Angularjs 可能未处理的拒绝,使用$q.reject
我正在用Angular和jasmine,得到了错误:Angularjs 可能未处理的拒绝,使用$q.reject,angularjs,jasmine,es6-promise,Angularjs,Jasmine,Es6 Promise,我正在用Angular和jasmine,得到了错误: createAccount() { // Return some ES6 Promise/async call with return value from test. return this.request({method: 'POST', body: {}}) .catch(error => { if (error.flaky) { // Handle a specific
createAccount() {
// Return some ES6 Promise/async call with return value from test.
return this.request({method: 'POST', body: {}})
.catch(error => {
if (error.flaky) {
// Handle a specific error with another call
return this.createDifferentAccount();
}
console.log('Passed the specific case');
// Generically fail for other errors.
this.$q_.reject(error);
});
}
可能未处理的拒绝:{“body2”:{}}抛出
引发错误的我的代码:
createAccount() {
// Return some ES6 Promise/async call with return value from test.
return this.request({method: 'POST', body: {}})
.catch(error => {
if (error.flaky) {
// Handle a specific error with another call
return this.createDifferentAccount();
}
console.log('Passed the specific case');
// Generically fail for other errors.
this.$q_.reject(error);
});
}
我的jasmine测试用例(失败):
如何处理此错误/我做错了什么?您只是拒绝错误,而不是返回拒绝 解决办法很简单,改变一下
this.$q.拒绝(错误)代码>
到
返回此项。$q\uu.拒绝(错误)代码>
请记住始终从.then()或.catch()块中返回值强>