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将值赋给mysql命令中的参数,但未返回正确的ID

[英]the Value are given to the parameter in the mysql command but it doesn't return the right id

这是程序中执行的查询的示例:

        Int64 kooid = new Int64();
        myCommand.CommandText = ("SELECT kooiid FROM kooien where NummerExtern =?NummerExtern");
        myCommand.Parameters.Add("?NummerExtern", MySqlDbType.Int64).Value = kooi.Kooinr;
        kooid = Convert.ToInt64(myCommand.ExecuteScalar());
        return kooid;

这是餐桌化妆的一个例子

kooiid(int)| kooidlistppg(int)|quarantaine(tiny int) |idvogelsoort(int)|idvogelondersoort(int)| vasteoflossekooie(tiny int)| bezetofniet(tiny int)| idsponsor(int)| NummerExtern(int)

在您问我何时手动运行查询之前,它可以正常运行,但是当程序正确排序时,即使kooinr变量正确,查询也会向kooiid返回零

找到了解决方案

Int64 kooid = new Int64();
        myCommand.CommandText = ("Select kooiid from kooien where NummerExtern = ?NummerExtern");
        myCommand.Parameters.Add("?NummerExtern", MySqlDbType.Int64).Value = kooi.Kooinr;
        string kooidstring = myCommand.ExecuteScalar().ToString();
        kooid = Convert.ToInt64(kooidstring);
        return kooid;

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