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php - 使用jquery php mysql提交数据？

转载 作者：行者123 更新时间：2023-11-30 00:01:49 26 4
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创建新注册时我无法写入数据库......!我有一个 javascript，它同时具有登录和注册部分，如下所示...这是用于登录和注册的脚本和 php 脚本的代码的更新版本

    <!DOCTYPE html>
<html>
<head>
<title>Load </title>
<meta name="viewport" content="width=device-width, height=device-height, initial-scale=1.0"/>
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.3.2/jquery.mobile-1.3.2.min.css" />
<script src="http://code.jquery.com/jquery-1.9.1.min.js"></script>
<script src="http://code.jquery.com/mobile/1.3.2/jquery.mobile-1.3.2.min.js"></script>
<script src="js/index.js"></script>
</head>
<body>
<div data-role="page" id="login" data-theme="b">
    <div data-role="header" data-theme="a">
        <h3>Login Page</h3>
    </div>

    <div data-role="content">
        <form id="check-user" class="ui-body ui-body-a ui-corner-all" data-ajax="false">
            <fieldset>
                <div data-role="fieldcontain">
                    <label for="username">Enter your username:</label>
                    <input type="text" value="" name="username" id="username"/>
                </div>                                 
                <div data-role="fieldcontain">                                     
                    <label for="password">Enter your password:</label>
                    <input type="password" value="" name="password" id="password"/>
                </div>
                <input type="button" data-theme="b" name="submit" id="submit" value="Submit">
            </fieldset>
            <a href="#registerp" data-role="button">Register</a>
        </form>                             
    </div>

    <div data-theme="a" data-role="footer" data-position="fixed">

    </div>
</div>
<div data-role="page" id="registerp">
    <div data-theme="a" data-role="header">
        <h3>Register</h3>
    </div>
     <div data-role="content">
         <form id="registerform" class="ui-body ui-body-a ui-corner-all" data-ajax="false">
            <fieldset>
                <div data-role="fieldcontain">
                    <label for="fname">First Name:</label>
                    <input type="text" value="" name="fname" id="fname"/>
                </div>         
                <div data-role="fieldcontain">
                    <label for="lname">Last Name:</label>
                    <input type="text" value="" name="lname" id="lname"/>
                </div>  
                <div data-role="fieldcontain">
                    <label for="uname">User Name:</label>
                    <input type="text" value="" name="uname" id="uname"/>
                </div>                  
                <div data-role="fieldcontain">                                     
                    <label for="pwd">Enter your password:</label>
                    <input type="password" value="" name="pwd" id="pwd"/>
                </div>

                <div data-role="fieldcontain">
                    <label for="email">Email:</label>
                    <input type="text" value="" name="email" id="email"/>
                </div>
                <input type="button" data-theme="b" name="submit" id="register" value="Register">
            </fieldset>
        </form>      
    </div>

    <div data-theme="a" data-role="footer" data-position="fixed">
        <h3>Page footer</h3>
    </div>
</div>
<div data-role="page" id="second">
    <div data-theme="a" data-role="header">
        <h3>Welcome Page</h3>
    </div>

    <div data-role="content">
        Welcome
    </div>

    <div data-theme="a" data-role="footer" data-position="fixed">
        <h3>Page footer</h3>
    </div>
</div>
<script type="text/javascript">
    $(document).on('pageinit', '#login', function(){ 
    $(document).on('click', '#submit', function() { // catch the form's submit event
        if($('#username').val().length > 0 && $('#password').val().length > 0){
            // Send data to server through the ajax call
            // action is functionality we want to call and outputJSON is our data
                $.ajax({url: 'check.php',
                    data: "action=login&" + $('#check-user').serialize(),
                    type: 'post',                  
                    async: 'true',
                    dataType: 'json',
                    beforeSend: function() {
                        // This callback function will trigger before data is sent
                        $.mobile.showPageLoadingMsg(true); // This will show ajax spinner
                    },
                    complete: function() {
                        // This callback function will trigger on data sent/received complete
                        $.mobile.hidePageLoadingMsg(); // This will hide ajax spinner
                    },
                    success: function (result) {
                        if(result.status) {

                            $.mobile.changePage("#second"); 

