我有使用MySQL表中的一個創建的這個下拉列表。下拉列表中工作正常,但由於某種原因,我不能在這裏呼應選擇的值是我的代碼:PHP:從下拉列表中回顯選定的值
<?php
require_once('config.php');
// CONNECT
mysql_connect('localhost', 'root', 'password');
mysql_select_db('Database');
?>
// other.php is another php file
<form action="other.php" method="POST">
<label>Quantity:</label>
<input type="number" min="1" name="quantity" value="1"/>
<br/>
<hr/>
<?php
echo makeFormEntry('Product Type', 'type', $types);
echo makeFormEntry('Product Occasion', 'occasion', $occasions);
echo makeFormEntry('Product Size', 'size', $sizes);
$sql = "SELECT * FROM Table";
$result = mysql_query($sql);
echo "<b>Name : </b>" . "<select id='Name' name='Name'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['Name'] . "'>" . $row['Name'] . "</option>";
}
echo "</select><br>";
echo "<input type='submit'/><input type='reset'/>";
?>
</form>
這裏是我曾嘗試:
$n=$_POST['Name'];
echo $n;
'回聲 「<選項」($ _ POST [ '名稱'] == $行[ '名稱'] '選擇': '')。」值=' – JustOnUnderMillions
$ sql =「SELECT * FROM TABLE」;什麼表?是你的表的名字?你的表單名是否存儲在數據庫中? – Tony
是的,它的表名只是爲了說明 –