如何获取另一个页面的id并使其显示基于id的值

时间:2016-04-13 06:05:57

标签: php html mysql html5

我已经提到过这篇文章How to get row ID in button click?,但仍然无法使其加载值,我不知道为什么。我很确定id已经在profile.php中被正确捕获。以下是我的代码:

hanj.php

  style type="text/css">
      .buttonize {
        text-decoration: none;
        border: 1px solid #ccc;
        background-color: #efefef;
        padding: 10px 15px;
        -moz-border-radius: 11px;
        -webkit-border-radius: 11px;
        border-radius: 11px;
        text-shadow: 0 1px 0 #FFFFFF;
      }
     <?php require_once('Connections/conn.php');?>
    </style>
    <table border="1" cellpadding="5" cellspacing="2" width="600">
    <tr>
      <th>Vendor Id</th>
        <th>Kategori</th>
        <th>Nama Vendor</th>
        <th>Alamat</th>
        <th>Poskod</th>
        <th>Bandar</th>
         <th>Negeri</th>
         <th>No</th>
         <th>Email</th>
    </tr>



     <?php
     session_start();
     $_SESSION['profile']= $row ['v_id'];
    mysql_select_db ($database_conn,$conn);
    $query="SELECT v_id,type,companyName,address,code,city,state,contact,email FROM vendor";
    $result=mysql_query($query) or die(mysql_error());

    while($row=mysql_fetch_array($result))
    {
         echo "</td><td>";
        echo $row['v_id'];
        echo "</td><td>";
        echo $row['type'];
        echo "</td><td>";
        echo $row['companyName'];
        echo "</td><td>";
        echo $row['address'];
        echo "</td><td>";
        echo $row['code'];
        echo "</td><td>";
         echo $row['city'];
        echo "</td><td>";
         echo $row['state'];
        echo "</td><td>";
        echo $row['contact'];
        echo "</td><td>";
        echo $row['email'];
        echo "</td><td>";
        print '<center><a href="profile.php?id='.$row['v_id'].'" class="buttonize">View</a></center>';
        echo "</td></tr>";
    }
    ?>



 profile.php
    <?php require_once('Connections/conn.php'); ?>
    <html>
      <head>
        <meta charset="utf-8">
        <meta name="viewport" content="width=device-width, initial-scale=1">
        <title>Casado</title>
          <script type="text/javascript" src="http://cdnjs.cloudflare.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script>
          <script type="text/javascript" src="http://netdna.bootstrapcdn.com/bootstrap/3.3.4/js/bootstrap.min.js"></script>
          <link href="http://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.3.0/css/font-awesome.min.css" rel="stylesheet" type="text/css">
          <link href="http://pingendo.github.io/pingendo-bootstrap/themes/default/bootstrap.css" rel="stylesheet" type="text/css">

    <div class="container">
    <div class="col-md-12">
    <div class="col-md-2"></div>

    <div class="col-md-8">
    <table class="table table-bordered">
                  <thead>
                    <tr>
                      <th>Pakej</th>
                      <th>Image</th>
                      <th>Harga</th>
                      <th>Vendor Id</th>
                    </tr>
                  </thead>
                  <tbody>

                  <?php
                  session_start();            
                   mysql_select_db ($database_conn,$conn);
                   $vid = $_SESSION['profile'];
                   $sql = mysql_query("SELECT * from item where v_id = '$vid' ") or die (mysql_error());
                   while($res = mysql_fetch_array($sql)) {
                    ?>




                    <tr>
                      <td><?php echo $res['pakej'] ?></td>
                      <td><?php echo '<img src="data:image;base64,'.$res['image'].'" class="img-thumbnail">' ?></td>
                      <td><?php echo $res['harga'] ?></td>
                      <td><?php echo $res['v_id'] ?></td>
                    </tr>
                   <?php }
                   ?>

                  </tbody>
                </table>
                </div>

                <div class="col-md-2"></div>
                </div>
                </div>

    </body>
    </html>

btw没有语法错误

4 个答案:

答案 0 :(得分:0)

在profile.php中使用$ _GET来引用传入的'id'。即 替换以下行

$vid = $_SESSION['profile'];

这个。

$vid = $_GET['id'];

希望这有帮助。

答案 1 :(得分:0)

这个问题由Bikash Paul回答,他建议将$ _SESSION [&#39;个人资料&#39;]更改为$ _REQUEST [&#39; id&#39;]

答案 2 :(得分:0)

而不是$vid = $_SESSION['profile'];尝试$vid = $_GET['profile'];,如果不起作用,则print_r($ _ GET)或print_r($ _ REQUEST)&amp;向我们展示详细信息

答案 3 :(得分:0)

而不是$_SESSION['profile']在您的$_REQUEST['id']中使用profile.php而应该有效。