使用entityframework时，“设置值”不会更新现有行。 [英] Updating existing rows is not working by 'set values' when using entityframework.

问题描述

为什么setvalues对我不起作用？它对模型对象不起作用吗？

Why setvalues is not working for me? Does it not work for model objects?

using (var connection = new TReportEntitiesConnection())
        {

            var header = connection.THeaders.Include("TReports").SingleOrDefault(f => f.ID == model.ID);

            if (header != null)
            {
                header.THeaderTitle = model.THeaderTitle; //update parent
            }
            foreach (var existingChild in header.TReports.ToList())
            {

                if (!model.TReports.Any(c => c.ID == existingChild.ID))
                    connection.TReport.Remove(existingChild);

            }
            foreach (var url in model.TReports)
            {
                var existingChild = header.TReports
                    .Where(c => c.ID == url.ID)
                    .SingleOrDefault();


                if (existingChild!= null)
                { //update child
                    connection.Entry(existingChild).CurrentValues.SetValues(url);
                }
                else
                {
                    var newChild = new TReport
                    {
                        TReportName = url.name,
                        URL = url.url,
                    };

                    header.TReports.Add(newChild);
                }
            }
            connection.SaveChanges();
        }

我尝试过：



我尝试将其更改为但后来我意识到它不断添加新行。

What I have tried:

I tried changing it to but then I realized it is keep adding new rows.

header.TReports.Add(new TReport()
                       {
                           TReportName=url.name,
                           URL=url.url

                       });

推荐答案

如何让它更容易。不要使用SetValues，只需设置要更改的属性的值。

How about making it easier. Don't use SetValues, instead, just set the value of the property you want to change.

existingChild.property = value;
connection.Entrry(existingChild).State = EntityState.Modified;

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