php - 警告: odbc_fetch_row(): 4 is not a valid ODBC result错误

Question

我正在做2个查询

CALCULA_DEPARTAMENTO.PHP

$consulta=     "SELECT * FROM Dept INNER JOIN Userinfo 
        ON Userinfo.Deptid = Dept.Deptid
        where Dept.DeptName = '$departamento'";

我选择这个人组成一个部门。然后我要确保它能正常工作

$rs=odbc_exec($conn,$consulta); 
if (!$rs)
{exit("Connection Failed: " . $rs);}

然后检查所有结果，我这样做:

    while (odbc_fetch_row($rs))
    { 
       session_start(); 
   ob_start();
   $_SESSION['departamento'] = $departamento;

       include_once("calcula_cono.php");
      calcularr();
    }
   odbc_close($conn);

在“很多事情”中，y调用另一个具有此查询的php，以便能够查看一个人的所有来龙去脉:

CALCULA_CONO.PHP

function calcularr()
 { 
     $departamento = $_SESSION['departamento'];

$consultaa = "SELECT  *  FROM Checkinout, Userinfo
         where Checkinout.Userid = '$userid'     AND 
             Userinfo.userid = '$userid' AND 
             Checkinout.Checktime BETWEEN   CDate('$fecha_inicio') AND 
             CDate('$fecha_fin')";

       $rss=odbc_exec($conn,$consultaa);
        if (!$rss)
        {exit("Connection Failed: " . $rss);}



        while (odbc_fetch_row($rss))
         { //more things here         
         }
  }  

它完成了第一次迭代，并且做得很好...但是当他必须进行第二次迭代时，我得到此错误:

Warning: odbc_fetch_row(): 4 is not a valid ODBC result resource

Answer 1

It finishes the first iteration and it does it well... but when he has to do the second one i get this error:

这表明 while循环中的某些代码行正在破坏结果变量 $rss(或可能是 $rs，具体取决于哪个 odbc_fetch_row()调用失败)。仔细检查代码，以确保您不会在循环内无意中将值分配给结果变量。