Ausnahmebehandlung in Azure Easy Table-Back-End

Question

Mein Frontend ist iOS Mobile App und als Backend-Datenquelle verwende ich die easy-Tabelle. Skriptcode einfügen/update wird in js-Dateien geschrieben.

Aber wenn irgendwas in meinem Backend-Code bricht, gibt es jetzt, wie die mobile App es weiß. kann mir jemand helfen, die Ausnahmebehandlung zu implementieren, damit ich die Ausnahme an die mobile App senden kann?

 var table = module.exports = require('azure-mobile-apps').table(); 

var insertMiddleware = function(req,res,next) { 
const util = require('util'); 
var bPromiseRejected = false; 
var promises = [], promisesInsert = [], promisesUpdate = []; 


var requestItem = req.body; 
var tableRecord = req.azureMobile.tables('Table1'); 
var itemsToInsert = requestItem.ItemsToInsert; 
var itemsToUpdate = requestItem.itemsToUpdate; 


console.log('Step 1.'); 

//delete records array from feature dict 
delete requestItem["ItemsToInsert"]; 
delete requestItem["itemsToUpdate"]; 
//context.item = requestItem; 

if (itemsToInsert) { 
    promisesInsert = itemsToInsert.map(function(item) { 

     return tableRecord.insert(item).then (
       // Log the fulfillment value 
       function(val) { 
        console.log('Step 5: val: ' + val); 
        return(val); 
       } 
      ) 
      .catch (
       // Log the rejection reason 
       function(reason) { 
        console.log('Step 6. Handle rejected promise ('+reason+') here.'); 
        bPromiseRejected = true; 
        return("Error: " + reason); 
       } 
      ); 
     }); 
} 

if (itemsToUpdate) { 
    promisesUpdate = itemsToUpdate.map(function(item) { 
     return tableRecord.update(item).then (
       // Log the fulfillment value 
       function(val) { 
        console.log('Step 5: val: ' + val); 
        return(val); 
       } 
      ) 
      .catch (
       // Log the rejection reason 
       function(reason) { 
        console.log('Step 6. Handle rejected promise ('+reason+') here.'); 
        bPromiseRejected = true; 
        return("Error: " + reason); 
       } 
      ); 
     }); 
} 

promises = promisesInsert.concat(promisesUpdate); 
console.log('Promises: ' + util.inspect(promises, false, null)); 

    var result = Promise.all(promises) 
    .then(function(arrVals) { 
     //found an error in a promise 
     console.log('Promise completed. Vals: ' + JSON.stringify(arrVals)); 
     var arrRet = []; 
     for(var ind in arrVals) { 
      var val = arrVals[ind]; 
      //if val is a string error, don't do anything 
      console.log('Typeof Val: ' + typeof(val)) 
      if (typeof(val) === 'string') { 
       console.log("Already Inserted. Skipping.Val: " + val); 
       continue; 
      } else { 
       console.log("Item Inserted. Adding to Array. Val: " + val); 
       arrRet.push(val) 
      } 
     } 
     var result = {id:"Success", items: arrRet}; 
     console.log('All Records Inserted: ' + JSON.stringify(result)); 
     return res.end(JSON.stringify(result)); 

    } 
) 
.catch(
    function(reason) { 
     var retVal = { id: 'ERROR', reason: reason }; 
     console.error("Error: Promise Final.In catch. Reason: " + reason); 
     res.end(JSON.stringify(retVal)); 
     console.log('After 400 submitted'); 
    } 
); 
return result; 
} 

table.insert.use(insertMiddleware, table.operation); 
table.insert(function (context) { 
    return context.execute(); 
}); 

Answer 1

Wenn Sie eine Ausnahme in der Middleware zu fangen, können Sie versuchen, diese Codezeile verwenden

res.status(500).send({ error: 'something blew up' }); 

statt return("Error: " + reason); die Fehler an den Client zu senden.