在PHP中执行其他查询之前，如何查看SQL查询的结果是否为空

Marko Vidalis 发表于 Dev

马尔科·维达利斯（Marko Vidalis）

我有以下用于应用程序投票系统的PHP代码。它是一个问答应用程序，用户可以对发布的问题和答案进行投票。

在我的php代码中，我首先检查用户是否对特定问题进行了投票。这将存在于QVOTES表中，其中电子邮件和问题的ID被投票。

执行此检查时，我不确定如何查看$ result是否为空集，以便在用户尚未对问题进行投票时提交用户的投票。

我该如何工作？非常感谢所有帮助。

<?php

$con=mysqli_connect("127.2.1.1","S837","887","D887");
if (mysqli_connect_errno($con))
{
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$qid = $_POST['qid'];
$email = $_POST['email'];
$result = mysqli_query($con, "SELECT * FROM QVOTES WHERE QID = $qid AND EMAIL = '$email'");
if (!mysqli_num_rows($result) ){
    if ($result = mysqli_query($con, "INSERT INTO QVOTES (QID, EMAIL) VALUES     ($qid, '$email')")) {
            mysqli_query($con, "Update QUESTIONS SET VOTES = VOTES +1 WHERE     QID = $qid");
            echo "Update successful";
        } else{
            echo "Update unsuccessful";
        }
            } else{
        echo "null";        
    }
    mysqli_close($con);

生死

其实您做错了。请尝试这样做：

<?php

$con=mysqli_connect("127.2.1.1","S837","887","D887");
if (mysqli_connect_errno($con))
{
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$qid = $_POST['qid'];
$email = $_POST['email'];
$result = mysqli_query($con, "SELECT * FROM QVOTES WHERE QID = $qid AND EMAIL = $email") or die(mysqli_error($con)); // no need of extra quote
if ($result->num_rows == 0 ){ // means no vote-up done till now
    $result = mysqli_query($con, "INSERT INTO QVOTES (QID, EMAIL) VALUES ($qid, $email)")or die(mysqli_error($con)); // insert
    if($result){
        echo "Vote Added successfully.";
    } else{
       echo "Error occur while adding vote.Please try again.";
    }
} else{
    $result = mysqli_query($con, "Update QUESTIONS SET VOTES = VOTES +1 WHERE  QID = $qid AND EMAIL = $email")or die(mysqli_error($con)); // upddate
       if($result){
        echo "Vote updated successfully.";
    } else{
       echo "Error occur while updating vote.Please try again.";
    }
}
    mysqli_close($con);