PHP：从下拉列表中回显选定的值

Question

我有使用MySQL表中的一个创建的这个下拉列表。下拉列表中工作正常，但由于某种原因，我不能在这里呼应选择的值是我的代码：

<?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> 

这里是我曾尝试：

$n=$_POST['Name']; 
    echo $n; 

Answer 1

下拉列表必须包含在表单中。如果未设置提交表单，则无法从$_POST阵列获取值。
您应使用此代码：

$sql = "SELECT * FROM TABLE"; 
$result = mysql_query($sql); 
echo "Options" . "<form method='post'><select id='name' name='name'>"; 
while ($row = mysql_fetch_array($result)) { 
    echo "<option value='" . $row['Name'] . "'>" . $row['Name'] . " 
    </option>"; 
} 
echo "</select><input type='submit' name='submit'></form><br>"; 

if(isset($_POST['submit'])) { 
    echo $_POST['name']; 
} 

我希望这会帮助你！

Answer 2

尝试这样

echo "<option if($row["Name"]=="Doe") ? "selected='selected'":"" value='".$row['Name']."'>".$row["Name"]."</option>";