我有一个包含下拉列表的表单。我希望用户应该能够从该下拉列表中选择多个值
<form class="form-horizontal" role="form" action="add.php" enctype="multipart/form-data" method="post">
<div class="form-group">
<label class="col-lg-4 control-label">Name</label>
<div class="col-lg-6">
<input class="form-control" value="" type="text" name="name" >
</div>
</div>
<div class="form-group">
<label class="col-lg-4 control-label">Address</label>
<div class="col-lg-6">
<input class="form-control" value="" type="text" name="address" >
</div>
</div>
<div class="form-group">
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db";
// Create connection
$con = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$con) {
die("Connection failed: " . mysqli_connect_error());
}
echo "<label class='col-lg-4 control-label'>Student</label>";
echo "<div class='col-lg-6'>";
$sql = "SELECT student FROM student";
$result = $con->query($sql);
echo "<select class='form-control' name='student' multiple>";
while($row = $result->fetch_assoc()) {
echo "<option value='" . $row['student'] . "'>" . $row['student'] . "</option>";
}
echo "</select>";
echo"</div>";
?>
</div>
<div class="form-group">
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db";
// Create connection
$con = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$con) {
die("Connection failed: " . mysqli_connect_error());
}
echo "<label class='col-lg-4 control-label'>Subject</label>";
echo "<div class='col-lg-6'>";
$sql = "SELECT subject FROM subject";
$result = $con->query($sql);
echo "<select class='form-control' name='subject' multiple>";
while($row = $result->fetch_assoc()) {
echo "<option value='" . $row['subject'] . "'>" . $row['subject'] . "</option>";
}
echo "</select>";
echo"</div>";
?>
</div>
<div class="form-group">
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db";
// Create connection
$con = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$con) {
die("Connection failed: " . mysqli_connect_error());
}
echo "<label class='col-lg-4 control-label'>Hobby</label>";
echo "<div class='col-lg-6'>";
$sql = "SELECT hobby FROM hobby";
$result = $con->query($sql);
echo "<select class='form-control' name='hobby' multiple>";
while($row = $result->fetch_assoc()) {
echo "<option value='" . $row['hobby'] . "'>" . $row['hobby'] . "</option>";
}
echo "</select>";
echo"</div>";
?>
</div>
<div class="form-group">
<label class="col-md-3 control-label"></label>
<div class="col-md-8">
<input class="btn btn-primary" value="Save Changes" type="submit" name="submit">
</div>
</div>
</form>
添加.php
<?php
include('admin_session.php');
$con=mysqli_connect("localhost","root","","db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$name = mysqli_real_escape_string($con, $_POST['name']);
$address = mysqli_real_escape_string($con, $_POST['address']);
$student = mysqli_real_escape_string($con, $_POST['student']);
$subject = mysqli_real_escape_string($con, $_POST['subject']);
$hobby = mysqli_real_escape_string($con, $_POST['hobby']);
$sql="INSERT INTO class (name,address,student,subject,hobby) VALUES ('$name','$address','$student','$subject''$hobby')";
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
header("Location: list.php");
mysqli_close($con);
exit;
?>
我希望从下拉列表中选择多个值后,它们应该保存在具有名为
class
的表的数据库中.名为类的表的 View 是id name address student subject hobby
1 a s t y j
2 b d i g d
问题是,虽然我可以选择多个值,但只有一个值存储在数据库中
P.S 我想我应该以更好的方式解释我的问题,所以我更新了我的帖子
最佳答案
将值存储在选择选项中的数组中
echo "<select class='form-control' name='student[]' multiple>"//it will store selected value in array
在循环内运行您的查询
foreach ($_POST['student'] as $students)
{
$student = mysqli_real_escape_string($con, $students);//use mysqli escape here
$sql="INSERT INTO class (students) VALUES ('$student')";
}
在关闭
form
之前添加此行在您的 while loop
之后<input type="submit" name="submit" value="submit">
并记住您正在使用
mysqli
使用 bindparam
以及 mysqli
不会自动保护您。检查此链接 MySQL vs MySQLi when using PHP
关于php - 将下拉菜单中的多个值保存到数据库中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26776423/