application/octet-stream"
Warning: unknown mime-type for "varible." -- using "application/octet-stream"
Error: no such file "This"
Error: no such file "text"
Error: no such file "string"
Error: no such file "should"
Error: no such file "be"
Error: no such file "a"
Error: no such file "varible."

当我输入cat answer.txt它打印出来时,这个文本字符串应该像它应该的变量但是,我仍然无法通过变量获得bash.

Answer 1

在bash中$(< answer.txt)是一个内置的简写$(cat answer.txt)

我怀疑你是在运行这个print:

NAME  
    run-mailcap, see, edit, compose, print ? execute programs via entries in the mailcap file

Answer 2

你需要反引号来捕获命令的输出(你可能想要echo而不是print):

file1=`cat answer.txt`
echo $file1

Answer 3

该$()建筑返回stdout从命令.

file_contents=$(cat answer.txt)