如何在单击按钮时将数据插入数据库？

Posted 2023-02-24

【中文标题】如何在单击按钮时将数据插入数据库？

【英文标题】：How to insert data in to database on button click?

【发布时间】：2015-03-30 11:28:53

【问题描述】：

我有一个模式，它读取如下代码之类的值。当我单击模式中的提交按钮时，是否可以将 $row3['first_name'] 等值插入数据库？我对此进行了研究，但我没有看到太多关于它的信息，因为它们大多使用文本字段。

<a href="#" class="btn btn-success hidden" id="request" data-toggle="modal" name="req" data-target="#basicModal">Request from Lender</a>
   </br></br>  

   <form action="" method="post">
<div class="modal fade" id="basicModal" tabindex="-1" role="dialog" aria-labelledby="basicModal" aria-hidden="true">
    <div class="modal-dialog">
        <div class="modal-content">
            <div class="modal-header">
                <button type="button" class="close" data-dismiss="modal" aria-hidden="true">x</button>           
                <h4 class="modal-title" id="myModalLabel"><?php
print $row3['title'] . ' from ' . $row3['first_name'] . ' ' . $row3['last_name'];
?></h4>
            </div>
            <div class="modal-body">
                <p>Are you sure you want to borrow <b> 1 x <?php
print $row3['title'];
?></b> from <b> <?php
print $row3['first_name'];
?> <?php
print $row3['last_name'];
?></b> for <b id="total"></b> cogs? 
                    </br></br> You will be required to collect and return this item to <?php
print $row3['houseno'];
?> <?php
print $row3['address1'];
?> <?php
print $row3['postcode'];
?>.
                    </br></br> <?php
print $_GET[$from];
?> Your cogs will not be transferred to <?php
print $row3['first_name'];
?> until you have confirmed you wish to proceed following the lenders decision!</p>
            </div>

            <div class="modal-footer">
                <button type="submit" value="submit" class="btn btn-success">Register</button>
                <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
        </div>


    </div>
  </div>
</div>
       </form>

【参考方案1】：

您需要调用一个方法，将要插入到数据库中的值传递给。

例子：

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) 
    die("Connection failed: " . $conn->connect_error);
 

$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', '[e-mail]')";

if ($conn->query($sql) === TRUE) 
    echo "New record created successfully";
 else 
    echo "Error: " . $sql . "<br>" . $conn->error;


$conn->close();
?>

【讨论】：

我该怎么做呢？ :S

谢谢 - 我最终还是这样做了... if(isset($_POST['button1'])) $q = mysql_query("INSERT INTO table (keys) VALUES (' $values')") 或死(mysql_error()); if($q) echo "成功"; else 回声“错误”；