php - html选项数据显示外出

Question

我想要<option>在产品名称字段旁边，但 <option>而信息却远离了选择。我在这里做错了什么？

echo '<form action="" method="post" id="pform"><select>';


echo 'Product name: <input type="text" name="pname"><br>';

$q = mysql_query("SELECT * FROM menu");
    while ($row = mysql_fetch_array($q)) {
    $menu_name = $row['menu_name'];
    echo '<option value="'.$menu_name.'">'.$menu_name.'</option>';
}

echo '<textarea rows="8" cols="60" name="pdes" form="pform">';
echo 'Enter product description here...</textarea><br><br>';


echo '</select><input type="submit" name="addp" value="Add this product">'.'</form>';

Answer 1

生成的 HTML 看起来像这样:

<form ...>
    <select>
        Product name: <input ...>
        <option ...> ... </option>
        <option ...> ... </option>
        <option ...> ... </option>
        ...
        <textarea ...> ... </textarea>
    </select>
    <input>
</form>

这是错误的。 <select>应该只包含<option>元素，没有别的。换句话说，您应该更改代码以生成以下结构:

<form ...>
    Product name: <input ...>
    <select>
        <option ...> ... </option>
        <option ...> ... </option>
        <option ...> ... </option>
        ...
    </select>
    <textarea ...> ... </textarea>
    <input>
</form>

所以在你的情况下:

echo '<form action="" method="post" id="pform">';
echo 'Product name: <input type="text" name="pname"><br>';
echo '<select>';
$q = mysql_query("SELECT * FROM menu");
while ($row = mysql_fetch_array($q)){
    $menu_name = $row['menu_name'];
    echo '<option value="'.$menu_name.'">'.$menu_name.'</option>';
}
echo '</select>';
echo '<textarea rows="8" cols="60" name="pdes" form="pform">';
echo 'Enter product description here...</textarea><br><br>';
echo '<input type="submit" name="addp" value="Add this product">'.'</form>';