使用bash -c与循环和标准输出重定向 [英] Using bash -c with loops and standard output redirection

问题描述

为什么从以下命令输出的标准有何不同？

命令1：

 为（（i = 0;我小于10，I = I + 1））;做回声$ I; DONE
 

命令2：

 的bash -c为（（i = 0;我小于10，I = I + 1））;做回声$ I;完成
 

命令1打印整数1至10在单独一行，符合市场预期。命令2 10打印在每行

什么我需要做的就是命令2打印同样的事情命令1？

长的故事：

我经常需要运行大量文件的可执行和捕捉的全过程的输出。我使用了一个类似的循环称为my_script.sh一个单独的文件为Command。

我然后可以使用得到的输出：

 庆典my_script.sh＆GT; RESULTS.TXT
 

由于这是一个痛苦位创造平凡的循环单独my_script.sh的文件，我希望用的bash -c来达到同样的事情，但遇到了这个问题。

感谢您的任何建议。期待回复。结果
米莎

解决方案

  

什么我需要做的就是命令2打印同样的事情命令1？

您需要修改该行：

 的bash -c为（（i = 0;我小于10，I = I + 1））;做回声$ I;完成
 

要，而不是这样说：

 的bash -c为（（i = 0;我小于10，I = I + 1））;做回声\\ $ I;完成
 

这是怎么回事的是， $ I 顺序由您从运行bash外壳PTED是间$ P $，它取代的自己的的 I 变量到串其传递到新的bash实例之前执行。因此由时间的新的bash看到串，它是：

 为（（i = 0;我小于10，I = I + 1））;做呼应10; DONE
 

假设 I 包含的值10。

（当我第一次跑第二个脚本，我做到了运行第一之前，写了10空行，因为我没有 I 变量定义的，因此它被代入空字符串）。

Why does the standard output from the following commands differ?

command1:

for ((i=0; i<10; i=i+1)); do echo $i; done

command2:

bash -c "for ((i=0; i<10; i=i+1)); do echo $i; done"

command1 prints the integers 1 to 10 on a separate line, as expected. command2 prints 10 on each line.

What do I need to do to get command2 to print the same thing as command1?

The long story:

I often need to run an executable on a large number of files and capture the output of the entire process. I use a for loop similar to command1 in a separate file called my_script.sh.

I can then get the output using:

bash my_script.sh > results.txt

Since it's a bit of a pain to create a separate my_script.sh file for trivial for loops, I was hoping to achieve the same thing by using "bash -c", but ran into this problem.

Thanks for any suggestions. Looking forward to replies.
Misha

解决方案

What do I need to do to get command2 to print the same thing as command1?

You need to fix this line:

bash -c "for ((i=0; i<10; i=i+1)); do echo $i; done"

To say this instead:

bash -c "for ((i=0; i<10; i=i+1)); do echo \$i; done"

What's happening is that the $i sequence is being interpreted by the shell you are running bash from, and it is substituting its own i variable into the string before passing it off to the new bash instance to be executed. So by the time the new bash sees the string, it is:

for ((i=0; i<10; i=i+1)); do echo 10; done

Assuming that i contained the value "10".

(When I first ran the second script, which I did before running the first, it wrote out 10 blank lines, because I had no i variable defined, and so it was substituting in the empty string.)

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