使用數據庫值填充下拉列表

Question

我想使用數據庫中特定字段的數據填充下拉列表。這裏是我的示例代碼

<?php 

$con = mysql_connect("localhost","root",""); 
if (!$con) 
    { 
    die('Could not connect: ' . mysql_error()); 
    } 

mysql_select_db("disertation ", $con); 
$results = mysql_query("SELECT name FROM user_parent;"); 
?> 

<select name="name"> 
        <option value="name">Select one</option> 
        <?php 
         while($row=mysql_fetch_array($results)) 
           { echo '<option value=" ' . $row['name'] . ' ">' . $row['name'] . '</option>'; } 
         ?> 
       </select> 

它目前沒有顯示任何數據庫，任何幫助？

Answer 1

讓您mysql_fetch_array通話閱讀：

mysql_fetch_array($results, MYSQL_ASSOC)

沒有MYSQL_ASSOC你不能引用列名$row。

另外，考慮使用MYSQLI或PDO。 MYSQL相當過時。

Answer 2

在你的代碼試試這個，一些白色的空間mysql_select_db("disertation", $con);

mysql_select_db("disertation", $con); 
$results = mysql_query("SELECT name FROM user_parent") or die (mysql_error()); 

Answer 3

我會建議你使用的mysqli爲mysql

<?php 
    $con=mysqli_connect("localhost","root","","disertation "); 
    // Check connection 
    if (mysqli_connect_errno()) { 
     echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

    $resource= mysqli_query($con,"SELECT * FROM user_parent"); 

    echo "<select class="name"><option value="name">Select one</option>"; 
    while($result = mysqli_fetch_array($resource)){ 
     echo '<option value="'.$result["name"].'">'.$result["name"].'</option>'; 
    } 
    echo "</select>"; 

    mysqli_close($con); 
?> 

Answer 4

要回答你的問題，直接，你應該首先檢查是否有任何錯誤（mysql_error()），然後檢查是否有一些結果（mysql_num_rows） - 這些使得代碼更容易調試，因爲它會告訴你什麼是錯誤的。

試試這個;

<?php 
// Connect with user 
$con = mysql_connect("localhost","root",""); 
if (!$con) 
{ 
    die('Could not connect: ' . mysql_error()); 
} 
// Select database 
mysql_select_db("disertation", $con); 

// Run query 
$results = mysql_query("SELECT `name` FROM `user_parent`;") or die (mysql_error()); 
// Check for no results 
if (mysql_num_rows($results) == 0) 
{ 
    echo 'There are no options for you to select.'; 
} 
else 
{ 
    // If results, loop them. 
    // If the names are user input, make sure they're displayed in non-raw form 
    echo '<select name="name"> 
     <option value="name">Select one</option>'; 

     while($row = mysql_fetch_assoc($results)) 
     { 
      $name = htmlentities($row['name'], ENT_QUOTES, "UTF-8"); 
      echo '<option value=" ' . $name . ' ">' . $name . '</option>'; 
     } 

    echo '</select>'; 
} 

將與mysqli_解決方案編輯，如果這是你的一個選擇，因爲mysql_已被棄用，並將從PHP的支持下降，遲早的事。

---

MySQLi解決方案;

<?php 
// Connect to database; 
$mysqli = new mysqli("localhost", "my_user", "my_password", "data_base"); 

if (mysqli_connect_errno()) 
{ 
    die("Connect failed: " . mysqli_connect_error()); 
} 

$result = $mysqli->query("SELECT `name` FROM `user_parent`"); 
if ($result->num_rows > 0) 
{ 
    echo '<select name="name"> 
     <option value="name">Select one</option>'; 

     while($row = $result->fetch_assoc) 
     { 
      $name = htmlentities($row['name'], ENT_QUOTES, "UTF-8"); 
      echo '<option value=" ' . $name . ' ">' . $name . '</option>'; 
     } 

    echo '</select>'; 
    $result->close(); 
} 
else 
{ 
    echo 'There are no options for you to select.'; 
}