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Post方法不返回值
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Stack Overflow用户
提问于 2019-07-09 15:11:23
回答 1查看 139关注 0票数 0

我只想分享我的细节到另一个php文件使用POST与Ajax,在我的代码中的post值没有启动共享,

下面是我的代码:

代码语言:javascript
复制
<?php
include('database_connection.php');

if(isset($_POST["action"]))
{
    // some logic





    // some logic

    $statement = $connect->prepare($query);
    $statement->execute();
    $result = $statement->fetchAll();
    $total_row = $statement->rowCount();
    $output = '';
    if($total_row > 0)
    {
        foreach($result as $row)
        {
            $output .= '
            <div class="col-sm-3 col-lg-4 col-md-3">
                <div style="border:1px solid #ccc; border-radius:5px; padding:10px; margin-bottom:16px; height:300px;">
                    <audio controls controlsList="nodownload" style="padding: 10px 10px 10px 10px;margin-left: -21px;">
                        <source src="audio_sample/'. $row['voice_audio_file'] .'" alt="" class="img-responsive">
                    </audio>
                    <p align="center"><strong> '. $row['voice_name'] .'</strong></p>

                    <p style="font-size: 12px;">

                    Voice Id          : '. $row['voice_id'].' <br />
                    Voice Name        : '. $row['voice_name'].' <br />
                    Gender            : '. $row['voice_gender'].' <br />
                    Genres            : '. $row['voice_genres'].' <br />
                    Voice Modulation  : '. $row['voice_voice_modulation'].' <br />
                    Languages         : '. $row['voice_languages'].' <br />
                    Jingle Moods      : '. $row['voice_jingle_moods'].' <br />
                    Ivr               : '. $row['voice_ivr'].' <br /> </p>

                    <button type="button" class="btn btn-primary" style="padding: 5px 83px 5px 83px;
                    "data-voice-id="'.$row["voice_id"].'
                    " data-voice-name="'.$row["voice_name"].'
                    "data-voice-id="'.$row["voice_gender"].'">Add to Playlist</button>

                </div>
            </div>
            ';
        } 
    }
    else 
    {
        $output = '<h3>No Data Found</h3>';
    }
    echo $output;
}
?>
<script>
    $('.btn').on('click',function() {

        var voice_id = $(this).data("voice_id");
        var voice_name = $(this).data("voice_name");
        var voice_gender = $(this).data("voice_gender");

        $.ajax({
            method   : "POST",
            url      : "manage_cart.php",
            datatype : "text",
            data     : 
            { 
                voice_id: voice_id, voice_name: voice_name, voice_gender: voice_gender 
            },
            success: function(data)
            {   
                // console.log(data);
            console.log('success',data);


            }
        });
    });

</script>

我想要将voice_id、voice_name、voice_gender共享到manage_cart.php,我的manage_cart.php文件,

代码语言:javascript
复制
 <?php 
ini_set('display_startup_errors', 1); 
error_reporting(E_ALL); 

$voice_id = '';   
$voice_name = ''; 
$voice_gender = ''; 

if(isset($_POST['voice_id']) && isset($_POST['voice_name']) && isset($_POST['voice_gender']) )
{
    $voice_id   = $_POST['voice_id'];
    $voice_name = $_POST['voice_name'];
    $voice_name = $_POST['voice_gender'];
}
    echo $voice_id ."&". $voice_name ."&". $voice_gender;

 ?> 

在这里,我的post值没有传递给manage_cart.php。

我该如何解决这个问题?

EN

回答 1

Stack Overflow用户

发布于 2019-07-09 15:16:27

您的按钮属性没有data-voice-gender。有重复的data-voice-id,并使用与data-voice-id相同的变量样式,但在javascript中使用的是$(this).data('voice_id')

票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/56947118

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