如何让Python Discord Bot接收多词输入并搜索JSON数据匹配ID

我有如下JSON数据片段：

{"draw":0,"recordsTotal":1261,"recordsFiltered":1261,"data":[{"id":"1","ClassID":null,"Name":"Default Sword","Description":"Default weapon for Legends!","Type":"Sword","Element":"None","File":"items/swords/GasparianBlade.swf","Link":"GasparianBlade","Icon":"iwsword","Equipment":"Weapon","Level":"1","DPS":"25","Range":"10","Rarity":"3","Quantity":"1","Stack":"1","Cost":"0","Coins":"0","Crystal":"0","Sell":"1","Market":"1","Temporary":"0","Upgrade":"0","Founder":"0","Vip":"0","Staff":"0","EnhID":"1","FactionID":null,"ReqReputation":"0","ReqClassID":null,"ReqClassPoints":"0","ReqQuests":"","QuestStringIndex":"-1","QuestStringValue":"0","Meta":null,"Color":"ffffff"},{"id":"2","ClassID":"1","Name":"Vagabond","Description":"A Vagabond that doesn't have much of experience, we must keep moving and learn more about our Adventure!","Type":"Class","Element":"None","File":"CyberPeasant.swf","Link":"CyberPeasant","Icon":"cclass","Equipment":"ar","Level":"1","DPS":"25","Range":"50","Rarity":"3","Quantity":"1","Stack":"1","Cost":"0","Coins":"0","Crystal":"0","Sell":"1","Market":"1","Temporary":"0","Upgrade":"0","Founder":"0","Vip":"0","Staff":"0","EnhID":"1","FactionID":null,"ReqReputation":"0","ReqClassID":null,"ReqClassPoints":"0","ReqQuests":"","QuestStringIndex":"-1","QuestStringValue":"0","Meta":null,"Color":"ffffff"},{"id":"3","ClassID":null,"Name":"Balrog Blade (Beta)","Description":...

我开发了一个Python Discord Bot，预期用户输入“Default Sword”时返回对应ID：1，但Bot只识别第一个单词“Default”，提示“ID not found”。现有代码如下：

@bot.command()
async def item(ctx, item):

 data = "* data *"

 info_dict = json.loads(data)

 # Taking user input
 user_input_name = item

 # Find and print the ID if Name matches
 for entry in info_dict['data']:
     if entry.get("Name") == user_input_name:
         await ctx.send(f"Name : {item} - ID : {entry['id']}")
         break
 else:
     await ctx.send("ID not found. try writing full name.")

请问如何修改脚本，让它能接收完整的多词输入并匹配JSON中的名称，返回对应ID？

问题根源是Discord.py默认的命令参数解析逻辑：位置参数只会捕获第一个空格前的内容。要支持多词输入，有两种实用方案：

方案1：用*接收所有剩余参数 #

将命令参数改为*item，此时item会被解析为包含所有输入单词的元组，再拼接成完整字符串即可：

@bot.command()
async def item(ctx, *item):
    # 把元组拼接成完整名称
    user_input_name = " ".join(item)
    
    data = "* data *"
    info_dict = json.loads(data)
    
    for entry in info_dict['data']:
        if entry.get("Name") == user_input_name:
            await ctx.send(f"Name : {user_input_name} - ID : {entry['id']}")
            break
    else:
        await ctx.send("ID not found. try writing full name.")

方案2：用str转换器强制接收字符串（需引号包裹输入） #

通过str转换器明确指定参数为字符串类型，但用户需要用引号包裹多词内容（比如输入!item "Default Sword"）：

@bot.command()
async def item(ctx, item: str):
    user_input_name = item
    
    data = "* data *"
    info_dict = json.loads(data)
    
    for entry in info_dict['data']:
        if entry.get("Name") == user_input_name:
            await ctx.send(f"Name : {user_input_name} - ID : {entry['id']}")
            break
    else:
        await ctx.send("ID not found. try writing full name.")

优先推荐方案1，无需用户额外操作引号，使用体验更流畅。

内容的提问来源于stack exchange，提问作者Daemon