Question

我无法在不离开当前页面的情况下尝试将数据发送到php页面，如果成功，则会显示从数据库加载的外部php页面的内容。我在下面列出了我目前的代码。请大家，我需要帮助，我将感激不尽。

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<form class="myform">
	  <input type="text" name="username" placeholder="Username" class="a" id="x" required ></input>
	  <input type="password" name="password" placeholder="Password" class="a" id="y" required></input>
	  <button class="b" id="n">Submit</button>
</form>
	
<div id="display"></div>

＆＃13;

以下是我的javascript

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$(document).ready(function() {
	
	$("button#n").click(function() {
		
	    var $username = $('input#x').val();
	    var $password = $('input#y').val();
		var $mydata = $('.myform').serialize();
	    
		if ($username != "" && $password != "") {
			
            $.ajax({url: 'http://localhost/products/s1.php'
			    type: 'POST',
		        data: $mydata,
				dataType: 'text',
		        success: function(data) {
			
			        $("#display").html(data);
		        }
		});
        }		 
	});
});
        		 
  });
});

＆＃13;

以下是我发布到`

的php页面的内容

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<?php
 require_once("data-var.php");

$mysqli = new mysqli($servername, $username, $password, $dbname);
if ($mysqli->connect_errno) {
	echo "Failed to connect to database: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}

if (!($stmt = $mysqli->prepare("SELECT username, password from Administrator WHERE password = (?)"))) {
	echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
if (isset($_POST['password'])) {
    $password = password_hash($_POST['password'], PASSWORD_DEFAULT);
}

if (!$stmt->bind_param("s", $password)) {
	echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
}

if (!$stmt->execute()) {
	echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}
if (!$stmt->store_result()) {
	echo "Failed to store result: (" .$stmt->errno . ") " . $stmt->error;
}
if ($stmt->num_rows > 0) {
	echo "welcome";
} else {
	echo 0;
}

// free result
$stmt->free_result();

// close statement
$stmt->close();

// close connection
$mysqli->close();

?>v

＆＃13;

`

Answer 1

将ajax请求数据设置为像这样的对象

data: {my data:$mydata,password:$password},

您的代码还存在许多其他问题

如何在不加载其他页面的情况下将数据发布到php

1 个答案: