将来自ajax调用的PHP链接数据发送到不同的页面onclick

Question

我忙着学习ajax.我有一个预订页面和一个结账页面.

我想做的事

用户将租用者名称键入搜索框,匹配的名称将从db中获取并动态显示.

像这样:

reservations.php

当用户点击突出显示的链接时,我希望将预订ID传递到结帐页面.那么,我正在寻找一种方法将预订ID传递到另一个/不同的页面,我可以在结帐页面上进行数据处理.

checkout.php

码

SEARCH/JQUERY

$(document).ready(function () {
        $("#searchbox").on('keyup',function () {
            var key = $(this).val();

            $.ajax({
                url:'fetch.php',
                type:'GET',
                data:'keyword='+key,
                beforeSend:function () {
                    $("#results").slideUp('fast');
                },
                success:function (data) {
                    $("#results").html(data);
                    $("#results").slideDown('fast');
                }
            });
        });
    });
</script>
<div id="main">
    <div id="header"><h1>Find Names</h1></div>
    <div id="content">
        <input type="search" name="keyword" placeholder="Search Names" id="searchbox">
        <div id="results"></div>
    </div>
</div>

FETCH.PHP

  if($_GET['keyword'] && !empty($_GET['keyword']))
    {
        $conn = mysqli_connect('localhost','root','','*****'); //Connection to my database
        $keyword = $_GET['keyword'];
        $keyword="%$keyword%";
        $query = "select renter_name from reservations where renter_name like ?";
        $statement = $conn->prepare($query);
        $statement->bind_param('s',$keyword);
        $statement->execute();
        $statement->store_result();
        if($statement->num_rows() == 0) // so if we have 0 records acc. to keyword display no records found
        {
            echo '<div id="item">Ah snap...! No results found :/</div>';
            $statement->close();
            $conn->close();

        }
        else {
            $statement->bind_result($name);
            while ($statement->fetch()) //outputs the records
            {
                echo "<div id='item'>$name</div>";
            };
            $statement->close();
            $conn->close();
        };
    };

D B

任何有关实施的帮助或建议都非常感谢!

Answer 1

只需使用链接和查询.首先回显链接而不是div:

echo "<a href='yourpage.php?id=$name'>$name</a>";

在您的页面上只需:

<?php
echo $_GET[" id"];
?>