我意识到这是与one非常相似的问题。但在我的情况下,我仍然不清楚如何做到这一点。只需要一些成功回调的帮助。
这是有效的:
function getStuff(accountNumber) {
var logMessage = 'POST GetStuff';
return $http.post(GetStuff, { custId: accountNumber })
.then(log);
}
function log(response) {
logger.debug(response);
return response;
}
这就是我想要完成的事情:
function getStuff(accountNumber) {
var logMessage = 'POST GetStuff';
return $http.post(GetStuff, { custId: accountNumber })
.then(log(response, logMessage);
}
function log(response, logMessage) {
logger.debug(logMessage, response);
return response;
}
答案 0 :(得分:8)
您可以使用:
function getStuff(accountNumber) {
var logMessage = 'POST GetStuff';
return $http.post(GetStuff, { custId: accountNumber })
.then(
function success(response) {
return log(response, logMessage);
}
);
}
答案 1 :(得分:0)
看看它是否可以帮到你:
function getStuff(accountNumber) {
var logMessage = 'POST GetStuff';
var deferred = $q.defer();
return $http.post(GetStuff, { custId: accountNumber })
.success(function(response){
deferred.resolve(response);
log(response, logMessage);
}).error(function(){
deferred.reject();
})
return deferred.promise;
}