【问题标题】:html multiple select statements not workinghtml多个选择语句不起作用
【发布时间】:2014-04-20 01:47:12
【问题描述】:

我有一个用于向我的每个数据库表添加事件的表单。

为了使数据和时间正确,我使用了下拉菜单。但是我使用 5 2 作为时间,3 作为日期。在第二个选择语句的 Dream weaver 中,它以黄色突出显示,就像我有错误一样。 http://pastebin.com/NJB84hed

我想确保这也保存了 id,以便可以将 id 发布到该表中

$results=mysqli_query($con, "select * from Users where `userName` ='$username'");
    $id = 'id_cust';

【问题讨论】:

    标签: php html sql mysqli


    【解决方案1】:

    (1) 您的if(isset($_POST['insert'])){ 缺少右括号}

    if(isset($_POST['insert'])){
    $artist = $_POST['artist'];
    $place = $_POST['place'];
    $hour = $_POST['hour'];
    $minute = $_POST['min'];
    $year = $_POST['year'];
    $month = $_POST['month'];
    $day = $_POST['day'];
    $price = $_POST['price'];
    $open = $_POST['open'];
    $time = $hour.':'.$minute;
    $date = $year.'-'.$month.'-'.$day;
    $result=mysqli_query($con,"insert into Concert values('$id','$artist','$date','$time','$place','$price','$open')");
                    if($result)
                    {
                    echo 'Values updated successfully';
                    }
    ?>
    

    应该是

    if(isset($_POST['insert'])){
    $artist = $_POST['artist'];
    $place = $_POST['place'];
    $hour = $_POST['hour'];
    $minute = $_POST['min'];
    $year = $_POST['year'];
    $month = $_POST['month'];
    $day = $_POST['day'];
    $price = $_POST['price'];
    $open = $_POST['open'];
    $time = $hour.':'.$minute;
    $date = $year.'-'.$month.'-'.$day;
    $result=mysqli_query($con,"insert into Concert values('$id','$artist','$date','$time','$place','$price','$open')");
                    if($result)
                    {
                    echo 'Values updated successfully';
                    }
    } // MISSING THIS CLOSING BRACKET
    ?>
    

    (2) 在第 73 行,您拼错了您的选择结束标签

    </selct>
    

    应该是

    </select>
    

    (3) 另外,我假设

    $results=mysqli_query($con, "select * from Users where `userName` ='$username'");
    $id = 'id_cust';
    

    应该是这样的

    $results=mysqli_query($con, "select * from Users where `userName` ='$username'");
    $row = mysqli_fetch_array($result);
    $id = $row['id_cust'];
    

    【讨论】:

    • 谢谢,多一双眼睛来解决问题总是好的
    • 乐于助人。确保将其标记为已回答(以及您提出的任何其他尚未选择答案的问题)stackoverflow.com/about
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