关于excel文件打开的问题

Question

请教,使用
excelapplication1.workbooks.open(opendialog1.filename,
null,null,null,null,
null,null,
null,null,null,null,
null,null,0);
总是报错,怎么解决???
使用
excelapplication1.workbooks.open(opendialog1.filename,
EmptyParam,EmptyParam,EmptyParam,
EmptyParam,EmptyParam,
EmptyParam,EmptyParam,EmptyParam,EmptyParam,
EmptyParam,EmptyParam,0);
出现同样问题
出现的错误信息是:
except or finally expected
too many actual parameters

Answer 1

ExcelWorkbook1.ConnectTo(ExcelApplication1.Workbooks.Open('D:\123.xls', EmptyParam, EmptyParam,
    EmptyParam, EmptyParam, EmptyParam, EmptyParam, EmptyParam, EmptyParam, EmptyParam, EmptyParam,
    EmptyParam, EmptyParam, EmptyParam, EmptyParam, 0));