如何运行的提交按钮无刷新/重新加载页面的PHP code [英] how to run php code on submit button without refreshing/ reloading the page

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问题描述

我要执行的提交按钮点击无刷新/重新载入我的网页一些PHP code。可能吗?另外我有在页面加载javascript函数,这就是为什么我不希望我的刷新网页。 在此先感谢。

I want to execute some php code on submit button click without refreshing/reloading my page. Is it possible? also i have javascript function on page load that's why i don't want to refresh my page. thanks in advance.

<?php
if(isset($_POST["search"]))
{
$show = "SELECT * FROM data";
$rs = mysql_query($show) or die(mysql_error());
 $add_to_textbox = "<input type='button' name='btn' value='add' />";
#****results in Grid****
    echo "<table width='360px' border='1' cellpadding='2'>";
    $rowID=1;
while($row = mysql_fetch_array($rs))
{
    echo "<tr>";
    echo "<td width='130px' id='name.$rowID.'>$row[Name]</td>";
    echo "<td width='230px' id='link.$rowID.'><a href = '$row[Link]'>$row[Link]</a></td>";
    echo "<td width='130px' onclick='Display($rowID);'>$add_to_textbox</td>";
    echo "</tr>";
    $rowID++;
}
    echo "</table>";
#**********************
mysql_free_result($rs);
}
?>

<script type="text/javascript">
function Display(rowID){
    var linkVal = document.getElementById('link'+rowID+'').innerHTML.replace(/<\/?[^>]+(>|$)/g, "\n");
    document.getElementById("name").value = document.getElementById('name'+rowID+'').innerHTML;
    document.getElementById("link").value = linkVal; 
    }
</script>

这是我的code

here is my code

推荐答案

那么,你需要使用的JavaScript / AJAX。

Well, you need to use the javascript / ajax.

例如:您提交的链接(A HREF为的exaple),加叩至js函数 submitMe 并传递任何变量,你需要

Example: on your submit link (a href for exaple), add call-in-to js function submitMe and pass on whatever variables you need

function submitMe() {
    jQuery(function($) {    
        $.ajax( {           
            url : "some_php_page.php?action=getsession",
            type : "GET",
            success : function(data) {
                alert ("works!"); //or use data string to show something else
                }
            });
        });
    }

如果你想要动态地改变一些内容,这是简易,你只需要创建标签,并指定ID对他们说:&LT; D​​IV ID =动态&GT; &LT; / DIV&GT;

IF you want to change some content dynamically, it is easy- you just need to create tags, and assign ID to them : <div id="Dynamic"> </div>

然后使用加载这两个标记之间的内容

Then you load ANYTHING between those two tags using

document.getElementById("Dynamic").innerHTML = "<b>BOOM!</b>";

意思是说你们两个标记之间的呼叫区域和装载的东西放进去。用同样的方法,你从那个地方的数据:

Meaning that you calling area between two tags and loading something into them. The same way you GET data from that place:

alert(document.getElementById("Dynamic").innerHTML);

请阅读本: <一href="http://www.tizag.com/javascriptT/javascript-getelementbyid.php">http://www.tizag.com/javascriptT/javascript-getelementbyid.php

此外,游戏和实验DOM元素,并了解它们是如何相互作用的。也不是很难,只是需要一些时间来掌握所有的概念。

In addition, play and experiment with DOM elements and learn how they interact. It is not hard, just takes some time to grasp all concepts.

这篇关于如何运行的提交按钮无刷新/重新加载页面的PHP code的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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