Question

我无法从表单中获取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>

我想用它来检查我的类表中的类容量，如果类是满容量则会引发错误，否则将成功添加学生数据。但是，我的动作代码似乎无法获取给定的类输入，当我提交按钮添加学生数据时，它总是会抛出一个错误，表示该类已满，即使它尚未完整。以下是我的行动代码，到目前为止我所做的：

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() 返回数组... ，而不是字符串/值。

所以当你这样做时

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

在比较2个数组时，总是返回false（类如果已满），即

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

而不是2个值，即

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

您需要做的是使用列名来获取要评估的实际值，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"])) {

无法从表单中获取选项值

1 个答案: