def check_file(x): #x = 根文件夹路径
    for d in next(os.walk(x))[1]: #如果我理解正确，[1] 将是项目文件夹
        for root, directories, files in os.walk(x):
            for name in files:
                if "file.xml" not in name:
                    print("找到")
                else:
                    print("文件不存在")

for name in files:
    if "file.xml" in name:
        print("找到")
    else:
        pass

import glob
def check_file(x):  # 其中 x 是搜索的根目录
    files = glob.glob('**/file.xml', root_dir=x, recursive=True)
    if files:
        print(f"找到 {len(files)} 个匹配的文件")
    else:
        print("未找到匹配的文件")

#!/usr/bin/env python
import os
import sys
def search_file(root_dir, base_name):
    found = 0
    for root, dirs, files in os.walk(root_dir):
        if base_name in files:
            print("Found: {:s}".format(os.path.join(root, base_name)))
            found += 1
    if not found:
        print("Not found")
# @TODO - cfati: Only care if file is found once
def search_file_once(root_dir, base_name):
    for root, dirs, files in os.walk(root_dir):
        if base_name in files:
            print("Found: {:s}".format(os.path.join(root, base_name)))
            break
    else:
        print("Not found")
def main(*argv):
    root = os.path.dirname(os.path.abspath(__file__))
    files = (
        "once.xml",
        "multiple.xml",
        "notpresent.xml",
    )
    for file in files:
        print("\n在 {:s} 中递归搜索 {:s}".format(root, file))
        search_file(root, file)
if __name__ == "__main__":
    print("Python {:s} {:03d} 位于 {:s}\n".format(" ".join(elem.strip() for elem in sys.version.split("\n")),
                                                64 if sys.maxsize > 0x100000000 else 32, sys.platform))
    rc = main(*sys.argv[1:])
    print("\n完成。\n")
    sys.exit(rc)

[cfati@CFATI-5510-0:e:\Work\Dev\StackExchange\StackOverflow\q076383189]> sopr.bat
### Set shorter prompt to better fit when pasted in StackOverflow (or other) pages ###
[prompt]> tree /a /f
Folder PATH listing for volume SSD0-WORK
Volume serial number is AE9E-72AC
E:.
|   code00.py
|
\---dir0
    +---dir00
    +---dir01
    |       multiple.xml
    |       once.xml
    |
    \---dir02
        \---dir020
                multiple.xml
[prompt]>
[prompt]> "e:\Work\Dev\VEnvs\py_pc064_03.10_test0\Scripts\python.exe" ./code00.py
Python 3.10.9 (tags/v3.10.9:1dd9be6, Dec  6 2022, 20:01:21) [MSC v.1934 64 bit (AMD64)] 064 位于 win32
在 e:\Work\Dev\StackExchange\StackOverflow\q076383189 中递归搜索 once.xml
Found: e:\Work\Dev\StackExchange\StackOverflow\q076383189\dir0\dir01\once.xml
在 e:\Work\Dev\StackExchange\StackOverflow\q076383189 中递归搜索 multiple.xml
Found: e:\Work\Dev\StackExchange\StackOverflow\q076383189\dir0\dir01\multiple.xml
Found: e:\Work\Dev\StackExchange\StackOverflow\q076383189\dir0\dir02\dir020\multiple.xml
在 e:\Work\Dev\StackExchange\StackOverflow\q076383189 中递归搜索 notpresent.xml
Not found
完成。

我以前写过这样的函数，以及其他一些函数，想要提供它们以供参考，其中一些可能需要最小或没有修改就可以用于您的情况。
## 查找所有匹配项（不仅仅是一个）：
## 示例用法：findAll（'*.txt'，'/path/to/dir'）
def findAll(name, path):
    result = []
    for root, dirs, files in os.walk(path):
        if name in files:
            result.append(os.path.join(root, name))
    return result
## 一个持续查找，直到找到所有目标文件的函数）
def findProjectFiles(Folder, targetFiles):
    import os
    os.chdir(Folder)
    filesFound = []
    while len(targetFiles) > len(filesFound):
        for root, dirs, files in os.walk(Folder):
            for f in files:
                current = os.path.join(Folder, f)
                if f in TargetFiles:
                    filesFound.append(f)
            for d in dirs:
                Folder = os.path.join(Folder, d)
            break;
    filePaths = os.path.abspath(filePaths)
    return filePaths
# 在文件夹中查找所有文件路径：
def findPaths(name, path):
    import os
    for root, dirs, files in os.walk(path):
        if name in files:
            return os.path.join(root, name)
## 可以轻松搜索返回的对象以查找您想要找到的字符串
## 类似，但这将匹配模式（即不必是精确的文件名匹配）。
import os, fnmatch
def findMatch(pattern, path):
    result = []
    for root, dirs, files in os.walk(path):
        for name in files:
            if fnmatch.fnmatch(name, pattern):
                result.append(os.path.join(root, name))
    return result