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在下面的代码中，我显示了一个 javascript 验证。但是当我点击moduleSubmit按钮时，它在if (isset($_POST['moduleSubmit'])) {. 我的问题是，当我点击按钮时，javascript验证已经通过，那么提交后如何显示表单？

JavaScript

function validation() {

    var isDataValid = true;

    var courseTextO = document.getElementById("coursesDrop");
    var moduleTextO = document.getElementById("modulesDrop");

    var errModuleMsgO = document.getElementById("moduleAlert");

    if (courseTextO.value == "") {
        $('#targetdiv').hide();
        $('#assessmentForm').hide();
        $('#updateForm').hide();
        $('#submitupdatebtn').hide();
        errModuleMsgO.innerHTML = "Please Select a Course";
        isDataValid = false;
    } else {
        errModuleMsgO.innerHTML = "";
    }

    if(isDataValid){
        $("#myForm").submit();
    }

}​

PHP/HTML

       <?php

// connect to the database
include('connect.php');


/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    die();
}


$sql = "SELECT CourseId, CourseNo, CourseName FROM Course ORDER BY CourseId"; 

$sqlstmt=$mysqli->prepare($sql);

$sqlstmt->execute(); 

$sqlstmt->bind_result($dbCourseId, $dbCourseNo, $dbCourseName);

$courses = array(); // easier if you don't use generic names for data 

$courseHTML = "";  
$courseHTML .= '<select name="courses" id="coursesDrop">' . PHP_EOL; 
$courseHTML .= '<option value="">Please Select</option>' . PHP_EOL;  

$outputcourse = "";

while($sqlstmt->fetch()) 
{

    $course = $dbCourseId;
    $courseno = $dbCourseNo;
    $coursename = $dbCourseName; 

    $courseHTML .= "<option value='" . $course . "'>" . $courseno . " - " . $coursename . "</option>" . PHP_EOL;  

    if (isset($_POST['courses']) && ($_POST['courses'] == $course)) {
        $outputcourse = "<p><strong>Course:</strong> " . $courseno .  " - "  . $coursename . "</p>";
    }

} 

$courseHTML .= '</select>'; 

?>

<form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" onsubmit="return validation();">
    <table>
        <tr>
            <th>Course: <?php echo $courseHTML; ?></th>
        </tr>
    </table>
    <p>
        <input id="moduleSubmit" type="submit" value="Submit Course and Module" name="moduleSubmit" />
    </p>
    <div id="moduleAlert"></div>
    <div id="targetdiv"></div>
</form>


<?php

if (isset($_POST['moduleSubmit'])) {    

    $sessionquery = "
    SELECT SessionId, SessionName, SessionDate, SessionTime, CourseId, SessionActive
    FROM Session
    WHERE (CourseId = ? AND SessionActive = ?)
    ORDER BY SessionName 
    ";

    $active = 1;

    $sessionqrystmt=$mysqli->prepare($sessionquery);
    // You only need to call bind_param once
    $sessionqrystmt->bind_param("si",$course, $active);
    // get result and assign variables (prefix with db)

    $sessionqrystmt->execute(); 

    $sessionqrystmt->bind_result($dbSessionId,$dbSessionName,$dbSessionDate,$dbSessionTime, $dbCourseId, $dbSessionActive);

    $sessionqrystmt->store_result();

    $sessionnum = $sessionqrystmt->num_rows();   

    if($sessionnum == 0) {
        echo "<p><span style='color: red'>Sorry, You have No Assessments under this Module</span></p>";
    } 
    else 
    { 
        echo "";
    }

    ...

}
4

2 回答 2

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恐怕输入的值type=button不会以表格形式发布。您可以使用 Firefox 的 firebug 插件或 chrome 中的开发者工具来检查 http 请求中发布的值。或者您也print_r($_POST)可以查看在 php 端发布的值。

尝试使用隐藏输入<input type="hidden" value="1" name="issubmit"/>来检查它是否是帖子。

于 2012-11-27T02:25:13.163 回答
0

您必须以 html 形式使用重定向操作。这会将您重定向到您想要的任何地方.. 喜欢，

if $valid=$form->validate();
{
  $this->redirect('form url')
}
于 2012-11-27T02:22:16.667 回答