选择下拉选项后,它会显示错误的信息

时间:2013-01-17 14:10:41

标签: php html mysqli

在下面的代码中,我有一个查询,它从db中检索所有模块,然后在下拉菜单中将它们列为选项。现在假设发生的是用户从下拉菜单中选择一个模块并提交。提交后,它将显示在用户选择的模块下方。

问题在于,无论用户从下拉菜单中选择了哪个模块,它始终指出用户已选择Advanced Web programming模块。我的问题是为什么总是说明这一点而不是显示选择的正确模块?

以下是VIEW SOURCE代码:

<form action="" method="post">Module:
  <select name="module" id="modulesDrop">
    <option value="">Please Select</option>
    <option selected='selected' value='CHI2513_Systems Strategy_1'>CHI2513 - Systems Strategy</option>
    <option value='CHI2565_Ecommerce, Business and Technology_3'>CHI2565 - Ecommerce, Business and Technology</option>
    <option value='CHT2220_Interactive Systems_4'>CHT2220 - Interactive Systems</option>
    <option value='CHT2520_Advanced Web Programming_5'>CHT2520 - Advanced Web Programming</option>
  </select>
  <input id="moduleSubmit" type="submit" value="Submit Module" name="moduleSubmit" />
</form>
<form action="" method="post">
  <p> <strong>Selected Module:</strong>CHT2520 - Advanced Web Programming
    <input type='text' value='CHI2513_Systems Strategy_1' />
  </p>
  Asessments: <span class='red'>Sorry, You have No Assessments under this Module</span>

以下是书面代码:

$moduleactive = 1;
            $sql = "SELECT ModuleId, ModuleNo, ModuleName FROM Module WHERE ModuleActive = ? ORDER BY ModuleNo"; 

            $sqlstmt=$mysqli->prepare($sql);
            $sqlstmt->bind_param("i", $moduleactive);
            $sqlstmt->execute(); 
            $sqlstmt->bind_result($dbModuleId,$dbModuleNo,$dbModuleName);
            $sqlstmt->store_result();
            $sqlnum = $sqlstmt->num_rows();
            ?>
            Module: 
            <select name="module" id="modulesDrop">
                <option value="">Please Select</option>
                <?php
                while($sqlstmt->fetch()) { 
                    $ov = $dbModuleNo . "_" . $dbModuleName . "_" . $dbModuleId; 
                    if($ov == $_POST["module"]) 
                        echo "<option selected='selected' value='$ov'>$dbModuleNo - $dbModuleName</option>" . PHP_EOL; 
                    else 
                        echo "<option value='$ov'>$dbModuleNo - $dbModuleName</option>" . PHP_EOL;
                 } 
                ?>
            </select>
            <input id="moduleSubmit" type="submit" value="Submit Module" name="moduleSubmit" />
        </form>
        <?php
        if(isset($_POST['moduleSubmit'])) 
{


if($_POST["module"] != "") { ?>
        <form action="" method="post"> 
            <p>
            <strong>Selected Module:</strong><?php echo $dbModuleNo .' - '. $dbModuleName;?> <input type='text' value='<?php echo $_POST["module"];?>'>
            </p>
            <?php } ?>

1 个答案:

答案 0 :(得分:0)

我的PHP有点生疏,但我相信这是因为模块Advanced Web Programming是列表中的最后一个。每次通过while循环从DB中获取行时,您的变量$dbModuleNo$dbModuleName都会更新,因此当您想要显示所选模块时,这些变量包含最近获取的数据,不是你想要的选定数据。

尝试这样的事情(我在现有代码中添加了几行):

<?php
$selectedText = "None";
while($sqlstmt->fetch()) { 
    $ov = $dbModuleNo . "_" . $dbModuleName . "_" . $dbModuleId; 
    if($ov == $_POST["module"]) 
        echo "<option selected='selected' value='$ov'>$dbModuleNo - $dbModuleName</option>" . PHP_EOL;
        $selectedText = $dbModuleNo .' - '. $dbModuleName;
    else 
        echo "<option value='$ov'>$dbModuleNo - $dbModuleName</option>" . PHP_EOL;
} 
?>

然后再......

<strong>Selected Module:</strong><?php echo $selectedText;?>