bash 读取命令不等待输入
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read command doesn't wait for input
提问by mohammadh montazeri
I have problem executing a simple script in bash. The script is like this:
我在 bash 中执行一个简单的脚本时遇到问题。脚本是这样的：
#! /bin/sh
read -p 'press [ENTER] to continue deleting line'
sudo sed -ie '$d' /home/hpccuser/.profile
and when I execute the script with ./script the output is like this:
当我用 ./script 执行脚本时，输出是这样的：
press [ENTER] to continue deleting line./script: 3: read: arg count
[sudo] password for user
I run the read command directly in terminal (copy and paste from script to terminal) and it works fine; it waits for an ENTER to be hit (just like a pause).
我直接在终端中运行读取命令（从脚本复制并粘贴到终端）并且它工作正常；它等待按下 ENTER 键（就像暂停一样）。
回答by Charles Duffy
Because your script starts with #!/bin/shrather than #!/bin/bash, you aren't guaranteed to have bash extensions (such as read -p) available, and can rely only on standards-compliant functionality.
由于您的脚本以#!/bin/sh而非开头#!/bin/bash，因此不能保证您有read -p可用的bash 扩展（例如），并且只能依赖符合标准的功能。
See the relevant standards documentfor a list of functionality guaranteed to be present in read.
请参阅相关的标准文件的要求被提供在功能列表read。
In this case, you'd probably want two lines, one doing the print, and the other doing the read:
在这种情况下，您可能需要两行，一行进行打印，另一行进行读取：
printf 'press [ENTER] to continue deleting...'
read _
回答by Pouya
You can do this with echo command too!:
您也可以使用 echo 命令执行此操作！：
echo "press [ENTER] to continue deleting line"
read continue
回答by hiten
Seems I'm late to the party, but echo -n "Your prompt" && sed 1qdoes the trick on a POSIX-compliant shell.
This prints a prompt, and grabs a line from STDIN.
似乎我迟到了，但是echo -n "Your prompt" && sed 1q在符合 POSIX 的 shell 上做到了这一点。这会打印一个提示，并从 STDIN 中抓取一行。
Alternatively, you could expand that input into a variable:
或者，您可以将该输入扩展为变量：
echo -n "Your prompt"
YOUR_VAR=$(sed 1q)
回答by chr
read -p " Ici mon texte " continue
it works on raspbian
它适用于 raspbian
