Question

我有一些表填充了一些数据，当我在我的控制器中单击“addTemplate”时，我没有从表中获取任何数据。

这是我的index.cshtml

public class FieldViewModel
{
    public TemplateField field { get; set; }
    public IEnumerable<TemplateField> IEnumerableField;
    public Fields fieldObj {get;set;}


    public Fields listOfFormats(string nameTemplate)
    {
        Fields obj = new Fields();
        var items = IEnumerableField.Where(i => i.TemplateName == nameTemplate).ToList();
        if (items != null)
        {
            foreach (var x in items)
            {
                obj.ColumnNamesDB.Add(x.ColumnNameDB);
                obj.Formats.Add(x.Format);
                obj.LengthColumns.Add(x.Length.ToString());
                obj.ColumnNamesUser.Add(x.ColumnNameUser);
            }
        }
        return obj;
    }
}

这是我的模特

public ActionResult addTemplate(FieldViewModel model)
{
    return ? // model is empty, there is nothing...
}

最后是我的控制器方法

public class Fields
{
    public List<string> ColumnNamesDB = new List<string>();
    public List<string> LengthColumns = new List<string>();
    public List<string> Formats = new List<string>();
    public List<string> ColumnNamesUser = new List<string>();
}

我认为我在模特中犯了一些错误，有人可以看一下吗？谢谢你的帮助！

编辑：

这是Fields类：

 public class TemplateField
{
    [Key]
    public int FieldId { get; set; }

    [Required(ErrorMessage = "Proszę podać nazwę kolumny z bazy danych.")]
    public string ColumnNameDB { get; set; }
    [Required(ErrorMessage = "Proszę podać długość rekordu.")]
    public int Length { get; set; }

    public string Format { get; set; }
    [Required(ErrorMessage = "Proszę podać nazwę szablonu.")]
    public string TemplateName { get; set; }
    [Required(ErrorMessage = "Proszę podać nazwę kolumny jaka będzie wyświetlana.")]
    public string ColumnNameUser { get; set; }
}

和TemplateField cass

        TransactionTemplate transactionTemplate = new TransactionTemplate(transactionManager);
        transactionTemplate.execute(new TransactionCallbackWithoutResult(){

        protected void doInTransactionWithoutResult(TransactionStatus status)
        {
            try 
            { 
                String SQL1 = "INSERT INTO Students (name, age) VALUES (?, ?)"; 
                jdbcTemplateObject.update( SQL1, name, age); 

                String SQL2 = "SELECT MAX(id) from Students"; 
                int sid = jdbcTemplateObject.queryForInt( SQL2 ); 

                String SQL3 = "INSERT INTO Marks(StudentId, marks, year) " + "VALUES (?, ?, ?)"; 
                jdbcTemplateObject.update( SQL3, sid, marks, year); 
                System.out.println("Created Name = " + name + ", Age = " + age); 

                transactionManager.commit(status);  // THIS IS THE LINE THAT CAUSES EXCEPTION. TRANSACTION DOES NOT ROLL BACK. 
            } 
            catch (Exception e) 
            { 
                System.out.println("Error in creating record, rolling back"); 
                status.setRollbackOnly();
                //throw e; 
            } 
        }           
    });

修改

好的，我在这里做了很多戏剧，所以这个问题没有我想的答案。

将动态模型传递给控制器

0 个答案: