使用带有php

时间:2016-08-02 17:07:37

标签: php jquery mysql ajax

我正在尝试使用日期类型获取有关日期更改的数据,从数据库中获取数据。但在此之后,获取的数据包含要单击的按钮。点击按钮再次获取数据。

这是代码

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Untitled Document</title>

</head>
<body>
    <form method="post">
        <input type="date" name="gdate" id="gdate" onchange="<?php echo $_SERVER['PHP_SELF'] ?>"/>
    </form>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
    <script>
        function makeAjaxRequest(dateval) {
            $.ajax({
                type: "GET",
                data: {dateva: dateval},
                url: "fetchbills.php",
                success: function (data) {
                    $('#showresult').html(data);
                    $('#det').on("click", function () {
                        var no = $(this).val();
                        makeAjaxRequest2(no);
                    });
                }
            });
        }
        function makeAjaxRequest2(billno) {
            $.ajax({
                type: "GET",
                data: {bno: billno},
                url: "fetchbilldetail.php",
                success: function (data) {
                    $('#showresult').html(data);
                }
            });
        }
        $("#gdate").on("change", function () {
            var id = $(this).val();
            makeAjaxRequest(id);

        });
    </script>
    <div id="showresult">

    </div>
</body>
</html>

fetchbills.php

<?php

include("./config.php");
$dt = $_REQUEST["dateva"];
$sql = "select bill_id,order_id from bill_master where order_date='$dt'";
echo $sql;
$result = mysqli_query($con, $sql) or die("error");
if ($result) {
    while ($row = mysqli_fetch_array($result)) {
        echo $row[0] . "<br>";
        echo "<form method='post' action=''><input type='submit' name='det' value='" . $row[1] . "' onclick='get(this.value)' /></form>
    <div id='show'></div>";
    }
}
?>

使用上面我得到的订单ID。身份证以按钮的形式显示。 点击按钮我想获得订单ID的详细信息。但是我做了什么sholud我无法理解。

我是ajax的新手,不知道该怎么做。

这里是订单详情的代码

<?php

include("./config.php");
$det = $_GET["bno"];
//echo $det;
$sql = "SELECT m.old_order_id,m.old_order_amount,l.item_id,l.old_order_quantity,i.item_name from old_order_master m,old_order_list l,item_list i where m.old_order_id=$det and m.old_order_id=l.old_order_id and l.item_id=i.item_id";
//echo $sql;
$result = mysqli_query($con, $sql);
echo "<table><tr><th>Item Name</th><th>Quantity</th></tr>";
while ($row = mysqli_fetch_array($result)) {

    echo "<tr><td>" . $row[4] . "</td><td>" . $row[3] . "</td></tr>";
}
echo "</table>";
?>

1 个答案:

答案 0 :(得分:1)

fetchbills.php 文件中进行以下更改,然后告诉我们

<?php

include("./config.php");

$dt = $_REQUEST["dateva"];
$sql = "select bill_id,order_id from bill_master where order_date='$dt'";

$result = mysqli_query($con, $sql) or die("error");

if ($result) {
    while ($row = mysqli_fetch_array($result)) {

        echo $row[0] . "<br>";
        echo "<form method='post' action=''>"
            ."<input type='button' name='det' value='" . $row[1] . "' onclick='makeAjaxRequest2(".$row[1].");' /></form>"
            ."<div id='show'></div>";
    }
}