如何在桐乡桑 shutil 中实现文件搜索?

代码示例：

import shutil

source_path = "/path/to/source/directory"
target_path = "/path/to/target/directory"

# Create a shutil.Path object for the source and target directories
source_path_obj = shutil.Path(source_path)
target_path_obj = shutil.Path(target_path)

# Get a list of files in the source directory
files = source_path_obj.listdir()

# Search for a specific file name in the source directory
file_name = "target_file.txt"
file_path = os.path.join(source_path, file_name)

# Check if the file exists in the source directory
if file_path in files:
    # If the file is found, copy it to the target directory
    target_path_obj.join(file_path)
    print("File found and copied to target directory.")
else:
    print("File not found in source directory.")

问题：

该代码示例中如何实现文件搜索?它使用 os.path.join() 函数来拼接文件路径，但该函数可能导致路径中的特殊字符（例如空格、引号等）导致文件路径无法被正确解析。

解决方案：

为了解决路径中的特殊字符问题，可以使用 os.path.join() 函数的第二个参数 sep 来指定分隔符。默认情况下，sep 参数设置为 os.path.sep，即 /。因此，可以使用以下代码来拼接文件路径：

file_path = os.path.join(source_path, file_name, sep="/")

修改后的代码示例：

import shutil

source_path = "/path/to/source/directory"
target_path = "/path/to/target/directory"

# Create a shutil.Path object for the source and target directories
source_path_obj = shutil.Path(source_path)
target_path_obj = shutil.Path(target_path)

# Get a list of files in the source directory
files = source_path_obj.listdir()

# Search for a specific file name in the source directory
file_name = "target_file.txt"
file_path = os.path.join(source_path, file_name, sep="/")

# Check if the file exists in the source directory
if file_path in files:
    # If the file is found, copy it to the target directory
    target_path_obj.join(file_path)
    print("File found and copied to target directory.")
else:
    print("File not found in source directory.")
```