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我有一个有效的 Google 索引 API 代码。代码如下：

from oauth2client.service_account import ServiceAccountCredentials 
import httplib2 

SCOPES = ["https://www.googleapis.com/auth/indexing"] 
ENDPOINT = "https://indexing.googleapis.com/v3/urlNotifications:publish" 
# service_account_file.json is the private key
# that you created for your service account. 
JSON_KEY_FILE = "/content/astute-tractor-329613-1baed60ec1c0.json" 
credentials = ServiceAccountCredentials.from_json_keyfile_name(JSON_KEY_FILE, scopes=SCOPES) 

http = credentials.authorize(httplib2.Http()) 

content = """{ 
"url": "https://sitename.com/index.php", 
"type": "URL_UPDATED" 
}""" 

response, content = http.request(ENDPOINT, method="POST", body=content)

if response.status == 200:
    print(f'The submission was successful. Google reported a {response.status} response code.')
else:
    print(f'The submission was not successful. Google reported a {response.status} response code, instead of 200.')

我可以用它一个一个地添加 URL，但我想给它一个CSV文件，以便它会逐行读取它并将其发送给 Google。

我创建了以下代码：

from oauth2client.service_account import ServiceAccountCredentials 
import httplib2 
import csv
 

SCOPES = ["https://www.googleapis.com/auth/indexing"] 
ENDPOINT = "https://indexing.googleapis.com/v3/urlNotifications:publish" 

# service_account_file.json is the private key
# that you created for your service account. 
JSON_KEY_FILE = "/content/astute-tractor-329613-1baed60ec1c0.json" 
credentials = ServiceAccountCredentials.from_json_keyfile_name(JSON_KEY_FILE, scopes=SCOPES) 

http = credentials.authorize(httplib2.Http()) 

with open('/content/indekseerida-1.csv') as csvfile:
    readCSV = csv.reader(csvfile, delimiter=',')
    for row in readCSV:
        content = """{ 
        "url": row[0], 
        "type": "URL_UPDATED" 
        }""" 

        response, content = http.request(ENDPOINT, method="POST", body=content)

        if response.status == 200:
            print(f'The submission was successful. Google reported a {response.status} response code.')
        else:
            print(f'The submission was not successful. Google reported a {response.status} response code, instead of 200.')

这只会给我返回错误响应代码 400。我对编码很陌生，所以不要对我苛刻:)

4

1 回答 1

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在你的 for 循环中放置一个print(content)语句，我想你会发现它不包含你认为它所做的事情。整个字符串只是您在此处键入的文字内容。

你需要类似的东西：

        content = f"""{{ 
        "url": {row[0]}, 
        "type": "URL_UPDATED" 
        }}"""

f"""创建一个字符串插值，其中内部的东西被{}评估。然后，您需要{{并}}获得文字大括号。

于 2021-10-20T16:46:37.880 回答