我似乎无法让这个简单的Parse查询在我的云代码then()中工作，但它在此之外工作但是当我将代码置于其中时，函数没有任何反应。变量现在只是测试方面的占位符，但是我从开始时开始Parse时会得到默认的TestObject类，但由于某种原因它只是不返回任何内容。
以下是我目前正在使用的完整功能。
// Function which will get the data from all the links passed into the function
Parse.Cloud.define("myNews", function (request, response) {
var promises = _.map(import_io_keys, function (news_api_key) {
return Parse.Cloud.httpRequest({
method: 'GET',
url: "https://api.import.io/store/connector/" + news_api_key + "/_query?input=webpage/url:https%3A%2F%2Fwww.designernews.co%2Fnew&&_apikey=xxxxxxxxxxxxxxxxxx",
headers: {
'Content-Type': 'application/json;charset=utf-8'
}
}).then(function (httpResponse) {
result = JSON.parse(httpResponse.text);
var success = false;
var news_icon = "";
var news_source_name = "";
var query = new Parse.Query("TestObject");
query.find({
success: function(results) {
success = true;
news_icon = results[0].get("foo");
news_source_name = results[0].get("foo");
response.success("done" + news_icon);
},
error: function() {
success = false;
response.error("Query lookup failed");
}
});
for (var story in result.results) {
if(story.length > 0){
if (story["article_link/_text"] !== "" && story["article_link"] !== "" && story["article_time"] !== "") {
if(success){
// Do the stuff later
}
}
}
}
});
});
Parse.Promise.when(promises).then(function () {
console.log("Got all the stories");
response.success(newsJsonData);
}, function () {
response.error("No stories");
console.log("API KEY IS: " + request.params.keys);
});
});