无法从表单中获得选项值

Question

我无法从表单中获得选项值，以将其传递给我的操作代码。

表格：

<select required name ="class" class="form-control">
    <option value="NULL">-select class-</option>
    <?php
    $link2 = query("select * from td_class");
    while ($list = mysql_fetch_array($link2)) {
    ?>
    <option value="<?php echo $list['class'] ?>"><?php echo $list['class']; ?></option>
    <?php } ?>
    <option value="Alumni">Alumni</option>
 </select>

我想使用它来检查我的类表中的类容量，如果这个类处于满负荷状态，它将抛出一个错误，否则将成功地添加学生数据。但是，我的操作代码似乎无法获得给定的类输入，当我提交按钮来添加学生数据时，它总是会抛出一个错误，说明类已经满了，尽管它还没有满。以下是我的行动守则，我迄今所做的：

if ($hal == 'managestudent') {
    $name = $_POST['name'];
    $image = $_FILES['image'];
    $class = $_POST['class'];
    $id = $_GET['id'];

    if ($aksi == 'addstudent') {
        $student = query("select name, count(name) from td_student where class='$class' ");
        $student2 = mysql_fetch_array($student);
        $capacity = query("select capacity from td_class where class='$class'");
        $cap = mysql_fetch_array($capacity);

        if (($student2) >= ($cap )) {
            echo"<script> alert ('the class is full'); "
            . "history.back();" . "</script>";
        } else {
            if (empty($image['tmp_name'])) {
                echo "Without image<br>";
                $sql = query("insert into td_student (name,class) values ('$name','$class)");
            } else {
                echo "With image<br>";
                $namafile = uploadBerkas($image);
                $sql = query("insert into td_student (name,class,image) values ('$name','$class,'$namafile')");
            }
            if ($sql == TRUE) {
                echo "SQL True<br>";
                echo"<script> alert('student data is added'); "
                . "window.location='datasiswa.php';"
                . "</script>";
            } else {
                echo "Sql Error " . mysql_error() . " ";
                echo"<script> alert ('Error'); "
                . "history.back();" . "</script>";
            }
        }
    }
}

谢谢你的帮助。

Answer 1

您的问题是mysql_fetch_array()返回一个数组.，而不是字符串/值。

所以当你这么做

if (($student2) >= ($cap ))

它总是返回false (类如果满)，因为它比较两个数组，即。

if((array(name, count(name)) >= (array(capacity)))

而不是两个价值，即。

if((count(name)) >= (capacity))

您需要做的是使用列名来获得要计算的实际值，即。$student2["count(name)"]/$cap["capacity"]

    $student = query("select name, count(name) from td_student where class='$class' ");
    $student2 = mysql_fetch_array($student);
    $capacity = query("select capacity from td_class where class='$class'");
    $cap = mysql_fetch_array($capacity);

    if (($student2["count(name)"]) >= ($cap["capacity"])) {