Javascript 设置并获取按钮id
我试图在将一项功能实现到我正在构建的网站之前,让它发挥作用 现在的想法是,当按下index.php页面中的按钮时,test.php页面将根据存储的按钮id(TableID)进行过滤。然后,存储的按钮id应该通过php会话传递到test.php页面,test.php中的mysql查询将获得传递的按钮id(TableID)。然后相应地过滤test.php页面 示例: 1) 用户点击测试按钮 2) 存储的按钮id将传递给$\u会话 3) 然后将用户重定向到test.php 4) 在test.php页面中,页面根据存储的按钮id进行过滤 因此,如果按下test 2按钮,下面的test.php页面图像将显示以下内容:Javascript 设置并获取按钮id,javascript,php,jquery,html,mysql,Javascript,Php,Jquery,Html,Mysql,我试图在将一项功能实现到我正在构建的网站之前,让它发挥作用 现在的想法是,当按下index.php页面中的按钮时,test.php页面将根据存储的按钮id(TableID)进行过滤。然后,存储的按钮id应该通过php会话传递到test.php页面,test.php中的mysql查询将获得传递的按钮id(TableID)。然后相应地过滤test.php页面 示例: 1) 用户点击测试按钮 2) 存储的按钮id将传递给$\u会话 3) 然后将用户重定向到test.php 4) 在test.php页面
- 测试2
- 测试2支架-如果未显示,则其为错误
<?php
session_start();
?>
<!DOCTYPE html>
<html class="full" lang="en">
<!-- Make sure the <html> tag is set to the .full CSS class. Change the background image in the full.css file. -->
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<meta name="description" content="">
<meta name="author" content="">
<title>Index Page </title>
<!-- Bootstrap Core CSS -->
<link href="css/bootstrap.min.css" rel="stylesheet">
<!-- Custom CSS -->
<link href="css/main.css" rel="stylesheet">
<link href="css/images.css" rel="stylesheet">
<link href="css/text.css" rel="stylesheet">
<link href="css/buttons.css" rel="stylesheet">
<!-- HTML5 Shim and Respond.js IE8 support of HTML5 elements and media queries -->
<!-- WARNING: Respond.js doesn't work if you view the page via file:// -->
<!--[if lt IE 9]>
<script src="https://oss.maxcdn.com/libs/html5shiv/3.7.0/html5shiv.js"></script>
<script src="https://oss.maxcdn.com/libs/respond.js/1.4.2/respond.min.js"></script>
<![endif]-->
</head>
<body>
<?php
include("php/connection.php");
?>
<div class="container">
<br/>
<div class="row">
<div class="col-lg-12">
<h2 align="center">Page Title</h2>
</div>
</div>
<BR/>
<!-- /.row -->
<div class="row">
<?php
$data = mysql_query(" SELECT *
FROM tableTest
") or die(mysql_error());
while($info = mysql_fetch_array( $data ))
{
echo "<div class=\"col-lg-3 col-md-3 col-sm-3\">\n";
echo " <div class=\"thumbnail well\">\n";
echo " <div class=\"thumbnail-pad\">\n";
echo " <br/>\n";
echo " <h4 align=\"center\">\n" . $info['testName'] . "</h4>\n";
echo " <br/>\n";
echo " <div align=\"center\"> <a href=\"test.php\" class=\"btn btn-success btn-md\" type=\"submit\">View League</a> </div>\n";
echo " </div>\n";
echo " </div>\n";
echo " </div>\n";
}
?>
</div>
</div>
<!-- /.container-->
<!-- jQuery Version 1.11.0 -->
<script src="js/jquery-1.11.0.js"></script>
<!-- Bootstrap Core JavaScript -->
<script src="js/bootstrap.min.js"></script>
</body>
</html>
索引页
页面标题
**
- test.php
- 测试页图像
测试页
将每个按钮的“按钮ID”以与按钮相同的形式存储在隐藏输入中
echo "<form action=\"test.php\" method=\"POST\"><div class=\"col-lg-3 col-md-3 col-sm-3\">\n";
echo "<div class=\"thumbnail well\">\n";
echo "<div class=\"thumbnail-pad\">\n";
echo "<br/>\n";
echo "<h4 align=\"center\">\n" . $info['testName'] . "</h4>\n";
echo "<br/>\n";
echo "<input type='hidden' value='".$info['tableID']."'>";
echo "<div align=\"center\"><input type=\"submit\" class=\"btn btn-success btn-md\" type=\"submit\" value=\"View League\"></div>\n";
echo "</div>\n";
echo "</div>\n";
echo "</div></form>\n";
希望这会有所帮助。请记住,从网页获取输入时,您需要使用表单。您的javascript或jquery代码在哪里?它是用这些标记的,所以我认为会有一些…隐藏字段可能会有帮助,但是请确保您在其中放置的任何内容都不需要让所有人看到。要查看隐藏字段,您需要做的就是查看HTML源代码。同意,请始终注意您向用户显示的内容。谢谢大家!但我还有一个问题。我是否只是将最底层的代码添加到test.php中?另外,我在php中没有被视为mysql,这使我在小组中遇到了错误。代码的最底层是您的代码,只做了一点修改。只要用我添加的代码替换
$data=
。
echo "<form action=\"test.php\" method=\"POST\"><div class=\"col-lg-3 col-md-3 col-sm-3\">\n";
echo "<div class=\"thumbnail well\">\n";
echo "<div class=\"thumbnail-pad\">\n";
echo "<br/>\n";
echo "<h4 align=\"center\">\n" . $info['testName'] . "</h4>\n";
echo "<br/>\n";
echo "<input type='hidden' value='".$info['tableID']."'>";
echo "<div align=\"center\"><input type=\"submit\" class=\"btn btn-success btn-md\" type=\"submit\" value=\"View League\"></div>\n";
echo "</div>\n";
echo "</div>\n";
echo "</div></form>\n";
$data = mysql_query(" SELECT testName , testBio FROM tableTest
WHERE tableID = ".$_POST['tableID']." GROUP BY tableID ") or die(mysql_error());