我有一个包含 id
、header
等的表格。想要从 header
列到下拉列表中指定一个名称每个值。现在它只显示值(value),这很不舒服:
<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>
最佳答案
(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;
}
每个 $option
在$array
将有 id 和 header,那么你可以简单地:
<select name="selectlink">
<?php foreach ($array as $option) {?>
<option value = "<?php echo $option['id'];?>"
} ?><?php echo $option['header'];?></option>
<?php } ?>
</select>
关于php - 为每个选项值命名,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42832503/