                        } else {
                            alert('Log on unsuccessful!');
                        }
                    },
                    error: function (request,error) {
                        // This callback function will trigger on unsuccessful action               
                        alert('Network error has occurred please try again!');
                    }
                });                  
        } else {
            alert('Please fill all necessary fields');
        }          
        return false; // cancel original event to prevent form submitting
    });   
    });

</script>
<script type="text/javascript">
    $(document).on('pageinit', '#registerp', function(){ 
    $(document).on('click', '#register', function() {
    if($('#uname').val().length > 0 && $('#pwd').val().length > 0){
            // Send data to server through the ajax call
            // action is functionality we want to call and outputJSON is our data
                $.ajax({url: 'insert.php',
                    data: "action=register&" + $('#registerform').serialize(),
                    type: 'post',                  
                    async: 'true',
                    dataType: 'json',
                    beforeSend: function() {
                        // This callback function will trigger before data is sent
                        $.mobile.showPageLoadingMsg(true); // This will show ajax spinner

                    },
                    complete: function() {
                        // This callback function will trigger on data sent/received complete
                        $.mobile.hidePageLoadingMsg(); // This will hide ajax spinner
                    },
                    success: function (result) {
                        if(result.status) {

                            $.mobile.changePage("#second"); 

                        } else {
                            alert(' Try again later ! Server is busy !');
                        }
                    },
                    error: function (request,error) {
                        // This callback function will trigger on unsuccessful action               
                        alert('Network error has occurred please try again!');
                    }
                });                  
        } else {
            alert('Please fill all necessary fields');
        }          
        return false; // cancel original event to prevent form submitting
    });
    });
    </script>
</body>
</html>

While my PHP Script is simple as shown below... please help


    <?php

    $con=mysqli_connect("...............", "...........", ".........","........");
    // Check connection
    if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    // escape variables for security
    $fname = mysqli_real_escape_string($con, $_POST['fname']);
    $lname = mysqli_real_escape_string($con, $_POST['lname']);
    $uname = mysqli_real_escape_string($con, $_POST['uname']);
    $email = mysqli_real_escape_string($con, $_POST['email']);
    $password = mysqli_real_escape_string($con, $_POST['pwd']);

    $action = $_POST['action'];
    // Decode JSON object into readable PHP object
     //$formData = json_decode($_POST['formData']);

    $sql="INSERT INTO userdb (username, fname, lname, password, email) VALUES ('$uname', '$fname', '$lname', '$password','$email')";

    if (!mysqli_query($con,$sql)) {
    die('Error: ' . mysqli_error($con));
    }
    echo "1 record added";

    mysqli_close($con);

    if($action == 'register'){
    $output = array('status' => true, 'message' => 'Registered');
    }

    echo json_encode($output);
    ?>

Insert php script doesnt work while the below register php script works fine.

<?php   
// We don't need action for this tutorial, but in a complex code you need a way to determine Ajax action nature
 $action = $_POST['action'];
// Decode JSON object into readable PHP object
 //$formData = json_decode($_POST['formData']);

// Get username
$username = $_POST['username'];
// Get password
$password = $_POST['password'];

$db = @mysql_connect('..........', '........', '..........') or die("Could not connect database");
@mysql_select_db('users', $db) or die("Could not select database");


$result = mysql_query("SELECT `password` FROM `userdb` WHERE `username`= '$username'");
$r = mysql_fetch_assoc($result);
$pass_ret = $r['password'];

// Lets say everything is in order
if($action == 'login' && $password == $pass_ret){
$output = array('status' => true, 'message' => 'Login');
}
else
{
$output = array('status' => false, 'message' => 'No Login');

}
echo json_encode($output);

？>

最佳答案

您使用插入语句的方式是错误的......并且它被错误地封装为 @rbcummings 所说的。

你必须改变

  $sql="INSERT INTO userdb (username, fname, lname, password, email) VALUES ('$uname', '$fname', '$lname', '$password','$email')";

到

$sql="INSERT INTO userdb (username, fname, lname, password, email) VALUES (".$uname.", ".$fname.", ".$lname.", ".$password.",".$email.")";

如果没有正确的意图，您可能会遇到错误。因此，意图您的代码可以解决您的问题。

关于php - 使用jquery php mysql提交数据？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24966371/

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