使用带有读取和返回语句的 bash 脚本

Question

我有以下脚本：

#!/bin/bash
#
# Example script for validating SVN credentials.

var_svn_user_name=
var_svn_password=

function get_svn_credentials()
{
  # First, get the credentials from the user
  read -r -p "Please enter SVN User Name: " var_svn_user_name
  echo -n "Please enter SVN Password:  "
  read -r -s var_svn_password
  echo ""
  echo "----------------"
  echo "The SVN User Name is:     ${var_svn_user_name}"
  echo "The SVN User Password is: ${var_svn_password}"

  # Next, validate provided credentials
  echo -n "Validating credentials... "
  #var_ret=$(svn list --username "${var_svn_user_name}" --password \
  #        "${var_svn_password}" ${var_url} ${var_cfg} ${var_opt}=${var_val} \
  #        --no-auth-cache --non-interactive 2>&1 | grep "Authentication failed")
  if [[ $var_ret == "" ]]
  then
    echo 'Success'
  else
    echo 'Failed'
  fi
}

function main()
{
  # EXAMPLE Call #1
  #result=$(get_svn_credentials)  # FAILURE
  # EXAMPLE Call #2
  get_svn_credentials           # SUCCESS

  echo "Return value is: $result"
  if [[ $var_ret == "Success" ]]
  then
    echo "SVN User Name and Password was validated."
  else
    echo "SVN User Name and Password was NOT validated."
  fi
}

main "$@"

为什么当我注释掉示例 2 并取消注释示例 1 时，密码的回显直到 read 执行才显示？

我试图弄清楚如何让 return 语句像 C 函数样式的 return 语句一样工作。

有人可以帮忙吗？

Answer 1

您正在将提示写入标准输出，该输出由命令替换捕获。而是将其写入标准错误（就像 read -p 一样）。

function get_svn_credentials()
{
  # First, get the credentials from the user
  read -r -p "Please enter SVN User Name: " var_svn_user_name
  echo -n "Please enter SVN Password:  " >&2
  read -r -s var_svn_password

  {
    echo ""
    echo "----------------"
    echo "The SVN User Name is:     ${var_svn_user_name}"
    echo "The SVN User Password is: ${var_svn_password}"
  } >&2

  # Next, validate provided credentials
  echo -n "Validating credentials... "
  #var_ret=$(svn list --username "${var_svn_user_name}" --password \
  #        "${var_svn_password}" ${var_url} ${var_cfg} ${var_opt}=${var_val} \
  #        --no-auth-cache --non-interactive 2>&1 | grep "Authentication failed")
  if [[ $var_ret == "" ]]
  then
    echo 'Success'
  else
    echo 'Failed'
  fi
}

也就是说，不要依赖输出来确定它是否成功；只需使用退出状态。

get_svn_credentials () {
  local user_name password
  # First, get the credentials from the user
  read -r -p "Please enter SVN User Name: " user_name
  read -r -p "Please enter SVN Password:  " -s password

  {
    echo ""
    echo "----------------"
    echo "The SVN User Name is:     ${user_name}"
    echo "The SVN User Password is: ${password}"
  } >&2

  # Next, validate provided credentials
  # Let the exit status of grep -q be the exit status
  # of the function
  printf '%s\n' "Validating credentials... " >&2
  svn list --username "${user_name}" \
           --password "${password}" \
           "${var_url}" ${var_cfg} "${var_opt}=${var_val}" \
          --no-auth-cache --non-interactive 2>&1 |
    grep -q "Authentication failed"
}

main () {
  if get_svn_credentials
  then
    echo "SVN User Name and Password was validated."
  else
    echo "SVN User Name and Password was NOT validated."
  fi
}

main

（注意：您应该可能引用$var_cfg，但它实际上可能是一个选项列表。在这种情况下，您应该使用数组，但因为不可能仅从这段代码来看，我没有引用它。）