无法在php中将数据加载到select标记的值中
我正在尝试从php中的无法在php中将数据加载到select标记的值中,php,Php,我正在尝试从php中的select标记中获取值。我从数组中获取foreach数据,但它不起作用。如何正确地工作 index.php <?php require 'app/config.php';?> <!-- connect database --> <?php require 'app/result.php'; ?> <?php require 'app/function.php'; ?> <!DOCTYPE html> <h
select
标记中获取值。我从数组中获取foreach
数据,但它不起作用。如何正确地工作
index.php
<?php require 'app/config.php';?> <!-- connect database -->
<?php require 'app/result.php'; ?>
<?php require 'app/function.php'; ?>
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<head>
<body>
<form action="" method="post">
Khối thi:
<?php $values = &OptionValues(); ?>
<select class="select">
<?php foreach($values as $value):?>
<option value="<?php $value['khoi_thi']; ?>"></option>
<?php endforeach;?>
</select>
<br/>
</form>
</body>
</html>
Khố我认为:
我想要的结果:
<select>
<option value="A1"></option>
<option value="A2"></option>
<option value="A3"></option>
<!-- 3 results if database has 3 rows -->
</select>
您应该知道您使用的驱动程序已被弃用: 我将用mysqli驱动程序(实际驱动程序)以这种简单的方式完成这项工作:
注意:未定义变量:第29行C:\xampp\htdocs\app\index.php中的mysqli致命错误:对第29行C:\xampp\htdocs\app\index.php中的非对象调用成员函数query()
第29行:
@Jr Jimnzadded了不推荐使用的函数
<select>
<option value="A1"></option>
<option value="A2"></option>
<option value="A3"></option>
<!-- 3 results if database has 3 rows -->
</select>
<?php
// inside config.php
$mysqli=new mysqli("localhost", "root","password_here","whatever_db");
if (mysqli_connect_errno()) {
echo 'Database Conection Failed: ', mysqli_connect_error();
exit;
}
// end config.php
$sql = "SELECT khoi_thi FROM 2014_khoathi";
$query = $mysqli->query($sql);
?>
<select class = "select">
<?php
while($row = $query->fetch_assoc()) {
echo "<option value = '{$row['khoi_thi']}'> {$row['khoi_thi']} </option>";
}
?>
</select>
<?php
// inside config.php
$mysql=mysql_connect('localhost', 'user_mysql', 'password_mysql');
if (!$mysql) {
echo 'Database Conection Failed: ';
exit;
}
// end config.php
$sql = "SELECT khoi_thi FROM 2014_khoathi";
$query = mysql_query($sql, $mysql);
?>
<select class = "select">
<?php
while($row = mysql_fetch_assoc($query)) {
echo "<option value = '{$row['khoi_thi']}'> {$row['khoi_thi']} </option>";
}
?>
</select>