Question

我的前端是iOS移动应用程序，作为后端数据源，我使用简易表。插入/更新脚本代码是用js文件编写的。

但是，如果我的后端代码中出现任何问题，那么移动应用程序现在就可以了解它。任何人都可以指导我如何实现异常处理，以便我可以将异常发送到移动应用程序？

      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

当你在中间件中捕获异常时，可以尝试使用这行代码

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

而不是return("Error: " + reason);将错误发送给客户端。

Azure Easy Table后端中的异常处理

1 个答案: