bash 在bash中为变量分配文件名
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Assigning a variable a filename in bash
提问by the_mamba
I am writing a script to generate an executable (arm executable) in Linux by taking in user-specified .s file. So the user enters an input file, say "input.s" and an output file name, say "output.axf" and the script generates the desired output (executable - .axf). Now I want an additional option wherein, if the user does not give an output filename in the arguments, I want to create a default output file myself. The script is as follows:
我正在编写一个脚本，通过接收用户指定的 .s 文件在 Linux 中生成一个可执行文件（arm 可执行文件）。所以用户输入一个输入文件，比如“input.s”和一个输出文件名，比如“output.axf”，脚本生成所需的输出（可执行文件 - .axf）。现在我想要一个额外的选项，如果用户没有在参数中提供输出文件名，我想自己创建一个默认输出文件。脚本如下：
#!/bin/bash
echo Enter the names of the input file and output file
read input_file output_file
if [ -z "$input_file" ]
then
echo "No input supplied"
elif [ -z "$output_file" ]
then
$output_file=brot.axf
elif [ -z "$input_file" && -z "$output_file" ]
then
echo "No input/output file supplied"
fi
ifilename=$(basename "$input_file")
ifilename="${input_file%.*}"
armasm -g --cpu=8-A.64 "$input_file"
armlink "$ifilename.o" -o "$output_file"
fromelf --test -c $output_file > disassembly.txt
Now my problem is that every time I run the script and do not specify anything for $output_file, I get this error:
现在我的问题是，每次我运行脚本并且没有为 $output_file 指定任何内容时，我都会收到此错误：
./script_test.sh: line 12: =brot.axf: command not found
./script_test.sh：第 12 行：=brot.axf：找不到命令
Fatal error: L3901U: Missing argument for option 'o'.
致命错误：L3901U：缺少选项“o”的参数。
However, when I do specify the input and output file names with extensions, it works as expected. How do I fix the error and assign a default name to the output file if the user doesn't assign one ?
但是，当我指定带有扩展名的输入和输出文件名时，它会按预期工作。如果用户没有分配一个默认名称，我该如何修复错误并为输出文件分配一个默认名称？
回答by Inian
Variable assignments don't take the $notation in bashshell. You just need below without the $
变量赋值不采用shell 中的$符号bash。你只需要下面没有$
output_file="brot.axf"
And later in the script if filenameis a variable and trying to construct a name with .oappended, enclose the variable name within {}, so that the variable is expanded properly
稍后在脚本中 iffilename是一个变量并尝试构造一个带有.o附加的名称，将变量名称括在 中{}，以便正确扩展变量
armlink "${filename}.o" -o "$output_file"
Also by the looks of it you have a likely typo in the filenameas variable ifilename. If you care trying to use it double-quote it as above.
此外，从它的外观来看，您可能在filenameas 变量中有一个错字ifilename。如果您想尝试使用它，请像上面一样将其双引号。
回答by Pierre Fran?ois
You have to remove the $ in front of output_file on line 12.
您必须删除第 12 行 output_file 前面的 $。
output_file=brot.axf
