如何在水晶报表中动态显示图像 [英] How to show image dynamically in crystal report

问题描述

如何在访问数据库中插入图像如何在水晶报表中动态显示图像c＃

how to insert image in access database How to show image dynamically in crystal report c#

推荐答案

此链接可以帮助您.. :)

示例1 [ ^ ]

示例2 [ ^ ]

示例3 [ ^ ]

Crystal Reports中的图像 [ ^ ]

如何使用Visual Studio 2005在Crystal Reports中动态加载图像 [ ^ ]

this link may help you.. :)
example 1[^]
example 2[^]
example 3[^]
Image in Crystal Reports[^]
How to dynamically load images in Crystal Reports using Visual Studio 2005[^]

private void Submit_Click(object sender, EventArgs e)
       {
           byte[] picbyte = System.IO.File.ReadAllBytes(textBox.Text);
           try
           {
               string constr = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=.\\PIS(ACU).mdb;";
               Con = new OleDbConnection(@constr);
               Con.Open();
               Com = new OleDbCommand();
               Com.Connection = Con;
               Com.CommandText = "INSERT INTO Pics(Patient_Id,ImageDate,Photo)VALUES(" + txtPatientId.Text + ",'" + txtImageDate.Value.ToString("yyyy/MM/dd HH:mm:ss") + "', @Photo)";
               OleDbParameter picParam = Com.Parameters.Add("@Photo", SqlDbType.Binary);
               picParam.Value = picbyte.ToArray();
               picParam.Size = picbyte.ToArray().Length;
               Com.ExecuteNonQuery();
           }
           catch (Exception ex)
           {
               MessageBox.Show(ex.Message);
           }
           finally
           {
               Con.Close();
               MessageBox.Show("Image Uploaded Successfully", "PIS System");
                DiseaseRecord D = new DiseaseRecord();
                D.Show();
                this.Close();
           }
       }


       private void btnBrowse_Click(object sender, EventArgs e)
       {
           OpenFileDialog fd = new OpenFileDialog();
           fd.InitialDirectory = "c:\\";
          // DialogResult Result = fd.ShowDialog();
           if (fd.ShowDialog() == DialogResult.OK)
           {
               textBox.Text = fd.FileName.ToString();
               pictureBox.ImageLocation = textBox.Text;
           }
           Console.ReadLine();
       }

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