NodeJS向对象数组返回响应输出
NodeJS向对象数组返回响应输出
提问于 2018-02-25 19:33:05
我在服务器端有一个从数据库返回结果的函数。下面是函数
const sql =
`SELECT addresses
FROM Addresses
WHERE JSON_EXTRACT(addresses, '$.address') LIKE ? `
const values = ['"'+beginning+'%']
const query = connection.query(sql, values, (error, results, fields) => {
connection.end()
if (error) {
return reject(error)
}
resolve(results)
})
})
.then((results) => {
return {
type: 'address',
attributes: results
}
})其输出如下所示
{
"data": {
"type": "address",
"attributes": [
{
"addresses": "{\"address\": \"5 street, auburn, nsw 2144\", \"display\": \"5-7 Mary Street, AUBURN, NSW 2144\"}"
},
{
"addresses": "{\"address\": \"5 street, lidcombe, nsw 2141\", \"display\": \"5 Street, LIDCOMBE, NSW 2141\"}"
}
]
}
}我是JS的新手，我需要按如下方式进行上述输出
{
"data": [
{
"type": "address",
"id": "5-13 aubrey street, granville, nsw 2142",
"attributes": {
"address": "5-13 aubrey street, granville, nsw 2142",
"display": "5-13 Aubrey Street, GRANVILLE, NSW 2142"
}
},
{
"type": "address",
"id": "5 street, lidcombe, nsw 2141",
"attributes": {
"address": "5 street, lidcombe, nsw 2141",
"display": "5 Street, LIDCOMBE, NSW 2141"
}
},
{
"type": "address",
"id": "5-7 mary street, auburn, nsw 2144",
"attributes": {
"address": "5-7 mary street, auburn, nsw 2144",
"display": "5-7 Mary Street, AUBURN, NSW 2144"
}
}
}
]
}有没有人可以帮我修改函数的.then((results) )来实现这一点。
在客户端，我在一个函数中获取结果，这就是我返回结果的方式
return $http.get(`${API_URL}/addresses/?beginning=${searchTerm || ''}`)
.then((result) => {
// Only want to return the address string
return result.data.data((address) => address.attributes.display)
})但是我发现result.data.data不是一个函数
回答 1
Stack Overflow用户
回答已采纳
发布于 2018-02-25 19:37:34
您可以将返回值更改为
return {
type: 'address',
attributes: results.map(result => JSON.parse(result.addresses))
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/48972923
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