我正在创建一个简单的表单，该表单具有"选择"菜单，需要使用SQL查询进行填充，该查询标识单个表和单个列中的所有唯一值。
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Bootstrap 101 Template</title>
<!-- Bootstrap -->
<link href="css/bootstrap.min.css" rel="stylesheet">
<link href="css/custom.css" rel="stylesheet">
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/font-awesome/4.3.0/css/font-awesome.min.css">
</head>
<body>
这是我想要的选择菜单的示例...
<form class="form-horizontal" role="form" method="post" action="db_test_bs.php">
<div class="form-group">
<label for="inputLocation" class="col-sm-2 control-label">Location</label>
<div class="col-sm-4">
<select class="form-control" id="inputLocation" name="inputLocation">
<option>Smith</option>
<option>Henry</option>
<option>Jackson</option>
<option>Hamilton</option>
</select>
</div>
</div>
这是我尝试使用查询填充的选择菜单...
<form class="form-horizontal" role="form" method="post">
<div class="form-group">
<label for="inputLocation" class="col-sm-2 control-label">Location</label>
<div class="col-sm-4">
<select class="form-control" id="inputLocation" name="inputLocation">
</select>
<?php
$servername = "localhost";
$username = "owner_1";
$password = "owner_1";
$dbname = "test1";
// Create Connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
trigger_error("Connection failed: " . mysqli_connect_error());
}
//Run Query
$stmt = "SELECT DISTINCT `CUST_NAME` FROM `customer` WHERE 1";
$result = mysqli_query($conn,$stmt) or die(mysqli_error());
while(list($category) = mysqli_fetch_row($result)){
$option = '<option value="'.$category.'">'.$category.'</option>';
echo ($option);
}
mysqli_close($conn);
?>
</div>
</div>
</form>
<!-- jQuery (necessary for Bootstrap's JavaScript plugins) -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
<!-- Include all compiled plugins (below), or include individual files as needed -->
<script src="js/bootstrap.min.js"></script>
</body>
</html>
上面代码的结果是两个选择菜单。 一个是硬编码的静态选项，另一个是"选择"菜单不包含任何选项。
使用 WAMP 和 BootStrap 3。
我是网络开发的新手，因此非常感谢任何帮助。
它应该是：
<select class="form-control" id="inputLocation" name="inputLocation">
...
//Your PHP Code
...
</select>
注意：在这种情况下，您实际上不需要在查询中WHERE 1。您可以直接回显结果，而无需将其转换为变量，如下所示：
echo '<option value="'.$category.'">'.$category.'</option>';
（如@Sean所述）你在PHP之前关闭了<select></select>标签，所以你的PHP循环在<select>元素的一侧运行。
此外，作为最佳实践，您应该通过将 $conn 变量传递到 mysqli_error 函数中来修复 mysqli_error() 语句，如下所示：
$result = mysqli_query($conn,$stmt) or die(mysqli_error($conn));
注意：您应该在开发环境中报告 PHP 错误。
这是更新的代码
<form class="form-horizontal" role="form" method="post">
<div class="form-group">
<label for="inputLocation" class="col-sm-2 control-label">Location</label>
<div class="col-sm-4">
<select class="form-control" id="inputLocation" name="inputLocation">
<?php
$servername = "localhost";
$username = "owner_1";
$password = "owner_1";
$dbname = "test1";
// Create Connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
trigger_error("Connection failed: " . mysqli_connect_error());
}
//Run Query
$stmt = "SELECT DISTINCT `CUST_NAME` FROM `customer` WHERE 1";
$result = mysqli_query($conn,$stmt) or die(mysqli_error($conn));
while(list($category) = mysqli_fetch_row($result)){
echo '<option value="'.$category.'">'.$category.'</option>';
}
mysqli_close($conn);
?>
</select>
</div>
</div>
</form>
</body>
</html>