如何编写传递给 printf 的参数数量可变的函数

Question

如何修复下面的 print_error 函数：

#!/bin/bash

script_name="tmp.bash"

function print_error {
    format=""
    shift
    printf "%s: ERROR: %s\n" "$script_name" "$format" "$@"
}

print_error "Today is %s; tomorrow is %s" "$(date)" "$(date -d "+1 day")"

这样它输出：

tmp.bash: ERROR: Today is Mon Nov 22 15:10:40 PST 2021; tomorrow is Tue Nov 23 15:10:50 PST 2021

目前输出：

tmp.bash: ERROR: Today is %s; tomorrow is %s
Mon Nov 22 15:10:40 PST 2021: ERROR: Tue Nov 23 15:10:50 PST 2021

print_error 应该能够接受可变数量的参数。

Answer 1

使用中间变量（此处substr）构建功能消息（日期消息）和技术消息（错误消息）的第二个buid：

#! /bin/bash

declare -r script_name="tmp.bash"

function print_error {
    local substr=
    printf -v substr "" "${@:2}"
    printf "%s: ERROR: %s\n" "$script_name" "$substr"
}

print_error "Today is %s; tomorrow is %s" "$(date)" "$(date -d "+1 day")"

然后，您可以将功能构建和技术构建分开：

#! /bin/bash

declare -r script_name="tmp.bash"

function print_tech_error () {
    printf "%s: ERROR: %s\n" "$script_name" ""
}
function print_fct_error {
    local substr=
    printf -v substr "" "${@:2}"
    print_tech_error "${substr}"
}

print_fct_error "Today is %s; tomorrow is %s" "$(date)" "$(date -d "+1 day")"

如何编写传递给 printf 的参数数量可变的函数

How to write a function with a variable number of arguments that are passed to printf

bash

printf

variadic-functions