修改使用PHP和AJAX在数据库中添加当前页面的ID - Modification to add in a DB using PHP and AJAX the ID of the current page

Modification to add in a DB using PHP and AJAX the ID of the current page

我使用一个表单使用AJAX发布到我的数据库的文本区域#inpitField的内容。当前url为domain.com/product.php?id=1

即使我正确地将数据添加到DB，我也想发布页面的ID。但是我遇到了一些问题。为了更好地理解，我已经发布了整个代码。

谢谢你的帮助。

在我的index.php我有这个表单。

<form id="tweetForm" action="submit.php" method="post">
<textarea name="inputField" id="inputField"></textarea>
<input class="submitButton inact" name="submit" type="submit" value="update"/>
</form>

我通过AJAX发送文本区域#inputField的内容。js

function tweet()
{
    var submitData = $('#tweetForm').serialize();
    $('.counter').html('<img src="img/ajax_load.gif" width="16" height="16" style="padding:12px" alt="loading" />');
    $.ajax({
        type: "POST",
        url: "submit.php",
        data: submitData,
        dataType: "html",
        success: function(msg){
            if(parseInt(msg)!=0)
            {
                $('ul.statuses li:first-child').before(msg);
                $("ul.statuses:empty").append(msg);
                $('#lastTweet').html($('#inputField').val());
                $('#inputField').val('');
                recount();
            }
        }
    });
}

和我的submit.php文件是

mysql_query("INSERT INTO offers SET tweet='".$_POST['inputField']."',dt=NOW(),company_id=".$myid.",product_id=".I WANT TO ADD THE PRODUCT ID." ");

function tweet()
{
    var urlid = window.location.href.split('?id=', 2)[1];
    $('#tweetForm').append('<input type="text" name="id" value="'+urlid+'" />');
    var submitData = $('#tweetForm').serialize();
    $('.counter').html('<img src="img/ajax_load.gif" width="16" height="16" style="padding:12px" alt="loading" />');
    $.ajax({
        type: "POST",
        url: "submit.php",
        data: submitData,
        dataType: "html",
        success: function(msg){
            if(parseInt(msg)!=0)
            {
                $('ul.statuses li:first-child').before(msg);
                $("ul.statuses:empty").append(msg);
                $('#lastTweet').html($('#inputField').val());
                $('#inputField').val('');
                recount();
            }
        }
    });
}

您应该能够像这样将您的页面id附加到您发布的数据:

submitData = submitData + "&pageid=" + <?php echo $_GET['id'] ?>;

对于你的代码:

function tweet(var pageID)
{
    var submitData = $('#tweetForm').serialize();
    submitData = submitData + "&pageid=" + pageID;

    $('.counter').html('<img src="img/ajax_load.gif" width="16" height="16" style="padding:12px" alt="loading" />');
    $.ajax({
        type: "POST",
        url: "submit.php",
        data: submitData,
        dataType: "html",
        success: function(msg){
            if(parseInt(msg)!=0)
            {
                $('ul.statuses li:first-child').before(msg);
                $("ul.statuses:empty").append(msg);
                $('#lastTweet').html($('#inputField').val());
                $('#inputField').val('');
                recount();
            }
        }
    });
}

称为:

tweet("<?php echo $_GET['id'] ?>")