我需要帮助，以从网站获取一些特定的锚点。网站具有这种结构
<li>
This has a link
<a href="#">1st link</a>
</li>
<li>
<a href="#">2nd link</a>
</li>
<li>
This also has a link
<a href="#">1st link</a>
</li>
我只想获取在li中没有文本的锚点。
使用BeautifulSoup最简单的方法是什么？
答案 0 :(得分：0)
您可以分析address: "Somewhere along PTI road, Effurun. Delta State"
name: "Effurun"
region: "South-South"
的元素以确定是否仅存在router.get('/:zipcode/:type/:rad', function (req, res) {
const rad = req.params.rad;
const zip = req.params.zipcode;
let zips = zipcodes.radius(zip, rad);
zips.push(zip);
let type;
if(req.params.type === 'bartenders') {
type = 0;
} else {
type = 1;
}
const params = {
'type': type,
'zips': zips
};
function userGroup(type, zip) {
return new Promise(resolve => {
connection.query(`SELECT * FROM bt9.users WHERE zip = ${zip} AND type = ${type} AND display = 1`, function (err, result) {
if(err) throw err;
resolve(result);
});
});
}
async function getUsers(params) {
let userList = [];
for (i = 0; i < params.zips.length; i++) {
const users = await userGroup(params.type, params.zips[i]);
for (u = 0; u < users.length; u++) {
userList.push(users[u]);
}
}
return userList;
}
function sendUsers(callback) {
getUsers(params).then( res => {
callback(null, res)
})
}
sendUsers(function(err, result) {
if(err) throw err;
res.send(result)
})
});个对象：
soup.contents输出：
soup