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[英]Is there a way to parse out HTML in a response from requests.get()?
[英]requests.get and parse response to json
作為標題,我想要“requests.get 並解析對 json 的響應”。 以下是我嘗試執行此操作的方法
requests.get(url = "https://ac.dict.naver.com/linedictweb/ac?q=%E6%B5%B7&st=010&r_lt=000&q_enc=UTF-8&r_format=json&r_enc=UTF-8&_t=1599558293560&_callback=jQuery111104358648061767311_1599558261058&_=1599558261060",verify = True).json()
=>Result: json.decoder.JSONDecodeError: Expecting value: line 1 column 1 (char 0)
2.另一種方式
r = requests.get(url = "https://ac.dict.naver.com/linedictweb/ac?q=%E6%B5%B7&st=010&r_lt=000&q_enc=UTF-8&r_format=json&r_enc=UTF-8&_t=1599558293560&_callback=jQuery111104358648061767311_1599558261058&_=1599558261060",verify = True)
json.loads(r.content.decode('utf-8'))
=> RESULT: json.decoder.JSONDecodeError: Expecting value: line 1 column 1 (char 0)
數據 Json在這里:
幫我解決這個問題。 謝謝
jquery 回調正在包裝輸出。 去掉 url 中的回調參數,您將獲得有效的 json:
r = requests.get(url = "https://ac.dict.naver.com/linedictweb/ac?q=%E6%B5%B7&st=010&r_lt=000&q_enc=UTF-8&r_format=json&r_enc=UTF-8&_t=1599558293560",verify = True)
r.json()
{
"query": [
"\u6d77"
],
"items": [
[],
[
[
[
"zh"
],
[
"597527adbb7248c3ae80d60b9acd7160"
],
[
"\u6d77"
],
[
" ocean; extra large"
],
...
}
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