rs.open "select 字段名 from 表名",cn,3,3 set gatagrid.DataSource = rs
rs.open "select 字段名 from 表名",cn,3,3 set gatagrid.DataSource = rs 同意此方法,不过解释一下CN是指一个打开的连接,即Connection对象
方法一:datagrid右键属性,列,绑定字段 方法二:select field1 from table
ssquery = "Select paymenttime as 缴费时间,annuities as 养老金,payagent as 代理费,name as 姓名,idcard as 身份证号码 from tblpaymentagent where idcard='" & Trim(TxtIDcard.Text) & "'order by paymenttime"
Set rsPaymentClone = cnnCMMS.Execute(ssquery)
Set dgPayment.DataSource = rsPaymentClone
这样是最好的,因为传递的数据量小rs.open "select 字段名 from 表名", CN, adOpenDynamic, adLockOptimistic set Datagrid1.DataSource = rs 如果是Rs包含多个字段 用右键Datagrid属性,列,绑定字段也可以做到,但传的数据量过大建议用第一种方法!
用ado控件,在其recordsourse属性里 ado.recordsource="select 字段 from 表"
同意~~所有人看法~rs.open "select 字段名 from 表名", CN, adOpenDynamic, adLockOptimistic set Datagrid1.DataSource = rs确实是最好的~无论在哪个角度上~
set gatagrid.DataSource = rs
set gatagrid.DataSource = rs
同意此方法,不过解释一下CN是指一个打开的连接,即Connection对象
方法二:select field1 from table
Set rsPaymentClone = cnnCMMS.Execute(ssquery)
Set dgPayment.DataSource = rsPaymentClone
set Datagrid1.DataSource = rs
如果是Rs包含多个字段
用右键Datagrid属性,列,绑定字段也可以做到,但传的数据量过大建议用第一种方法!
ado.recordsource="select 字段 from 表"
set Datagrid1.DataSource = rs确实是最好的~无论在哪个角度上~