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python - 从 Python 中的图像键获取 URL

问题描述

我需要获得 'image' ：值，但我尝试过的没有任何工作。

我试图转换为dict，但没有成功。尝试过json，但也没有成功。

[
    {
        "image": "https://scontent.cdninstagram.com/vp/.../61070103916_8222623480192096121_n.jpg?_nc_ht=scontent.cdninstagram.com",
        "thumbnail": "https://scontent.cdninstagram.com/vp/.../61070103916_8222623480192096121_n.jpg?_nc_ht=scontent.cdninstagram.com",
        "ts": "8382423"
    },
    {
        "image": "https://scontent.cdninstagram.com/..../7366801460_3370303770835888477_n.jpg?_nc_ht=scontent.cdninstagram.com",
        "thumbnail": "https://scontent.cdninstagram.com/.../7366801460_3370303770835888477_n.jpg?_nc_ht=scontent.cdninstagram.com",
        "ts": "8283861"
    }
]

预计输出如下：

https://scontent.cdninstagram.com/vp/.../61070103916_8222623480192096121_n.jpg?_nc_ht=scontent.cdninstagram.com
https://scontent.cdninstagram.com/..../7366801460_3370303770835888477_n.jpg?_nc_ht=scontent.cdninstagram.com```

标签： python

解决方案

您可以将整个内容放入列表中，每个列表项都将转换为字典，然后您可以遍历它并从每个字典中获取“图像”。

img_list = [
        {
            "image": "https://scontent.cdninstagram.com/vp/.../61070103916_8222623480192096121_n.jpg?_nc_ht=scontent.cdninstagram.com",
            "thumbnail": "https://scontent.cdninstagram.com/vp/.../61070103916_8222623480192096121_n.jpg?_nc_ht=scontent.cdninstagram.com",
            "ts": "8382423"
        },
        {
            "image": "https://scontent.cdninstagram.com/..../7366801460_3370303770835888477_n.jpg?_nc_ht=scontent.cdninstagram.com",
            "thumbnail": "https://scontent.cdninstagram.com/.../7366801460_3370303770835888477_n.jpg?_nc_ht=scontent.cdninstagram.com",
            "ts": "8283861"
        }
    ]

for i in img_list:
    print(i["image"])

输出：

https://scontent.cdninstagram.com/vp/.../61070103916_8222623480192096121_n.jpg?_nc_ht=scontent.cdninstagram.com
https://scontent.cdninstagram.com/..../7366801460_3370303770835888477_n.jpg?_nc_ht=scontent.cdninstagram.com

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