Question

我想使用AJAX在iframe中显示图片，但我的整页每5秒加载一次，如果状态表中有任何更改，我只想加载内容。

我想我已经对部分做了些什么。可能是我需要将PHP代码转移到某些地方或者我需要在DIV上进行一些更改。如果有人请帮助我？

main.php

<!DOCTYPE html>
<html>
<head>

<script src="http://code.jquery.com/jquery-1.11.1.min.js"></script>
<script>
    $(function(){
        function status(){
           $.ajax({
               type: "GET", 
               url: "main.php",
               success: function(data){
                  $('div.divGranite').html(data);
               }
           });
       };
       setTimeout(status, 5000);
    }); 
</script>


</head>
<body>

<div id="divGranite">

<?php

//Connect to Database
    $connection = mysqli_connect("localhost", "root", "pass");
    if (!$connection) {
        die("Error: " . mysqli_error());
    }

    $db_select = mysqli_select_db($connection, "database");
    if (!$db_select) {
        die("Error: " . mysqli_error());
    }

    $appID = "select * from table where state= '1' ";
    $result = mysqli_query($connection,$appID);

    while ($row = $result->fetch_assoc()) {
        $currentID = $row['id']."<br>";
    }


?>
</div>

<div id="display1">

</div>
</body>

<?php

if($currentID != $displayID){
function loadUnload($ID) {

    $connection = mysqli_connect("localhost", "root", "pass");
    if (!$connection) {
        die("Error: " . mysqli_error());
    }

    $db_select = mysqli_select_db($connection, "usr_web26_1");
    if (!$db_select) {
        die("Error: " . mysqli_error());
    }

        switch ($ID) {
            case $ID == 1:
                    echo '<div>';
                    echo '<iframe src="../imageview/index.php" id="imgiframe" style="position:fixed; top:0px; left:0px; bottom:0px; right:0px; width:100%; height:100%; border:none; margin:0; padding:0; overflow:hidden; z-index:999999;"></iframe>';
                    echo '</div>';
                    break;
                case $ID == 2:      
                    echo '<div>';
                    echo '<iframe src="../videoview/index.php" id="imgiframe" style="position:fixed; top:0px; left:0px; bottom:0px; right:0px; width:100%; height:100%; border:none; margin:0; padding:0; overflow:hidden; z-index:999999;"></iframe>';
                    echo '</div>';
                    break;
                default:
                    echo "Wrong ID";
                break;
            }
}
?>
</html>

Answer 1

对于ajax代码，请写这个

$(function(){
    var lastData = '';
        function status(){
           $.ajax({
               type: "GET", 
               url: "main.php",
               success: function(data){
                 if (data != prevData) {
                    $('div.divGranite').html(data);
                    prevData = data;
                }                  
               }
           });
       };
       setTimeout(status, 5000);
    });

其余的都很好！

Answer 2

这个main.php正在返回包括head-tag并将其放入divGranite的孔main.php，如果你想要一个ajax请求只返回main.php生成的一部分你需要制作两个文件：< / p>

main.php - 呈现默认页面
ajax.php - 渲染 iframe 仅

这会让你打电话给ajax.php而不是给你想要的结果您的 ajax.php 看起来像这样，

<?php
//Connect to Database
$connection = mysqli_connect("localhost", "root", "pass");
if (!$connection) {
    die("Error: " . mysqli_error());
}

$db_select = mysqli_select_db($connection, "database");
if (!$db_select) {
    die("Error: " . mysqli_error());
}

$appID = "select * from table where state= '1' ";
$result = mysqli_query($connection,$appID);

while ($row = $result->fetch_assoc()) {
    $currentID = $row['id']."<br>";
}

if($currentID != $displayID) {
    function loadUnload($ID) {

        $connection = mysqli_connect("localhost", "root", "pass");
        if (!$connection) {
            die("Error: " . mysqli_error());
        }

        $db_select = mysqli_select_db($connection, "usr_web26_1");
        if (!$db_select) {
            die("Error: " . mysqli_error());
        }

        switch ($ID) {
            case $ID == 1:
                echo '<div>';
                echo '<iframe src="../imageview/index.php" id="imgiframe" style="position:fixed; top:0px; left:0px; bottom:0px; right:0px; width:100%; height:100%; border:none; margin:0; padding:0; overflow:hidden; z-index:999999;"></iframe>';
                echo '</div>';
                break;
            case $ID == 2:      
                echo '<div>';
                echo '<iframe src="../videoview/index.php" id="imgiframe" style="position:fixed; top:0px; left:0px; bottom:0px; right:0px; width:100%; height:100%; border:none; margin:0; padding:0; overflow:hidden; z-index:999999;"></iframe>';
                echo '</div>';
                break;
            default:
                echo "Wrong ID";
            break;
        }
    }

AJAX加载整个页面而不是页面的一部分来从数据库加载数据

2 个答案: