Question

我有以下代码并且效果很好，我只想将其转换为实时，因此每10秒左右更新一次，无需刷新页面，我猜测我需要使用AJAX或Jquery但我缺乏如何这样做的知识。

=====VIA <?php include("database.php"); ?>====
<?php
// Create connection
$con=mysqli_connect("ip/host","user","pass","db");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>

====ON THE PAGE====
<? php

$result = mysqli_query($con, "SELECT * FROM sql347511.1 ORDER BY ID DESC LIMIT 1;");

while ($row = mysqli_fetch_array($result)) {
    echo "<div class='infobox_data'>Temperature: ".$row['TEMP']."&deg;C</div>";
    echo "<div class='infobox_data'>Humidity: ".$row['HUMID']."%</div>";
    echo "<div class='infobox_time'>Captured: ".date("g:i:s a F j, Y ", strtotime($row["TIME"]))."</div>";
}

mysqli_close($con); ?>

Answer 1

让它工作，感谢大家的帮助。

<强>的Javascript

$(document).ready(function(){    
    loadstation();
});

function loadstation(){
    $("#station_data").load("station.php");
    setTimeout(loadstation, 2000);
}

<强> station.php

<?php
include ("database.php");

$result = mysqli_query($con, "SELECT * FROM sql347511.1 ORDER BY ID DESC LIMIT 1;");

while ($row = mysqli_fetch_array($result))
    {
    echo "<div class='infobox_data' id='infobox_temp'>" . $row['TEMP'] . "&deg;C</div>";
    echo "<div class='infobox_data' id='infobox_humid'>" . $row['HUMID'] . "%</div>";
    echo "<div class='infobox_time'>At " . date("g:i:s a F j, Y ", strtotime($row["TIME"])) . "</div>";
    }

mysqli_close($con);
?>

放置数据的位置

<div id="station_data"></div>

Answer 2

您可以双击来自div的输入，然后通过jquery获取此输入值：

$().val;

然后使用ajax将此值发送到php：

$.ajax({
    url: 'url_to_php_which_update_mysql',
    data: {'data': 'value_from_input'},
    cache: false,
    success: function(response){
        $(input).val(response);
    }
});

在php文件中你需要在数据库中上传$ _GET [＆＃39;数据＆＃39;]

实时更新MySQL数据

2 个答案: