[英]Inserting data into MySQL using AJAX and PHP
我试图使用此代码在mysql中的表中插入名称和密码。 它会生成成功警报,但不会将数据添加到表中。 有人可以解决这个问题吗?
我的html文件-
<html> <body> <table border="1"> <tr> <td align="center">Test Form</td> </tr> <tr> <td> <table> <form onsubmit="return false" id="frm"> <tr> <td>Name</td> <td><input type="text" id="name" name="name" size="50"> </td> </tr> <tr> <td>Password</td> <td><input type="text" id="password" name="password" size="50"> </td> </tr> <tr> <td></td> <td align="right"> <input type="submit" id="btnSubmit" name="submit" value="Send"> </td> </tr> </form> </table> </td> </tr> </table> </body> </html>
和javascript-我想在不加载任何其他页面的情况下进行插入(停留在同一html页面上)。 此ajax部分会弹出警报,但数据不会添加到表中。
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js"></script> <script> $("#frm").submit(function() { var name= $("#name").val(); var password= $("#password").val(); $.ajax({ type: "POST", url: "db.php", data: "name=" + name+ "&password=" + password, success: function(data) { alert("success!"); } }); }); </script>
最后是PHP
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dynamic";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$order = "INSERT INTO userdata (name, password) VALUES ('$_POST[name]','$_POST[password]')";
$result = $conn->query($order);
if($result){
echo("Successfully added!");
} else{
echo("Input failed!");
}
?>
试试这个$ frm到#btnSubmit
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js"></script>
<script>
$("#btnSubmit").click(function() {
var data = {name: $("#name").val(), password: $("#password").val()'}
$.ajax({
type: "POST",
url: "db.php",
data: data,
success: function(data) {
alert("success!");
}
});
});
</script>
更改HTML代码:
<form onsubmit="return false" id="frm" method="POST">
</form>
添加表单方法POST
更改您的PHP代码:
$name = $_POST["name"];
$password = $_POST["password"];
$sql = "INSERT INTO userdata (name, password)
VALUES ($name, $password)";
并关闭您的连接
$conn->close();
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