Question

下面的代码从数据库动态生成json数据。生成数据时，会将其添加到特定div。如何每隔30秒重新加载JSONObject的内容。这将使内容显示接近实时的变化。

<script>

JSONObject = <?php echo include_once('../includes/dashboard-stats.php'); ?>;

document.getElementById("today_visits").innerHTML=JSONObject.todayVisits;

</script>

以下是JSONObject = <?php echo include_once('../includes/dashboard-stats.php'); ?>;

的输出

JSONObject = {"todayEarnings":"2.60","todayVisits":"212","todayClicks":"36","todayLeads":"3","todayCalculateCR":"12%","todayEPC":"0.08","todayCTR":"17%","yesterdayEarnings":"0.40","yesterdayClicks":"35","yesterdayVisits":"148","yesterdayLeads":"1","yesterdayCalculateCR":"35%","yesterdayEPC":"0.03","yesterdayCTR":"24%","monthEarnings":"3.00","monthClicks":"75","monthVisits":"392","monthLeads":"4","monthCalculateCR":"19%","monthEPC":"0.05","monthCTR":"19%"}

    1;

我尝试使用它来尝试重新加载json数据。

<script>
 function load(){
JSONObject = <?php echo include_once('../includes/dashboard-stats.php'); ?>
document.getElementById("today_visits").innerHTML=JSONObject.todayVisits;
 setTimeout("load()",9000);
      }
</script>

Answer 1

您的PHP代码仅运行一种类型这就是为什么结果显示相同。每隔30秒使用Ajax调用并通过PHP文件从DB获取新数据。

// Jquery语法

$.post("Your PHP SCRIPT FILE PATH HERE", { PARAMS you want to pass }, function( Get DATA FROM PHP FILE ) {

// HERE IS YOU Operation

},"DATA FORMAT");

==================================

//代码示例

function loadStats(){
 $.post( "../includes/dashboard-stats.php", { get:"stats" }, function(data) {
    $("#today_visits").html(data.todayVisits);
 }, "json");
}
$(function(){
     loadStats();
    setInterval(loadStats,9000);
}):

Answer 2

更新回答：

<script>      
   function load() {
        var colors = ["#CCCCCC","#333333","#990099"]; 
        var rand = Math.floor(Math.random()*colors.length); 

        $.getJSON("../includes/dashboard-stats.php", { get:"stats" },function(data) {
             $("#today_visits").fadeOut().fadeIn().html(data.todayVisits).css("background-color", colors[rand]);
        });
    }
    $(function() {
        load();//on the page load.
        setInterval(load,9000);
    });
</script>

../includes/dashboard-stats.php的网址是相对于其所在脚本所在页面的位置

Answer 3

使用jQuery library，我们可以做类似的事情;

setInterval(load, 5000);

function load() {
  $.get('../includes/dashboard-stats.php', function(ReturnData) {
    JSONObject = ReturnData
  });
}

使用setInterval
在函数load中，创建一个获取数据的ajax请求。

ReturnData保存触发../includes/dashboard-stats.php时生成的内容的输出（HTML） - 尝试在浏览器中运行它，然后找到您想要使用的数据的位置。

编辑

在Moonwave评论

"load()"

如何重新加载JSONObject内容

3 个答案:

编辑