Question

我对制作bash脚本非常陌生，但我的目标是获取我拥有的.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

当我运行时，我得到了

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"

我可以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时，它打印出来的文本字符串应该是应该的变量，但是，我仍然无法通过varible来获取bash。

Answer 1

在bash中$(< answer.txt)是内置的 $(cat answer.txt)

的简写

我怀疑你正在运行此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

$()构造从命令返回stdout。

file_contents=$(cat answer.txt)

需要将文本文件的内容分配给bash脚本中的变量

3 个答案: