Question

我有一个名为'fishdata'的MySQL数据库，里面有一个名为'Fish'的表，如下所示：

FishType | X_B | X_SB |

|沙丁鱼| 1＃2＃3 | 7＃4＃6 |

|鳕鱼| 3＃6＃8 | 7＃4＃6 |

|鲶鱼| 4＃3＃8 | 7＃4＃6 |

我正在尝试将“FishType”列中的所有条目显示为选择框中的选项。这是我的代码：

<html>
<body>
    <select> 
        <?php
            $con=mysqli_connect("localhost","root","root","fishdata");
            // Check connection
            if (mysqli_connect_errno()) {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
            }
        $sql = "SELECT * FROM Fish";
        $result = mysqli_query($con,$sql)or die(mysqli_error());
            while($row = mysqli_fetch_array($result)) {
            {
            ?>
            <option value = "<?php echo($row['FishType'])?>" >
                <?php echo($row['FishType']) ?>
            </option>
            <?php
            }               
        ?>
    </select>
</body>
</html>

但是，选择框显示时没有选项。有什么想法吗？

Answer 1

while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
 echo "<option value='" . $row['FishType'] . "'>" . $row['FishType'] . "</option>";
}

你的while循环旁边有一个额外的括号。

修改

<html> <head></head> <body> <?php $con = mysqli_connect("localhost","root","root","fishdata"); if (mysqli_connect_errno()){ echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con, "SELECT FishType FROM Fish"); if (mysqli_num_rows($result) > 0){ echo "Records found"; echo "<select>"; while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)){ echo "<option value='Test'>Test</option>"; echo "<option value=\"" . $row['FishType'] . "\">" . $row['FishType'] . "</option>"; } echo "</select>"; } mysqli_close($con); ?> </body> </html>

在选择框中显示数据库列

1 个答案: