MVC保存到两个表

传感器已经存在

您不必再将Sensor添加到数据库中，只需添加一些资产，每个响应一个资产。

Asset.Value等于相应的响应值，并且每个资产都属于主键等于StartAddress的Sensor。外键SensorId引用此Sensor，因此SensorId需要设置为StartAddress。

int[] assetValues = client.ReadHoldingRegisters(startAddress, quantity);

// for every assetValue create an Asset with Value equal to assetValue
// and a foreign key equal to startAddress:
var assetsToAdd = assetValues
    .Select(assetValue => new Asset()
    {
        Value = assetValue,
        SensorId = startAddress,
    });

using (var dbContext = new MyDbContext(...))
{
    // Add the Assets in one call to DbSet.AddRange(IEnumerable)
    dbContext.Assets.AddRange(assetsToAdd);

    // and save the changes:
    dbContext.SaveChanges();
}

如果对序列使用正确的标识符和正确的复数，对单个项目使用正确的复数，您是否看到代码容易得多？

传感器尚不存在

在这种情况下，您必须添加一个传感器。您无法分配主键，因此无法填写Asset.SensorId。幸运的是，实体框架足够聪明，可以为您解决此问题：

int[] assetValues = client.ReadHoldingRegisters(startAddress, quantity);

using (var dbContext = new MyDbContext(...))
{
    // Add one Sensor with all its Assets in one go
    dbContext.Sensors.Add(new Sensor()
    {
        // fill the Sensor properties that you know.
        // don't fill the Id
        ...

        Assets = assetValues
                 .Select(assetValue => new Asset()
                 {
                      Value = assetValue,
                      // no need to fill SensorId, Entity framework will do that for you
                      ... // if needed: fill other Asset properties
                 })
                 .ToList(),
    });
    dbContext.SaveChanges(); 
}

如果需要，可以分两个步骤进行：

using (var dbContext = new MyDbContext(...))
{
    // Add one Sensor without Assets
    var addedSensor = dbContext.Sensors.Add(new Sensor()
    {
        // fill the Sensor properties that you know.
        // don't fill the Id, nor property Assets
        ...
    });

    // Add the Assets in one call to DbSet.AddRange(IEnumerable)
    dbContext.Assets.AddRange(assetValues.Select(assetValue => new Asset()
                     {
                         Value = assetValue,
                         Sensor = addedSensor,
                     }));
    dbContext.SaveChanges(); 
}