Question

我有一个包含id，header等的表格。想要从header列中为每个值的下拉列表指定一个名称。现在它只显示价值，这非常不舒服：

<form method="POST" enctype="multipart/form-data" action ="uploadext.php">
    <?php require_once('uploadext.php'); ?>
    <div class="col-md-6"><input type="file" name="fileToUploadgp"></div>
    <div class="col-md-6"><input type="file" name="fileToUploadpro"><br>
        <?php
        $dbc = mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME);
        $query = "SELECT id, header FROM done_add";
        $query1="SELECT header FROM done_add";
        $data = mysqli_query($dbc, $query);
        $data1=mysqli_query($dbc, $query1);
        $array=[];

        while ($row = mysqli_fetch_array($data)) {
            $arrayid[] = $row['id'];
            $arrayhead[]=$row['header'];
        }
        ?>
        <select name="selectlink">
            <?php foreach ($arrayid as $arr) {?>    
            <option value = "<?php print($arr)?>"
            } ?><?php print($arr) ?></option>
            <?php } ?>
        </select>
        <input type="submit" value="Отправить файлы" name="submita">
    </div>

</form>

Answer 1

（抱歉英语不好）

更改您在$array中存储数据的方式：

while ($row = mysqli_fetch_array($data)) {
    $option = [];
    $option['id'] = $row['id'];
    $option['header'] = $row['header']
    $array[] = $option;
    }

$option中的每个$array都会有ID和标题，您可以简单地说：

<select name="selectlink">
    <?php foreach ($array as $option) {?>    
    <option value = "<?php echo $option['id'];?>"
    } ?><?php echo $option['header'];?></option>
    <?php } ?>
</select>

为每个选项值命名

1 个答案: