使用Flask-Script/Python从标准输入中读取 [英] Reading from standard input using Flask-Script / Python

问题描述

现在我有flask-script命令，该命令将路径作为参数，然后从路径中读取:

Right now I have flask-script command that takes a path as an argument, then reads from the path:

@manager.option('-f', '--file', dest='file_path')
def my_command(file_path):
     open(file_path)
     ...

我希望它也能够从标准中读取.(我经常需要在剪贴板上传递文本，每次都必须创建一个文件很烦人.)

I'd want it to be able to read from standard in as well. (I frequently need to pass it text on the clipboard, and it's annoying to have to create a file each time.)

我该怎么做?

我尝试通过 https://stackoverflow.com使用 fileinput.input()/a/1454400/1164573 ，并通过以下方式调用:

I've tried using fileinput.input(), via this https://stackoverflow.com/a/1454400/1164573, invoked with the following:

cat << EOF | ./manage.py my_command
abc
def
ghi
EOF

但是 fileinput.input()为空.这是因为flask-script包装了我的函数，并且没有直接向其中公开标准吗?我该如何解决?

But fileinput.input() is empty. Is this because flask-script is wrapping my function and not exposing standard in to it directly? How can I get around this?

推荐答案

您几乎可以像您的示例那样进行操作，但是可以使用进程替换而不是管道:

You could do it almost like your example, but using process substitution instead of a pipe:

./manage.py my_command <(cat <<EOF
abc
def
ghi
jkl
EOF
)

适用于我的简单测试...假设您至少在外壳上使用bash.我只使用bash，所以不知道此语法是否适用于其他shell.

works for my simple test. . . assuming you're using bash for your shell at least. I only use bash, so don't know if this syntax works for other shells.

或者，您可以测试文件名的值是否为特殊值，通常为-，如果这是要读取的文件名，则使用 sys.stdin .

Alternately, you could test the value of the filename for a special value, typically - and use sys.stdin if that's the name of the file to read.

if(sys.argv[1] == '-'):
    f = sys.stdin
else:
    f = file(sys.argv[1])

for line in f:
    print line

以此类推

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