Php mysql\u查询返回错误的结果

考虑以下代码

if ( isset( $_SESSION['FBID'] )   ) {
    $uid      = $_SESSION['FBID'];
    $sql      = "SELECT *, count(member_nr) AS notifyMe 
                 FROM poolWinners 
                 WHERE member_nr = '$uid'   AND notification ='1'";
    $result = mysql_query($sql);
    while($row=mysql_fetch_array($result)){
       $notification = $row['notifyMe'];
    }//while
      if ( $notification > 0 ) {
        echo '<span class="badge">' . $notification . '</span>';
    } //if
    var_dump($notification);
} //isset( $_SESSION['FBID'] )

上面的脚本返回一个成员有多少通知，如下图所示

我的问题

脚本返回的结果错误，通知数量错误。查看下表，表中的成员编号出现3次，因此： $notification=$row['notifyMe']应该=3而不是1

我在这里错过了什么或做错了什么？感谢阅读

使用

$sql      = "SELECT *, count(*) AS notifyMe 
             FROM poolWinners 
             WHERE member_nr = '$uid'   AND notification ='1'";

注意count*，它将获取符合条件的记录数

---

并初始化$notification=0；开始时。

你试过从这个角度接近它吗

$sql = "SELECT * FROM poolWinners WHERE member_nr = '$uid'   AND notification ='1'";

$result = mysql_query($sql);
$notification = array();
    while($row=mysql_fetch_array($result)){
       $notification[] = $row['notifyMe'];
    }
//an array count on notification should give you the number of elements in the array i.e those that matched the query

$total_count = count($notification);

---

在您的代码中，通知总是一个，因为它只接受结果集中最后一行的notifyMe字段

如果要获取通知数，请尝试以下操作

if ( isset( $_SESSION['FBID'] )   ) {
    $uid      = $_SESSION['FBID'];
    $sql      = "SELECT *, count(member_nr) AS notifyMe 
                 FROM poolWinners 
                 WHERE member_nr = '$uid'   AND notification ='1'";
    $result = mysql_query($sql);
    $notification = 0;
    while($row=mysql_fetch_array($result)){
       $notification++; 
       /*
       OR $notification += $row['notifyMe'];
       */
    }//while
      if ( $notification > 0 ) {
        echo '<span class="badge">' . $notification . '</span>';
    } //if
    var_dump($notification);
} //isset( $_SESSION['FBID'] )

---

谢谢，伙计，但我也遇到了同样的问题谢谢你的努力，但我不能为我的生活找出原因，但它仍然返回一个，即使编辑，我认为可以解决它，var_dump确认$notification值为1……任何其他建议定义$notification=0之前，同时还有相同的问题…：-猜测它返回到绘图板上，用于meEdit问题，并为表结构粘贴SQL如果您处理的是少量行，您可以尝试mysql_num_rows，以查看返回了多少行求和通知列中的值？因为它们看起来每个都是1，就像这样，从poolWinners中选择sumNot，其中member_nr='$uid'和notification=1；不需要对结果进行循环，直接获取总和，或者从poolWinners中选择countmember\u nr，其中member\u nr='$uid'和notification=1；并使用计数而不是总和，以避免通知值大于1。再次直接获取完整值，无需在结果集上循环

$sql      = "SELECT *  
             FROM poolWinners 
             WHERE member_nr = '$uid'   AND notification ='1'";

$result = mysql_query($sql);
$notification = mysql_num_rows($result);

$sql = "SELECT * FROM poolWinners WHERE member_nr = '$uid'   AND notification ='1'";

if ($result=mysqli_query($con,$sql))
  {
  // Return the number of rows in result set
echo "Toal notification".mysqli_num_rows($result);
  }

mysqli_close($con);
?>