数据库值在php中查找Json格式

时间:2016-01-11 13:43:58

标签: php jquery json gridview

我在网格中使用了jquery网格值查找值。值从数据库中检索。我使用了我的Sample静态数据,我以前的设计json格式。看起来像

 $(document).ready(function() {

        var jqgrid_data = [{
            id : "1",
            date : "2007-10-01",
            name : "test",
            note : "note",
            amount : "200.00",
            tax : "10.00",
            total : "210.00"
        }]

}

但我希望从数据库Json格式中检索值。我的代码看起来像这样

                                <div class="form-group">

                                    <label for="pn">Title</label>

                                    <input type="text" class="form-control" id="Title<?php echo $row['ID']; ?>" value="<?php echo $row['Title']; ?>" />

                                </div>

                                <div class="form-group">

                                    <label for="al">Tag Line</label>

                                    <input type="text" class="form-control" id="TageLine<?php echo $row['ID']; ?>" value="<?php echo $row['TagLine']; ?>" />

                                </div>

                                <div class="form-group">

                                    <label for="al">Banner URL</label>

                                    <input type="text" class="form-control" id="BannerURL<?php echo $row['ID']; ?>" value="<?php echo $row['BannerURL']; ?>" />

                                </div>

                                <div class="form-group">

                                    <label for="al">Icon URL</label>

                                    <input type="text" class="form-control" id="IconURL<?php echo $row['ID']; ?>" value="<?php echo $row['IconURL']; ?>" />

                                </div>

                                <div class="form-group">

                                    <label for="al">Video Announcement URL</label>

                                    <input type="text" class="form-control" id="VideoAnnouncementURL<?php echo $row['ID']; ?>" value="<?php echo $row['VideoAnnouncementURL']; ?>" />



get Data.php 



  </div>

                            <div class="form-group">

                                <label for="al">Video Description</label>

                                <input type="text" class="form-control" id="VideoDescription<?php echo $row['ID']; ?>" value="<?php echo $row['VideoDescription']; ?>" />

                            </div>

                            <div class="form-group">

                                <label for="al">Order</label>

                                <input type="text" class="form-control" id="Order<?php echo $row['ID']; ?>" value="<?php echo $row['Order']; ?>" />

                            </div>

                        </form>

这个值我发现我看起来像脚本

  function viewdata() {

        $.ajax({

            type: "GET",

            url: "inc/getdata.php"

        }).done(function (data) {

            $('#viewdata').html(data);

        });

    }

    $('#save').click(function () {



        var nm = $('#nm').val();

        var gd = $('#gd').val();

        var pn = $('#pn').val();

        var al = $('#al').val();



        var datas = "nm=" + nm + "&gd=" + gd + "&pn=" + pn + "&al=" + al;



        $.ajax({

            type: "POST",

            url: "inc/newdata.php",

            data: datas

        }).done(function (data) {

            $('#info').html(data);

            viewdata();

        });

    });

    function updatedata(str) {



        var id = str;

        var nm = $('#nm' + str).val();

        var gd = $('#gd' + str).val();

        var pn = $('#pn' + str).val();

        var al = $('#al' + str).val();



        var datas = "nm=" + nm + "&gd=" + gd + "&pn=" + pn + "&al=" + al;



        $.ajax({

            type: "POST",

            url: "inc/updatedata.php?id=" + id,

            data: datas

        }).done(function (data) {

            $('#info').html(data);

            viewdata();

        });

    }

    function deletedata(str) {



        var id = str;



        $.ajax({

            type: "GET",

            url: "inc/deletedata.php?id=" + id

        }).done(function (data) {

            $('#info').html(data);

            viewdata();

        });

    }

update.php

 <?php

require("inc/mysqli_connect.php");

if(isset($_GET['ID'])){

    $stmt = $conn->prepare("UPDATE video_parent SET Title=?, TagLine=?, BannerURL=?, IconURL=?, VideoAnnouncementURL=?, VideoDescription=?, Order=? WHERE ID=?");

    $stmt->bind_param('sssss', $title, $tagLine, $bannerURL, $iconURL, $videoAnnouncementURL, $videoDescription, $order, $id);



    $title = $_POST['Title'];

    $tagLine = $_POST['TagLine'];

    $bannerURL = $_POST['BannerURL'];

    $iconURL = $_POST['IconURL'];

    $videoAnnouncementURL = $_GET['VideoAnnouncementURL'];

    $videoDescription = $_GET['VideoDescription'];

    $order = $_GET['Order'];

    $id = $_GET['ID'];



    if($stmt->execute()){

?>

<div class="alert alert-success alert-dismissible" role="alert">

    <button type="button" class="close" data-dismiss="alert" aria-label="Close">

        <span aria-hidden="true">&times;</span>

    </button>

    <strong>Success!</strong>

    Anda berhasil mengubah data.

</div>

<?php

    } else{

?>

<div class="alert alert-danger alert-dismissible" role="alert">

    <button type="button" class="close" data-dismiss="alert" aria-label="Close">

        <span aria-hidden="true">&times;</span>

    </button>

    <strong>Error!</strong>

    Maaf terjadi kesalahan, data error.

</div>

<?php

    }

} else{

?>

<div class="alert alert-warning alert-dismissible" role="alert">

    <button type="button" class="close" data-dismiss="alert" aria-label="Close">

        <span aria-hidden="true">&times;</span>

    </button>

    <strong>Warning!</strong>

    Maaf anda salah alamat.

</div>

<?php

}

?>

我想要价值Json格式?请任何人帮助我?

1 个答案:

答案 0 :(得分:0)

如果我理解正确,您希望以jSON格式将数据发送到PHP以及urlencode帖子格式吗?

function viewdata() {
    $.ajax({
        type: "GET",
        url: "inc/getdata.php"
    }).done(function (data) {
        $('#viewdata').html(data);
    });
}

$('#save').click(function () {
    var datas ={
        nm: $('#nm').val(),
        gd: $('#gd').val(),
        pn: $('#pn').val(),
        al: $('#al').val()
    } ;
    $.ajax({
        type: "POST",
        url: "inc/newdata.php",
        contentType: 'application/json',
        data: datas
    }).done(function (data) {
        $('#info').html(data);
        viewdata();
    });
});

function updatedata(str) {
    var datas ={
        id: str,
        nm: $('#nm' + str).val(),
        gd: $('#gd' + str).val(),
        pn: $('#pn' + str).val(),
        al: $('#al' + str).val()
    } ;
    $.ajax({
        type: "POST",
        url: "inc/updatedata.php",
        contentType: 'application/json',
        data: datas
    }).done(function (data) {
        $('#info').html(data);
        viewdata();
    });

}
在PHP方面

,使用$ _POST正文中的json_decode()

$inputJSON =file_get_contents ('php://input') ;
$input =json_decode ($inputJSON) ;

现在,如果您想将PHP的JSON回复返回到您的javascript代码:

header ('Content-Type: application/json') ;
echo json_encode (mydata) ;
exit ;

并在javascript中

}).done(function (data) {
    data =JSON.parse (data) ;
    ...
}) ;

我还对你的'url:'inc / updatedata.php“,'进行了更改,因为我不明白为什么要在帖子正文和查询参数之间拆分数据。他们可以保持一致,以保持一致性和代码可读性。

最后,当您构建编码参数字符串时(即var datas =“nm =”+ nm +“&amp; gd =”+ gd +“&amp; pn =”+ pn +“&amp; al =”+ al ;),确保使用encodeURIComponent()

编码值