Question

我有4个文本框，它们将url作为输入。如果用户全部跳过，则会抛出错误以填充至少一个输入。还要验证用户输入，例如它是否是有效的URL。如何实现这个Codeigniter 3.x？

Answer 1

你必须使用input type =“url”，并输入验证（为此添加validate.js文件），如下所示：

你可以从这里获得validation.js：Validation.js

<!DOCTYPE html>
<html lang="en">
<head>
    <title>Add User</title>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"> </script>
</head>
<body class="backgroundfff">
    <section>
        <div class="padding0 main_padding" id="main_container">
            <div class="container-fluid padding0 bg-clr">
                <div class="container backgroundfff user-porfile-padding">
                    <div class="col-lg-12 col-md-12 col-sm-9 col-xs-12 margintop20px marginbottom45px">
                        <div class="col-lg-3 col-md-3 col-sm-3 col-xs-12 padding0"></div>
                        <div class="cp-box col-lg-6 col-md-6 col-sm-6 col-xs-12">
                            <?php
                            $add_form  =
                            array(
                                'name'    =>  'advertiser-form',
                                "id"    =>  "validateForm",
                                'method'  =>  'post',
                                'class'     =>  "form-horizontal",
                                'role'  =>'form'
                                );
                            echo form_open_multipart(DOMAIN_URL."users/adduser",$add_form);
                            ?>
                            <div class="col-lg-12 col-md-12 col-sm-12 col-xs-12 cyp-sub-title margintop20px">
                                <input type="url" placeholder="Textbox 1" name="Textbox1" id="Textbox1" class="">
                            </div>

                            <div class="col-lg-12 col-md-12 col-sm-12 col-xs-12 cyp-sub-title margintop20px">
                                <input type="url" placeholder="Textbox 2" name="Textbox2" id="Textbox2" class="">
                            </div>

                            <div class="col-lg-12 col-md-12 col-sm-12 col-xs-12 cyp-sub-title margintop20px">
                                <input type="url" placeholder="Textbox 3" name="Textbox3" id="Textbox3" class="">
                            </div>

                            <div class="col-lg-12 col-md-12 col-sm-12 col-xs-12 cyp-sub-title margintop20px">
                                <input type="url" placeholder="Textbox 4" name="Textbox4" id="Textbox4" class="">
                            </div>

                            <div class="col-lg-12 col-md-12 col-sm-12 col-xs-12 padding0 margintop20px marginbottom20px text-center">
                                <button type="submit" class="cp-submit">Submit</button>
                            </div>

                            <?php echo form_close(); ?>

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

    </section>

</body>
<script src="<?=DOMAIN_URL?>/js/validate.js" type="text/javascript" charset="utf-8"></script>    
</html>

服务器端验证：

<?php    
function adduser($postData)
{
    if(empty($postData['Textbox1']) || empty($postData['Textbox2']) || empty($postData['Textbox3']) || empty($postData['Textbox4'])) {
        // redirect user to add form
    }
}
?>

您可以使用 FILTER_SANITIZE_URL 检查网址是否有效，如下所示：

filter_var($posted_url, FILTER_SANITIZE_URL);

codeigniter 3中的多个文本框验证

1 个答案: