【发布时间】:2016-08-23 03:07:42
【问题描述】:
嗯...我有两个名为“chefs”和“cust_Data”的表。
我想:
将 cust_Data 中正在发布的 cmets 放入 chefs 表的评论列中。
-
发布意味着-> M 提供评论部分,以便客户将他们的 cmets 放在“cust_Data”中的“comment”列中,并且一旦这些 cmets 在 cust_Data...
将这些评论放在“chefs”表的“cmets”列中。
剃刀代码:
@{
var db = Database.Open("vendors");
var selectCommand = "SELECT dish FROM chefs WHERE ID= @0";
var chefID = Request.QueryString["chefid"];
var selectedData = db.Query(selectCommand, chefID);
var dishName=Request.QueryString["dish"];
var chefName=Request.QueryString["chefName"];
var comments=Request["SELECT comments from chefs"]; //IMPORTANT
var ID=Request["SELECT ID from chefs"]; //IMPORTANT
var grid = new WebGrid(source: selectedData, rowsPerPage: 10);
Validation.RequireField("cust_fname", "You must enter your firstname");
Validation.RequireField("cust_lname", "You must enter your surname");
Validation.RequireField("cust_email", "You haven't entered a valid Email_ID");
Validation.RequireField("rating", "Rating is required");
Validation.RequireField("comment", "Please provide your Comment about Respective Homeschef");
var cust_fname = "";
var cust_lname= "";
var cust_email= "";
var rating= "";
var comment= ""; //IMPORTANT
if(IsPost && Validation.IsValid() && chefID.AsInt()!=0){
cust_fname = Request.Form["cust_fname"];
cust_lname = Request.Form["cust_lname"];
cust_email = Request.Form["cust_email"];
rating = Request.Form["rating"];
comment = Request.Form["comment"]; // IMPORTANT->TAKING COMMENT FROM USER
var insertCommand = "INSERT INTO cust_Data (cust_fname, cust_lname, cust_email, dish, chef_ID, rating, comment) Values(@0, @1, @2, @3, @4, @5, @6)";
db.Execute(insertCommand,cust_fname,cust_lname, cust_email, dishName, chefID, rating, comment); // EXECUTED SUCCESSFULLY IN cust_Data
var insert= "INSERT INTO chefs(comments) FROM cust_Data(comment) WHERE chefs(ID)=cust_Data(chef_ID) Values(@0,@1,@2,@3)"; //IMPORTANT
db.Execute(insert,comments,comment,ID,chefID); // IMPORTANT -> ERROR OCCURED ?????????
}
}
异常详情:System.Data.SqlServerCe.SqlCeException: There was an error parsing the query. [ Token line number = 1,Token line offset = 29,Token in error = FROM ]
我插入了重要标签(以节省您处理重要事情的时间),从中获取我需要解析的参数。 再次,提前非常感谢你......
【问题讨论】:
-
我为你的成就感到骄傲!顺便问一下,您是想问我们一个问题吗?
-
您最后一次插入语法时出错了。阅读w3schools.com > insert into select 以了解正确的语法。为什么要从 DB 的 2 个单独请求中获取 ID、评论?
-
感谢@Spidey 的回复... 仍然存在令牌行错误=79。 var insert= "INSERT INTO chefs(cmets) SELECT cust_Data(comment) WHERE chefs(ID)=cust_Data(chef_ID) Values(@0,@1,@2,@3)";
标签: sql asp.net database razor database-schema