编写可接收命令行参数的Bash脚本统计目录文件数量

Hey there! Let's break down what's wrong with your current approach and build a reliable script that does exactly what you need.

What's Wrong with Your Original Command? #

Your ls /$@/ | wc -l has a couple of key issues:

1. Incorrect path formatting: /$@/ adds unnecessary slashes around your directory argument. If you pass an absolute path like /home/my_project, this becomes ///home/my_project/ — while some shells might tolerate this, it's messy and error-prone. Since you only need one directory parameter, use "$1" (the first command-line argument) instead of $@ (which refers to all parameters).
2. Unreliable counting: ls | wc -l counts all entries in the directory, including subdirectories, not just regular files. Even worse, if any filename contains a newline character (yes, that's allowed in Unix!), ls will split it into multiple lines, making wc -l return an incorrect count.

Reliable Script Solutions #

Here are two robust approaches to get the job done right:

Option 1: Use find (Most Reliable)

This method handles all edge cases, including weird filenames, and explicitly counts only regular files (not directories, symlinks, or special files):

#!/bin/bash

# Check if exactly one directory argument is provided
if [ $# -ne 1 ]; then
    echo "Usage: $0 <directory>"
    exit 1
fi

TARGET_DIR="$1"

# Verify the directory exists
if [ ! -d "$TARGET_DIR" ]; then
    echo "Error: '$TARGET_DIR' is not a valid directory."
    exit 1
fi

# Count regular files (non-recursive, only current directory)
file_count=$(find "$TARGET_DIR" -maxdepth 1 -type f | wc -l)

echo "$TARGET_DIR has $file_count files"

• -maxdepth 1: Ensures we don't count files in subdirectories (remove this if you want recursive counting).
• -type f: Filters only regular files, excluding directories, symlinks, and other special file types.

Option 2: Use Shell Globbing (Faster for Simple Cases)

If you're working with a directory that doesn't have filenames with newlines, this approach is more lightweight:

#!/bin/bash

if [ $# -ne 1 ]; then
    echo "Usage: $0 <directory>"
    exit 1
fi

TARGET_DIR="$1"

if [ ! -d "$TARGET_DIR" ]; then
    echo "Error: '$TARGET_DIR' is not a valid directory."
    exit 1
fi

# Enable nullglob to handle empty directories properly
shopt -s nullglob
# Collect all entries in the directory
all_entries=("$TARGET_DIR"/*)

file_count=0
# Loop through entries and count only regular files
for entry in "${all_entries[@]}"; do
    if [ -f "$entry" ]; then
        ((file_count++))
    fi
done

# Disable nullglob (optional, but clean)
shopt -u nullglob

echo "$TARGET_DIR has $file_count files"

• shopt -s nullglob: Makes the glob return an empty list if the directory has no files, instead of the literal string $TARGET_DIR/* (which would mess up the count).

How to Use the Script #

1. Save either script to a file (e.g., count_files.sh).
2. Make it executable: chmod +x count_files.sh.
3. Run it with your target directory: ./count_files.sh my_project

You'll get output exactly like you wanted: my_project has 3 files

内容的提问来源于stack exchange，提问作者user16068074