I have this method in my controller that get a json file from my API.
我在我的控制器中有这个方法从我的API获取一个json文件。
$scope.get = function (code) {
api.get(code)
.then(onJsonPart, onError);
};
Doing this:
这样做:
function onJsonPart(json) {
console.log(json);
}
I can print out the JSON I received, but I want to pass multiple variables to the onJsonPart method. Something like:
我可以打印出我收到的JSON,但我想将多个变量传递给onJsonPart方法。就像是:
$scope.get = function (code) {
api.get(code)
.then(onJsonPart(code, data), onError);
};
Here I get ERROR: data is not defined obviously, question is how do I define the data variable with the data received.
这里我得到错误:数据没有明确定义,问题是我如何用收到的数据定义数据变量。
and then have two parameters for my onJsonPart function like this:
然后我的onJsonPart函数有两个参数,如下所示:
function onJsonPart(code, json) {
console.log('Code: ' + code);
console.log('json:');
console.log(json);
}
3 个解决方案
#1
2
The code below will pass code
and the API's answer res
to onJsonPart()
:
下面的代码将传递代码,API的答案将传递给onJsonPart():
$scope.get = function (code) {
api.get(code)
.then(
function(res) {
//...some place for logic
onJsonPart(code, res);
},
function (err) {
//...some place for logic
onError(err));
}
);
};
#2
0
Try
尝试
api.get(code)
.then(function (json) {
onJsonPart(code, data);
}, onError);
#3
0
You could do this:
你可以这样做:
$scope.get = function (code) {
api.get(code)
.then(function(data) {
onJsonPart(code, data)...
}
};
#1
2
The code below will pass code
and the API's answer res
to onJsonPart()
:
下面的代码将传递代码,API的答案将传递给onJsonPart():
$scope.get = function (code) {
api.get(code)
.then(
function(res) {
//...some place for logic
onJsonPart(code, res);
},
function (err) {
//...some place for logic
onError(err));
}
);
};
#2
0
Try
尝试
api.get(code)
.then(function (json) {
onJsonPart(code, data);
}, onError);
#3
0
You could do this:
你可以这样做:
$scope.get = function (code) {
api.get(code)
.then(function(data) {
onJsonPart(code, data)...
}
};