在这张图片中,我想创建一个从PHP脚本加载数据的框。新数据显示在顶部。我希望在顶部看到新数据。随着新数据的更新,旧数据会下降,您无法在4行后看到旧数据。我用css来实现它。我使用溢出来隐藏旧数据。相反,新数据位于底部。请帮我。谢谢。
我的代码位于下面
php脚本
<?php
header('Content-Type: text/event-stream');
header('Cache-Control: no-cache');
$dbhost = "localhost";
$dbusername = "root";
$dbpassword = "netwitness";
$dbname = "abdpractice";
$con = mysqli_connect ($dbhost, $dbusername, $dbpassword) or die ('Error in connecting: ' . mysqli_error($con));
//Select the particular database and link to the connection
$db_selected = mysqli_select_db($con, $dbname ) or die('Select dbase error '. mysqli_error());
//Make A SQL Query and link to the connection
$result = mysqli_query($con,"SELECT * FROM `countryattack` ORDER BY RAND() LIMIT 1");
while ($row = mysqli_fetch_assoc($result))
{
echo "data: [X] NEW ATTACK: FROM " . $row["countrysrc"]. " TO " . $row["countrydst"]. " \n\n";
}
mysqli_close($con);
?>
html代码
<!DOCTYPE html>
<html>
<head>
<style>
div.hidden {
background-color: #00FF00;
width: 500px;
height: 100px;
overflow: hidden;
}
</style>
</head>
<body>
<h1>Getting server updates</h1>
<div class="hidden" id="result"></div>
<script>
if(typeof(EventSource) !== "undefined") {
var source = new EventSource("shownewattack.php");
source.onmessage = function(event) {
document.getElementById("result").innerHTML += event.data + "<br>";
};
} else {
document.getElementById("result").innerHTML = "Sorry, your browser does not support server-sent events...";
}
</script>
</body>
</html>
输出结果是它们在底部显示数据。这不是我想要的......关于如何做的任何想法..
我的问题是如何创建一个框来观察数据。随着脚本的更新,旧数据会下降。在第四行之后你看不到任何东西。新数据将出现在顶部。你能帮我么。谢谢..
答案 0 :(得分:0)
$result = mysqli_query($con,"SELECT * FROM `countryattack` ORDER BY createdDate LIMIT 4");
while ($row = mysqli_fetch_assoc($result))
{
echo "data: [X] NEW ATTACK: FROM " . $row["countrysrc"]. " TO " . $row["countrydst"]. "";
}
echo "data: [X] NEW ATTACK: FROM " . $row["countrysrc"]. " TO " . $row["countrydst"]. "<br>";
那应该做你想要的。只需将行添加到表中,然后按行对数据进行排序,就可以将createdDate添加到数据库中。然后将记录限制为4.此外,如果您想要新行上的数据,您可以添加&lt; br&gt;到你的回声结束。祝你好运。
答案 1 :(得分:0)
确保您使用DESC或ASC对查询进行排序,以使其相应地到达。这肯定能解决问题。
答案 2 :(得分:0)
您可以使用Node.insertBefore()
替换父元素.innerHTML
的连接。
const result = document.getElementById("result");
source.onmessage = function(event) {
const node = document.createTextNode(event.data + "\n");
if (!result.firstChild) {
result.appendChild(node);
else {
result.insertBefore(node, result.firstChild);
}
};
<pre id="result"></pre>
<script>
const result = document.getElementById("result");
let n = 0;
const interval = setInterval(() => {
let node = document.createTextNode(++n + "\n");
if (!result.firstChild) {
result.appendChild(node);
} else {
result.insertBefore(node, result.firstChild)
}
}, 500);
</script>
&#13;