Shell脚本自定义函数传参报错：日期解析变量替换问题排查

Hey there! Let's break down what's causing that substitution error and get your date-to-weekday function working smoothly.

The Root Cause of the Error #

That bad substitution message is happening because you’re trying to nest variable references incorrectly in your function. When you write ${$1:0:2}, the shell gets confused—$1 already refers to the first argument passed to your function, and you don’t need an extra $ inside the curly braces. The correct syntax for substring extraction is ${1:0:2} (no extra dollar sign).

How to Fix Your Date Parsing Function #

Let’s start with a corrected version of your function. Assuming your goal is to take a date in MM/DD/YYYY format and return the corresponding weekday, here’s a robust approach using the date command (works in Bash, Zsh, and most POSIX-compliant shells):

#!/bin/bash

# Function to convert MM/DD/YYYY date to weekday name
get_weekday() {
    # Extract month, day, year from the input date (format: MM/DD/YYYY)
    local date_str="$1"
    local month="${date_str:0:2}"
    local day="${date_str:3:2}"
    local year="${date_str:6:4}"

    # Convert to YYYY-MM-DD format (required for date command's -d flag)
    local formatted_date="${year}-${month}-${day}"

    # Use date command to get the full weekday name
    date -d "$formatted_date" +"%A"
}

# Check if exactly two date arguments are provided
if [ $# -ne 2 ]; then
    echo "Usage: $0 MM/DD/YYYY MM/DD/YYYY"
    exit 1
fi

# Get weekdays for both input dates
weekday1=$(get_weekday "$1")
weekday2=$(get_weekday "$2")

# Output the results
echo "Date $1 is a $weekday1"
echo "Date $2 is a $weekday2"

Key Fixes Explained #

1. Removed the extra $ in substring extraction: Changed ${$1:0:2} to ${date_str:0:2} (or directly ${1:0:2} if you skip the date_str variable). Using a local variable like date_str just makes the code more readable.
2. Added input validation: The script now checks if exactly two arguments are passed, which helps avoid unexpected errors from missing or extra inputs.
3. Used date -d for reliable date parsing: Instead of manually calculating weekdays (which gets messy with leap years and calendar edge cases), leveraging the date command is far more accurate and less error-prone.

Testing the Script #

If you run:

./datematch.sh 01/03/1984 06/12/2008

You should get output like:

Date 01/03/1984 is a Tuesday
Date 06/12/2008 is a Thursday

Notes for Portability #

If you’re using a system where date -d isn’t available (like macOS), you can use gdate (from the GNU coreutils package) instead, or adjust the date line to work with macOS’s native date command:

# For macOS, replace the date line with:
date -j -f "%m/%d/%Y" "$date_str" +"%A"

内容的提问来源于stack exchange，提问作者Aidan Low