Question

我现在想要花2个小时来弄清楚为什么我的桌子不会更新，而对于上帝的爱，我无法弄明白。这是一个简单的民意调查设置。我已经取代＆＃34; UPDATE民意调查＆＃34;用＆＃34;更新评论＆＃34;它会毫无问题地更新评论。然而民意调查只是不起作用，所以我猜问题存在于数据库中，但就我所能得到的而言。也许有人可以帮我解决这个问题？

if (@$_POST['vote']) {
    $check_votes = mysql_query ("SELECT votes FROM votes WHERE voter='$user' AND vote_subject='$subject'");
    $numrows_votes = mysql_num_rows($check_votes);
    if ($numrows_votes == 1) {
        $error = "You already voted on this poll.";
    } 
    else {
        $check_polls = mysql_query ("SELECT * FROM polls WHERE poll_subject='$subject'");
        while($row_polls = mysql_fetch_array($check_polls)) {
            $vote1 = $row_polls['vote_one'];
            $vote2 = $row_polls['vote_two'];
            $numvote = 1;
            $newvote1 = $vote1 + $numvote;
            $newvote2 = $vote2 + $numvote;
            $polloption = $_POST['choice'];
            if ($polloption == "") {
                $error = "You didn't select any option.";
            } 
            else if ($polloption == "one") {
                mysqli_query("UPDATE polls SET vote_one='$newvote1' WHERE poll_subject='$subject'");
            } 
            else {
                mysqli_query("UPDATE polls SET vote_two='$newvote2' WHERE poll_subject='$subject'");
            }
        }
    }
}

这是表单代码：

<form action="" method="POST">
    <h4>What question would you like to ask?</h4>
    <input type='radio' name='choice' value='one'>
    <?php echo "$newvote1"; ?>
    <br>
    <input type='radio' name='choice' value='two'>
    <?php echo "$newvote2"; ?>
    <p class="buttons">
        <input type="submit" name="vote" value="VOTE">
    </p>
</form>

Answer 1

正如上面@jeroen的评论中所提到的，您正在混合数据库API。但是，你应该做的是抛弃它们并使用更高级别的API，如PDO。

mysql_*函数已经不支持5年了，并且将完全从下一版本的PHP中删除。如果使用正确，mysqli_*函数可以正常使用，但PDO提供了许多便捷方法，并且还允许您更轻松地更改基础数据库。

这是使用预处理语句快速用PDO重写的代码片段。假设$db为set up correctly，则完全未经测试但应该有效。如果您计划对数据库执行任何操作，则应阅读准备好的语句并不断使用它们。

if (isset($_POST['vote'])) {
    $stmt = $db->prepare("SELECT COUNT(*) FROM votes WHERE voter=? AND vote_subject=?");
    $stmt->execute(array($user, $subject));
    if ($stmt->fetchColumn() > 0) {
        $error = "You already voted on this poll.";
    } else {
        $stmt = $db->prepare("SELECT * FROM polls WHERE poll_subject=?");
        $stmt->execute(array($subject));
        while($row_polls = $stmt->fetch(PDO::FETCH_ASSOC)) {
            $vote1 = $row_polls['vote_one'];
            $vote2 = $row_polls['vote_two'];
            $numvote = 1;
            $newvote1 = $vote1 + $numvote;
            $newvote2 = $vote2 + $numvote;
            $polloption = $_POST['choice'];
            if ($polloption == "") {
                $error = "You didn't select any option.";
            } elseif ($polloption == "one") {
                $stmt2 = $db->prepare("UPDATE polls SET vote_one=? WHERE poll_subject=?");
                $stmt2->execute(array($newvote1, $subject));
            } else {
                $stmt2 = $db->prepare("UPDATE polls SET vote_two=? WHERE poll_subject=?");
                $stmt2->execute(array($newvote2, $subject));
            }
        }
    }
}

由于一些奇怪的原因，表格不会更新

1 个答案: