php - 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

Question

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，用户可以为发布的问题和答案投票。

在我的 php 代码中，我首先检查用户是否为特定问题投票。这将存在于 QVOTES 表中，其中包含电子邮件和所投票问题的 ID。

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投票时提交用户投票。

我怎样才能让它工作？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);

Answer 1

其实你做错了。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);

注意:- 我更改了消息以便更好地理解。您可以根据自己的意愿进行更改。谢谢。