windows之打开文件，另存文件，返回值为文件路径

今天药忘吃喽~ 2022-03-02 02:30 212阅读 0赞

public string OpenFile()
            {
                string sqlFile;
                OpenFileDialog openFileDialog = new OpenFileDialog
                {
                    AutoUpgradeEnabled = true,
                    FilterIndex = 1,
                    RestoreDirectory = true,
                    FileName = string.Empty,
                    Filter = "xml(*.xml)|*.xml",
                    Title = "加载文件"
                };
    
                if (openFileDialog.ShowDialog() == DialogResult.OK && openFileDialog.FileName != string.Empty)
                {
                    sqlFile = openFileDialog.FileName;
    
                    return sqlFile;
                }
                else
                {
                    return null;
                }
            }

public string SaveFile()
            {
                SaveFileDialog fileDialog = new SaveFileDialog
                {
                    AutoUpgradeEnabled = true,
                    OverwritePrompt = true,
                    FilterIndex = 1,//设置默认文件类型显示顺序 
                    RestoreDirectory = true, //保存对话框是否记忆上次打开的目录 
                    FileName = string.Empty,
                    Filter = "SqL(*.sql)|*.sql",
                    Title = "先建sql文件"
                };
                if (fileDialog.ShowDialog() == DialogResult.OK && fileDialog.FileName != string.Empty)
                {
                    string path = fileDialog.FileName;
                    return path;
                }
                else return null;
    
            }