Question

因此，我要向数据库中添加一个场所，然后向数据库中添加一个特殊项。一个场所可以有1到许多特惠，所以我试图在第三个.then()下创建一个关系。问题是，在建立并将特殊项添加到数据库之前，正在执行第三个.then()。我需要先将建立和特殊添加到数据库，因为我需要获取它们的objectIds var establishmentObjectId;和var specialObjectId;

我该如何确保在建立关系之前检索到场所和特殊的objectId？

    //Save the establishment objectId
    var establishmentObjectId;
    //Save the special objectID
    var specialObjectId;

    //Save establishment
    .then(() => {
        //Get the add form data
        var addName = document.getElementById('add_name');
        var addCountry = document.getElementById('add_country');
        var addEstablishmentType = document.getElementById('add_establishment_type');
        var addCuisineType = document.getElementById('add_cusine_type');

        //Create establishment from user's enteries
        var establishment = {
            Name: addName.value,
            Address: addAddress.value,
            Suburb: addSuburb.value,
            Country: addCountry.value,
            Cuisine_Type: addCuisineType.value,
            Establishment_Type: addEstablishmentType.value
        }

        //Save establishment to db
        Backendless.Data.of('Establishment').save( establishment )
        .then( function( savedObject ) {
            establishmentObjectId = savedObject.objectId;
            console.log( "new Establishment instance has been saved" );
          })
        .catch( function( error ) {
            console.log( "an error has occurred " + error.message );
          });
    })
    //Save special
    .then(() => {
        //Get the add form data
        var addCategory = document.getElementById('add_category');
        var addTypeOfSpecial = document.getElementById('add_type_of_special');
        var addDescription = document.getElementById('add_description');

        //Create special from user's enteries
        var special = {
            Category: addCategory.value,
            Type_Of_Special: addTypeOfSpecial.value,
            Description: addDescription.value
        }

        //Save special to db
        Backendless.Data.of('Special').save( special )
        .then( function( savedObject ) {
            specialObjectId = savedObject.objectId;
            console.log( "new Special instance has been saved" );
            })
        .catch( function( error ) {
            console.log( "an error has occurred " + error.message );
            });
    })
    //Add special to establishment/form relation
    .then(() => {

        //These are undefined even though they are declared above
        console.log(establishmentObjectId);
        console.log(specialObjectId);

        var parentObject = { objectId:establishmentObjectId };
        var childObject = { objectId:specialObjectId };
        var children = [ childObject ];

        Backendless.Data.of( "Establishment" ).addRelation( parentObject, "establishmentSpecials", children )
        .then( function( count ) {
          console.log( "relation has been set" );
        })
        .catch( function( error ) {
          console.log( "server reported an error - " + error.message );
        });

    })

非常感谢

Answer 1

无论何时，只要有99％的时间处于Promise链中，当您创建新的Promise（使用API调用或使用new Promise等）时，都应将其返回或以其他方式放入以及其他内容（例如Promise.all并返回 that ）。这将意味着可以在更高层次上处理可能的拒绝，并且意味着链中的 next .then仅在先前的.then完成后才开始。 >

每更改一次

Backendless.Data.of(

到

return Backendless.Data.of(

此外，您可能不想在每个级别上都catch，就像您目前正在做的那样-每当您catch时，您都会拒绝接受承诺到已解决的Promise中，随后的.then可能会运行在先前.then中的所有内容均已成功完成的假设下。对于此类顺序异步操作，如果任何地方错误，您可能希望停止整个过程，而不是尝试继续进行（可能会导致错误）。

例如，如果Backendless.Data.of('Establishment').save失败，则establishmentObjectId永远不会分配给它，而在最后的.then中尝试稍后使用它将无效。与specialObjectId一样。

对于此代码，还有一个更好的模式可以遵循：不要使用两个单独的.then，一个要获取establishmentObjectId，另一个要获取specialObjectId，请考虑让它们操作并行运行，并在通过.then完成后运行最后的Promise.all。类似于：

const getEstablishmentObjectId = () => {
  //Get the add form data
  // ...
  //Create establishment from user's enteries
  var establishment = {
    // ...
  }
  //Save establishment to db
  return Backendless.Data.of('Establishment').save(establishment)
    .then(savedObject => savedObject.objectId);
};
const getSpecialObjectId = () => {
    // ...
    return Backendless.Data.of('Special').save(special)
      .then(savedObject => savedObject.objectId);
};

Promise.all([
  getEstablishmentObjectId(),
  getSpecialObjectId(),
]).then(([establishmentObjectId, specialObjectId]) => {
    console.log(establishmentObjectId);
    console.log(specialObjectId);
    var parentObject = {
      objectId: establishmentObjectId
    };
    // etc
  })
  .catch((error) => {
    console.log('Error', error);
  });

如何使.then完全完成，然后再移至下一个？

1 个答案: