Re: something about"like"

Haitao -
I would recommend trying something like this:

  "select * from buyerinfo where BuyerName like concat('%', ?blurName, '%')"

Placing "?blurName" in the middle of a string literal will prevent the
parameter replacement from happening. A good way to debug a problem like
this is to enable query logging on the server and check what is actually
sent.

Jess

On Mon, Dec 03, 2007 at 09:27:38AM +0800, longhaitao wrote:
>  hi,guys:
> Now i am in this situation:I want to use the cause "like",i wrote:
> string commd = "select * from buyerinfo where BuyerName like '%?blurName%'";
> MySqlParameter para = new MySqlParameter("?blurName", MySqlDbType.VarChar, 45, "BuyerName");
> para.Value = blurName;
> It didn't work.
> I changed that like this:
> string commd = "select * from buyerinfo where BuyerName like '%"+"?blurName"+"%'";
> MySqlParameter para = new MySqlParameter("?blurName", MySqlDbType.VarChar, 45, "BuyerName");
> para.Value = blurName;
>  or:
> blurName="'%"+"blurName"+"%'";
> string commd = "select * from buyerinfo where BuyerName like ?blurName";
> MySqlParameter para = new MySqlParameter("?blurName", MySqlDbType.VarChar, 45, "BuyerName");
> para.Value = blurName;
> It also didn't work, when executed ,It returned nothing ,which definitely wrong.
> Any advise would be appreciated,thx!
> --
> Best regards??
> haitao

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