我想从ACCESS中读取图片的路径，然后在WINFORM中用PICTUREBOX来显示，怎么做

Question

我想从ACCESS中读取图片的路径，然后在WINFORM中用PICTUREBOX来显示，怎么做

Answer 1

用pictureBox1.Load("文件路径")就可以。

Answer 2

这个应该很好实现吧,将处理后的图片保存在固定的文件夹中,上传路径至数据库,然后在winform中读取路径赋值给picturebox的属性就可以了吧.
我这里有之前做的web的案例,希望对你有所帮助.
//上传图片的例子if (pic1.PostedFile.ContentLength > 0)
        {
            string tempFileName = "dm_" + DateTime.Now.ToString("yyyyMMddhhmmss");
            try
            {
                string temp = FileManager.UpImage("uploads/supplier/", pic1.PostedFile, tempFileName, "no");
                if (temp.Substring(0, 1) == "1")
                {
                    pic_1 = tempFileName + Path.GetExtension(pic1.PostedFile.FileName);
                }
            }
            catch
            {
            }
        }//在页面的asp控件中显示vp1.Src = "../uploads/supplier/" + model.CNSPIC;

Answer 3

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Data.OleDb;
using System.IO;namespace OleDbImageSample
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }        private void Form1_Load(object sender, EventArgs e)
        {        }        private string strConn = @"Provider=Microsoft.Jet.OLEDB.4.0;Data Source=D:\VisualStudioProject2010\Winform\OleDbImageSample\OleDbImageSample\DB\ImagesDB.mdb";        private void button1_Click(object sender, EventArgs e)
        {
            if (openFileDialog1.ShowDialog() != System.Windows.Forms.DialogResult.OK)
                return;            var file = openFileDialog1.FileName;            try
            {
                using (var conn = new OleDbConnection(strConn))
                {
                    var comm = new OleDbCommand();
                    comm.Connection = conn;
                    comm.CommandText = "insert into images (FileName, Data) values (?, ?)";
                    var p1 = new OleDbParameter();
                    p1.DbType = DbType.String;
                    p1.Value = Path.GetFileName(file);
                    comm.Parameters.Add(p1);
                    var p2 = new OleDbParameter();
                    p2.DbType = DbType.Binary;
                    p2.Value = File.ReadAllBytes(file);
                    comm.Parameters.Add(p2);
                    
                    conn.Open();
                    comm.ExecuteNonQuery();
                    MessageBox.Show("Insert Successfully");                    var adapter = new OleDbDataAdapter("select Id, FileName from images", conn);
                    var list = new DataTable();
                    adapter.Fill(list);
                    this.listBox1.DataSource = list;
                    this.listBox1.DisplayMember = "FileName";
                    this.listBox1.ValueMember = "Id";
                }
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }        }        private void button2_Click(object sender, EventArgs e)
        {
            if (listBox1.SelectedValue == null)
                return;
            var id = (int)listBox1.SelectedValue;
            var sql = "select data from images where id=" + id;
            using (var conn = new OleDbConnection(strConn))
            {
                var comm = new OleDbCommand();
                comm.CommandText = sql;
                comm.Connection = conn;
                conn.Open();
                var reader = comm.ExecuteReader();
                if (reader.Read())
                {
                    var data = (byte[])reader.GetValue(0);
                    using (var ms = new MemoryStream(data))
                        this.pictureBox1.Image = Image.FromStream(ms);
                }
            }
        }
    }
}