javascript - Ajax + php 不工作，php 工作但 javascript 不工作

Question

PHP 代码有效，但 javascript 无效。我无法理解他有什么问题。如何让它发挥作用？链接https://nice-host.com/domain

Javascript代码

             $("#src-btn").click(function() { 
                
                var domainname = $('#domain-name').val();
                
                $.ajax({
                    url: "domain-search.php",
                    type: 'post',
                    data: 'Domain '+domainname,
                    success: function(data)
                    {
                        output = '<div class="success">'+data.text+'</div>';
        
                        $("#result").hide().html(output).slideDown();           
                    }
                });
            });
            
            $("#domain-search input").keyup(function() { 
                $("#domain-search input").css('border-color',''); 
                $("#result").slideUp();
            });

        $domainname = $_POST["domain-name"];
        if (isset($_POST["domain-name"])) {
            if ( gethostbyname($domainname) != $domainname ) {
                echo "<script>alert('Found')</script>";
            }
            else {
                echo "<script>alert('Not found')</script>";
            }
        } else {
            $output = json_encode(array('type' => 'error', 'text' => 'Empty'));
            die($output);
        }

HTML 代码

                    <div id="result"></div>
                    <form class="domain-search-form parsley-validate" data-animate="fadeInUp" data-delay=".5" method="post" id="domain-search" name="domain-search">
                        <input type="text" placeholder="Domain name" id="domain-name" name="domain-name" required>
                        <button class="src-btn" name="src-btn" id="src-btn"><i class="fa fa-search"></i></button>
                    </form>

标签： javascriptphphtmlajax

Answer 1

您的请求没有发送正确的数据，domain-name是您需要发送的参数名称。

$.ajax({
    url: "domain-search.php",
    type: 'post',
    data: {'domain-name':domainname},
    success: function(data)
    {
        output = '<div class="success">'+data.text+'</div>';

        $("#result").hide().html(output).slideDown();           
    }
});