我需要有关TinyPaste API的帮助。 我做到了:
var profiles = {
firstProfile: {
name: "DemostanisYt",
avatar: "https://i.imgur.com/p3kb3ie.png"
}
}
profiles = JSON.stringify(profiles)
var response
$.ajax({
url: "http://www.penyacom.org/api/v1/paste.php",
type: "POST",
data: {
code: JSON.stringify(profiles)
},
success: function(e) {
response = JSON.parse(e)
$.ajax({
url: "http://" + response.raw_link,
type: "GET",
crossOrigin: "anonymous",
success: function(e) {
console.log(e)
}
})
}
})
一个问题:它对代码做了一些处理。而不是像这样:
{"firstProfile":{"name":"DemostanisYt","avatar":"https://i.imgur.com/p3kb3ie.png"}}
它是这样做的:
"{\"firstProfile\":{\"name\":\"DemostanisYt\",\"avatar\":\"https://i.imgur.com/p3kb3ie.png\"}}"
有人可以帮助我吗?
答案 0 :(得分:0)
您正在双重配置文件。声明配置文件时,应该可以删除JSON.stringify
函数,或者在ajax请求中发送数据时也可以删除它:
var profiles = {
firstProfile: {
name: "DemostanisYt",
avatar: "https://i.imgur.com/p3kb3ie.png"
}
}
profiles = JSON.stringify(profiles) // <--- You're stringifying the object here
var response
$.ajax({
url: "http://www.penyacom.org/api/v1/paste.php",
type: "POST",
data: {
code: JSON.stringify(profiles) // <--- Then restringifying it again here
},
success: function(e) {
response = JSON.parse(e)
$.ajax({
url: "http://" + response.raw_link,
type: "GET",
crossOrigin: "anonymous",
success: function(e) {
console.log(e)
}
})
}
})