Question

此代码有效：

#!/usr/bin/env python3

import urllib.request, json

url = urllib.request.urlopen('https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.xchange%20where%20pair%20in%20(%22DKKNOK%2CEURNOK%2CGBPNOK%2CISKNOK%2CNOKNOK%2CPLNNOK%2CSEKNOK%22)&format=json&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys')

data = json.loads(url.read().decode(url.info().get_content_charset('utf-8')))

print(data['query']['results']['rate'][:])

它会打印出data['query']['results']['rate']的所有7个元素。

所以我认为这段代码也应该有效：

#!/usr/bin/env python3

import urllib.request, json

url = urllib.request.urlopen('https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.xchange%20where%20pair%20in%20(%22DKKNOK%2CEURNOK%2CGBPNOK%2CISKNOK%2CNOKNOK%2CPLNNOK%2CSEKNOK%22)&format=json&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys')

data = json.loads(url.read().decode(url.info().get_content_charset('utf-8')))

for d in data['query']['results']['rate'][:]
    print(d)

使用for循环并打印出data['query']['results']['rate']中的每个元素。

但是这不起作用并且出错。

如何在Python中迭代json元素？

Answer 1

[:]无需迭代数据：

for d in data['query']['results']['rate']:
    print(d)

..应该有用。您错过了:。

您还可以直接对请求的响应使用.json()方法将JSON解码为python结构。

Answer 2

试试这个

for d in data['query']['results']['rate']：
    print(d)

而不是

for d in data['query']['results']['rate'][:]
    print(d)                       #here    ^ missed :

如何在Python中使用for迭代json元素

2 个答案: