原创 JE核心 [Python脚本]Folia插件识别，便于传统核心的腐竹快速找出哪些插件不支持Folia 0.0.1

import os
from zipfile import ZipFile

import yaml

folia_supported = []
no_folia_supported = []

for filename in os.listdir('.'):
    if not "jar" in filename[-4:]:
        continue
    with ZipFile(filename, 'r') as f:
        try:
            path = f.extract("plugin.yml")
            print(f"找到Spigot插件：{filename}")
        except KeyError:
            try:
                path = f.extract("paper-plugin.yml")
                print(f"找到Paper插件：{filename}")
            except Exception as e:
                print(f"{filename}这玩意真是插件嘛？")
                raise Exception("有个jar疑似不是插件") from e
        with open(path, 'r') as plugin_yml:
            dic = yaml.load(plugin_yml, Loader=yaml.FullLoader)
            # print(dic)
            try:
                if dic["folia-supported"]:
                    print(f"{filename}支持Folia")
                    folia_supported.append(filename)
                else:
                    print(f"{filename}不支持Folia")
                    no_folia_supported.append(filename)
            except KeyError:
                print(f"{filename}不支持Folia")
                no_folia_supported.append(filename)

print("Folia支持的插件：")
for filename in folia_supported:
    print(filename)
print("\n不支持Folia的插件：")
for filename in no_folia_supported:
    print(filename)

need_output = input("是否需要输出结果？(y/n)")
if need_output == "y":
    with open("folia_supported.txt", "w") as f:
        for filename in folia_supported:
            f.write(filename + "\n")
    with open("no_folia_supported.txt", "w") as f:
        for filename in no_folia_supported:
            f.write(filename + "\n")