2017-03-22 42 views
1

我想制作一个食谱书籍,用户可以制作新食谱,因此他可以从下拉字段中的预先输入的成分中进行选择。使用php下拉字段填充带有SQL数据的html表单

我的代码看起来像这样至今:

<html> 
    <body> 
    <form method="post" > 
    <?php 

    $servername = "localhost"; 
    $username = "root"; 
    $password = ""; 
    $dbname = "tagebuch1"; 

    // LOGIN TO DATABASE SCRIPT WRITTEN FOR MYSQLI 
    $mysqli = new mysqli($servername, $username, $password, $dbname); 
    if ($mysqli->connect_errno) { 
     echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " .   $mysqli->connect_error; 
    } 
    // END OF LOGIN TO DB SCRIPT 
    $query_ak='SELECT * FROM privnahrungsmittel'; 
    $result = $mysqli->query($query_ak); 
    ?> 
    <select name="exa" > 
     <?php 
     while ($row = mysqli_fetch_assoc($result)) { 
      echo '<option value="'.$row['idprivNahrungsmittel'].'">'.$row['brennwert'].'</option>'; 
     } 
     ?> 
    </select> 
    <input type="submit" value="GO!" name="go"/> 
    </form> 
    </body> 
    </html> 

但我只得到这样一个空的输出,虽然表中填充内容:

Dropdown field empty

任何人都可以请指教错误在哪里?

谢谢

+1

如果您的问题并不重要 - 将其删除。 –

回答

0

请更改$row = mysqli_fetch_assoc($result)$result->fetch_assoc()

while ($row = $result->fetch_assoc()) { 
     echo '<option value="'.$row['idprivNahrungsmittel'].'">'.$row['brennwert'].'</option>'; 
    } 
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