python - 在python中更改json文件中的值

Question

所以我在 vs 代码 python 中有一个 Json 文件，如下所示:

{
    "matches_details":[
            { 
            "MatchID": "Liverpool-Manchester United",
            "Score": "5-0",
             "Date": "17/12/2020",
              "Time": "16:00",
               "Yellow cards": "2",
                "Red cards": "1",
                 "State": "FT"},
             { 
            "MatchID": "Crystal Palace-Chelsea",
                "Score": "3-2",
                "Date": "20/1/2021",
                "Time": "12:00",
                    "Yellow cards": "1",
                    "Red cards": "0",
                    "State": "HT"
                    }]
}

我想更改比赛的“分数”，我该怎么做？
这是我尝试过的，但无法执行:

Time_client='12:00'
    Date_client='20/1/2021'
    Match_client='Crystal Palace-Chelsea'
    Score_client='10-0'
    f = open('datngu.json','r+')
    data = json.load(f)
    f.close()
    for item in data['matches_details']:
        if item['Name']==Match_client:
            if item['Time']==Time_client:
                if item['Date']==Date_client:
                    item['Score']=item['Score'].replace(Score_client,item['MatchID'])
    
    f2=open('datngu.json','r+')
    json.dump(data,f2)
    f2.close()

回溯(最近一次调用最后一次):
文件“c:\Users\DELL.vscode\extensions\ms-python.python-2021.2.633441544\json_reading.py”，第 40 行，在
更新分数()
文件“c:\Users\DELL.vscode\extensions\ms-python.python-2021.2.633441544\json_reading.py”，第 31 行，在 updateScore
如果 item['Name']==Match_client:
key 错误:'名称'

Answer 1

我想你的意思是输入 MatchID而不是 name因为没有 name JSON 中的属性:

Time_client='12:00'
    Date_client='20/1/2021'
    Match_client='Crystal Palace-Chelsea'
    Score_client='10-0'
    f = open('datngu.json','r+')
    data = json.load(f)
    f.close()
    for item in data['matches_details']:
        if item['MatchID']==Match_client:
            if item['Time']==Time_client:
                if item['Date']==Date_client:
                    item['Score']=item['Score'].replace(Score_client,item['MatchID'])
    
    f2=open('datngu.json','r+')
    json.dump(data,f2)
    f2.close()