本文介绍了将参数传递给promisejs $ q中的promise成功回调的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我意识到这是一个与此非常相似的问题。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。
I realize this is a very similar question to this one. But I'm still unclear on how to do it in my situation. Just need some help with a successful callback.
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:
This is what I want to accomplish:
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;
}
推荐答案
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);
}
);
}
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