OpenFileDialog 为空时返回异常答案

【问题标题】：OpenFileDialog return an exception when it is emptyOpenFileDialog 为空时返回异常
【发布时间】：2014-02-21 19:04:01
【问题描述】：

bool gGender = false;

if (radioBtnMale.Checked == true)
   gGender = true;

if (!string.IsNullOrEmpty(txtboxName.Text))
{
    if (imageToByteArray(Image.FromFile(openFileDialogPhoto.FileName)) == null)
    {
        repository.WorkflowsRepository.AddEmployee(txtboxName.Text, dateTimePickerBirthDate.Value, dateTimePickerHireDate.Value, gGender, txtboxMobile.Text, txtboxAddress.Text, txtboxEmail.Text,null);
    }

    repository.WorkflowsRepository.AddEmployee(txtboxName.Text, dateTimePickerBirthDate.Value, dateTimePickerHireDate.Value, gGender, txtboxMobile.Text, txtboxAddress.Text, txtboxEmail.Text, imageToByteArray(Image.FromFile(openFileDialogPhoto.FileName)));

    Information WFNameInfo = new Information("Massege", "The Employee " + txtboxName.Text + " was Added successfully");
    WFNameInfo.ShowDialog();
}
else
{
    Error err = new Error("The Employee must have a name, please try again");
    err.ShowDialog();
}

在此示例中，我想添加一名员工，但当员工没有照片时，我的意思是 OpenFileDialog 没有任何值，它在添加操作时返回异常。

我需要能够添加没有照片的员工！

【问题讨论】：

请出示您的代码
帮助我们帮助您。如果您不添加代码，我们将无法帮助您。我们应该怎么看问题是什么？您还需要向我们展示完整的异常消息
如果您的图片为空，您将添加两次员工
您是在另一个事件中调用openFileDialogPhoto.ShowDialog()，然后在txtboxName.Text 中设置文件名吗？
是的，我从另一个方法调用

标签： c# winforms openfiledialog

【解决方案1】：

我认为您的问题是缺少 else 语句：

if (imageToByteArray(Image.FromFile(openFileDialogPhoto.FileName)) == null)
{
    repository.WorkflowsRepository.AddEmployee(txtboxName.Text, dateTimePickerBirthDate.Value, dateTimePickerHireDate.Value, gGender, txtboxMobile.Text, txtboxAddress.Text, txtboxEmail.Text,null);
}
 repository.WorkflowsRepository.AddEmployee(txtboxName.Text, dateTimePickerBirthDate.Value, dateTimePickerHireDate.Value, gGender, txtboxMobile.Text, txtboxAddress.Text, txtboxEmail.Text, imageToByteArray(Image.FromFile(openFileDialogPhoto.FileName)));

if (imageToByteArray() == null) 您仍在执行第二条语句。试试：

if (imageToByteArray(Image.FromFile(openFileDialogPhoto.FileName)) == null)
{
    repository.WorkflowsRepository.AddEmployee(txtboxName.Text, dateTimePickerBirthDate.Value, dateTimePickerHireDate.Value, gGender, txtboxMobile.Text, txtboxAddress.Text, txtboxEmail.Text,null);
}
else
{
    repository.WorkflowsRepository.AddEmployee(txtboxName.Text, dateTimePickerBirthDate.Value, dateTimePickerHireDate.Value, gGender, txtboxMobile.Text, txtboxAddress.Text, txtboxEmail.Text, imageToByteArray(Image.FromFile(openFileDialogPhoto.FileName)));
}

【讨论】：

【解决方案2】：

问题：您没有调用openFileDialogPhoto.ShowDialog()，而是尝试访问openFileDialogPhoto.FileName，当然是null。

解决方案：您应该调用openFileDialogPhoto.ShowDialog()，并且只有在用户选择文件时才能继续进行。您可以在访问其FileName之前通过DialogResult返回值进行检查属性。

试试这个：

if(openFileDialogPhoto.ShowDialog()==DialogResult.OK)
{
   if (imageToByteArray(Image.FromFile(openFileDialogPhoto.FileName)) == null)
   {
     repository.WorkflowsRepository.AddEmployee(txtboxName.Text, dateTimePickerBirthDate.Value, dateTimePickerHireDate.Value, gGender, txtboxMobile.Text, txtboxAddress.Text, txtboxEmail.Text,null);
   }
   repository.WorkflowsRepository.AddEmployee(txtboxName.Text, dateTimePickerBirthDate.Value, dateTimePickerHireDate.Value, gGender, txtboxMobile.Text, txtboxAddress.Text, txtboxEmail.Text, imageToByteArray(Image.FromFile(openFileDialogPhoto.FileName)));
   Information WFNameInfo = new Information("Massege", "The Employee " + txtboxName.Text + " was Added successfully");
   WFNameInfo.ShowDialog();
}

【讨论】：