Question

如何从网页（php）上的mysql数据库获取实时数据，以显示传感器的数据（无需刷新网页）就像我有一个温度传感器，数据由传感器插入数据库。我想在网页上显示最新数据，并进一步处理数据

[注意]
1）我有一个带有查询的普通index.php

$conn = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);

if (!$conn) 
{
    die("Connection failed: " . mysqli_connect_error());
}

$sqlq="SELECT * FROM data WHERE id=(SELECT MAX(id) FROM data);

$row_result = mysqli_query($conn, $sqlq);
while($row = mysqli_fetch_assoc($row_result))
{   
    $id = $row['id'];
    $num = $row['num'];
    echo $id." | ".$num;
}

[输出]
12 | 36

但我必须刷新页面才能获得最新数据没有刷新页面的任何解决方案？

Answer 1

使用javascript ajax GET-request接收数据。

＆＃13;

function load_data() {
  var xhttp = new XMLHttpRequest();
  xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
      document.getElementById("data_output").text = this.responseText;
    }
  };
  xhttp.open("GET", "output.php", true);
  xhttp.send();
}

window.setInterval (load_data, 1000); //1 second interval

＆＃13;

<div id="data_output"></div>

＆＃13;

你的output.php看起来像是：

$conn = mysqli_connect ($dbhost, $dbuser, $dbpass, $dbname);

if (!$conn) 
    die ("Connection failed: ".mysqli_connect_error ());

$sqlq = "SELECT     * 
         FROM       data 
         ORDER BY   id DESC
         LIMIT      0,1";

$row_result = mysqli_query ($conn, $sqlq);
$row        = mysqli_fetch_assoc ($row_result);

echo $row['id']." | ".$row['num'];

如何从网页（php）上的mysql数据库中获取实时数据，以显示传感器的数据（无需刷新网页）

1 个答案: