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我想從一個或多個div中的默認內容的MySql數據庫中交換數據,而無需重新加載頁面。我已經在Dreamweaver CS6中使用'set text of container'javascript行爲插入PHP代碼,但沒有成功。Javascript,AJAX將MySQL內容轉換爲div
如果有人能告訴我如何做到這一點,我將不勝感激。
我想從一個或多個div中的默認內容的MySql數據庫中交換數據,而無需重新加載頁面。我已經在Dreamweaver CS6中使用'set text of container'javascript行爲插入PHP代碼,但沒有成功。Javascript,AJAX將MySQL內容轉換爲div
如果有人能告訴我如何做到這一點,我將不勝感激。
save your php code in a file named getinfo.php then call it with ajax
<?php
include("connectdatabase.php");
$sql="SELECT * FROM table ORDER BY id DESC";
$result=mysqli_query($con,$sql);
$i = 1;
while ($row = $result->fetch_assoc()) {
echo "<tr>";
echo "<td>" . $i . "</td>";
echo "<td>" . $row["Name"] . "</td>";
echo "<td>" . $row["Score"] . "</td>";
echo "</tr>";
$i++;
}
mysqli_close($con);
?>
在你的主文件,請使用此
<div class="scoreBoard"></div>
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
setInterval(refreshScoreBoard, 5000);
var inRequest = false;
function refreshScoreBoard() {
if (inRequest) {
return false;
}
inRequest = true;
var load = $.get('getinfo.php');
$(".scoreBoard").html('Refreshing');
load.error(function() {
console.log("Error");
// do something here if request failed
});
load.success(function(res) {
console.log("Success");
$(".scoreBoard").html('<table>'+res+'</table>');
});
load.done(function() {
console.log("Completed");
inRequest = false;
});
}
</script>
嘗試[** **阿賈克斯(https://www.google.com/search?q=ajax)。 –