承诺：忽略捕获并返回链

Question

是否可以忽略捕获并返回到链？

promiseA()        // <-- fails with 'missing' reason
  .then(promiseB) // <-- these are not going to run 
  .then(promiseC)
  .catch(function(error, ignore){
     if(error.type == 'missing'){
        ignore()  // <-- ignore the catch and run promiseB and promiseC
     }
  })

这样的事情可能吗？

Answer 1

这是同步类比：

try {
  actionA(); // throws
  actionB(); // skipped
  actionC(); // skipped
} catch (e) {
  // can't resume
}

但你想要：

try{
   try {
      actionA(); // throws
   } catch (e) {
      if(error.type !== 'missing'){
         throw error; // abort chain
      }
      // keep going
  }
  actionB(); // executes normally if error type was 'missing'
  actionC();
} catch (e) {
  // handle errors which are not of type 'missing'.
}

这是承诺版本：

asyncActionA()        // <-- fails with 'missing' reason
   .catch(error => {
      if(error.type !== 'missing'){
         throw error; // abort chain. throw is implicit Promise.reject(error); 
      }
      // keep going, implicit: return Promise.resolve();
   })
   .asyncActionB(promiseB) // <-- runs
   .asyncActionC(promiseC)
   .catch(err => {
      // handle errors which are not of type 'missing'.
   });

Answer 2

如果你需要忽略promiseA中的所有错误，你可以这样做：

promiseA()  
  .catch(function(error){
      //just do nothing, returns promise resolved with undefined
  })  
  .then(promiseB)
  .then(promiseC) 

如果你只需要在

error.type == 'missing'

时运行promiseB，你可以这样做：

promiseA()       
  .catch(function(error, ignore){
     if(error.type == 'missing'){
        return promiseB.then(promiseC)  
     }
  })