摘要: OpenFileDialog of1 = new OpenFileDialog(); if (of1.ShowDialog() == DialogResult.OK) { textBox1.Text = of1.FileName; } FileSt... 阅读全文
posted @ 2012-03-18 16:55 Uoolo 阅读(584) 评论(1) 推荐(1) 编辑
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