javascript 如何从 flickr API 获取正确的 JSON 对象
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how to get correct JSON object from flickr API
提问by ankitr
I used flickr photo search method to retrieve public photos. when I run it with jquery, it works fine, I get the json object in correct form.
我使用 flickr 照片搜索方法来检索公共照片。当我使用 jquery 运行它时，它工作正常，我以正确的形式获取了 json 对象。
{
"photos": {
"photo": [
{
.........
}
]
},
"stat": "ok"
}
But when I use it with AngularJs, I got the same object with a prefix jsonFlickrApi
但是当我将它与 AngularJs 一起使用时，我得到了带有前缀 jsonFlickrApi 的相同对象
jsonFlickrApi({
"photos": {
"photo": [
{
......
}
]
},
"stat": "ok"
})
what I used in AngularJs is:
我在 AngularJs 中使用的是：
myApp.service('dataService', function($http) {
delete $http.defaults.headers.common['X-Requested-With'];
this.flickrPhotoSearch = function() {
return $http({
method: 'GET',
url: 'https://api.flickr.com/services/rest/?method=flickr.photos.search&api_key=3f807259749363aaa29c2fa93945&tags=india&format=json&callback=?',
dataType: 'json'
});
}
});
Please tell me how can I convert the second JSON to the first one.
Is there anything I can do in $httpcall or have to alter JSON object.
请告诉我如何将第二个 JSON 转换为第一个。有什么我可以在$http通话中做或必须更改 JSON 对象的吗？
回答by ankitr
There was problem in the url. This Url works for me.
网址有问题。这个网址对我有用。
https://api.flickr.com/services/rest/?method=flickr.people.getPublicPhotos&api_key=3f807259749363aaa29c712fa93945&user_id=61495424@N00&format=json&nojsoncallback=?
回答by maxwellnewage
You have to add nojsoncallback=1 to the url, like this
你必须在 url 中添加 nojsoncallback=1，就像这样
https://api.flickr.com/services/rest/?method=flickr.photos.getRecent&api_key=XXX&format=json&nojsoncallback=1
And that remove the jsonFlickrApi function on response.
并删除响应中的 jsonFlickrApi 函数。
回答by Alan Scarpa
The above answer which says to add nojsoncallback=1is correct, but I'd like to provide more context and reasoning behind it.
上面说添加的答案nojsoncallback=1是正确的，但我想提供更多的上下文和背后的推理。
If you take a look at the documentation (which is poorly formatted in my opinion): https://www.flickr.com/services/api/response.json.htmland scroll all the way to the bottom you will find, tucked away:
如果你看一下文档（我认为它的格式很差）：https: //www.flickr.com/services/api/response.json.html并一直滚动到底部你会发现，隐藏离开：
Callback Function
If you just want the raw JSON, with no function wrapper, add the parameter nojsoncallback with a value of 1 to your request.
To define your own callback function name, add the parameter jsoncallback with your desired name as the value.
nojsoncallback=1 -> {...}
jsoncallback=wooYay -> wooYay({...});
The key part being: If you just want the raw JSON, with no function wrapper, add the parameter nojsoncallback with a value of 1 to your request
关键部分是：如果您只想要原始 JSON，没有函数包装器，请将值为 1 的参数 nojsoncallback 添加到您的请求中
Therefore, add this to your request and it will remove jsonFlickrAPIwrapper that you are seeing in your response: nojsoncallback=1
因此，将此添加到您的请求中，它将删除jsonFlickrAPI您在响应中看到的包装器：nojsoncallback=1
回答by Fabio Milheiro
You are getting a JSONPresponse (notice the P) which is basically a function call with an argument that is the response you are expecting.
你得到一个JSONP响应（注意 P），它基本上是一个函数调用，参数是你期望的响应。
Easy solution for you: Create a function named jsonFlickrApiwith a parameter responseand you can do your handing in there.
为您提供简单的解决方案：创建一个以jsonFlickrApi参数命名的函数response，您可以在那里进行处理。
At this point, I am not sure if you can define your function inside your service but try it. If not, you can define it outside.
在这一点上，我不确定您是否可以在您的服务中定义您的函数，但请尝试一下。如果没有，你可以在外面定义它。
Alternatively, try using $http.get function instead. Were you specifying the return accepted by the client in your jQuery original code?
或者，尝试改用 $http.get 函数。您是否在 jQuery 原始代码中指定了客户端接受的返回？
EDIT
编辑
Try this setting the data type of the request using $http or $http.getwithout forgetting to specify the data property in the settings object.
尝试使用 $http 或 $http.get 设置请求的数据类型，而不要忘记在设置对象中指定数据属性。
$http({
url: 'http://localhost:8080/example/teste',
dataType: 'json',
method: 'POST',
data: '',
headers: {
"Content-Type": "application/json"
}
}).success(function(response){
$scope.response = response;
}).error(function(error){
$scope.error = error;
});
