2016-08-13 37 views
0

的第二頁基於選擇檢索數據,所以我不知道如何在標題中正確表達它,但我會在這裏完成。我有兩個頁面,第一個添加了一些列表視圖(每個都有值)。我想要的是,當用戶點擊一個項目,他/她進入第二頁,它會向他/她顯示一個列表視圖(根據他們選擇的值檢索MySQL數據)。但我的問題是,當我點擊第一頁中的一個項目時,它在第二頁中顯示我什麼都沒有。JQuery Mobile PHP MySQL基於第一頁

我已經嘗試了幾種方法幾天,現在它變得令人沮喪!我真的需要幫助。非常感激。

這是我的第一個頁面代碼:

<?php 
session_start(); 
include "partials/connectDb.php"; 

if (isset($_POST["goverVal"])) 
{ 
    $_SESSION["goverVal"] = $_POST["goverVal"]; 
} 
?> 

<div data-role="page" id="homepage"> 
    <!-- HEADER INCLUDE --> 
    <?php include "partials/header.php"; ?> 
    <div data-role="main" class="ui-content" align="center"> 
     <h2>Select A Governorate</h2> 
      <form action="index.php" method="POST"> 
       <ul data-role="listview" id="goverVal" name="goverVal"> 
       <?php 
        include "partials/connectDb.php"; 

        $sql = "SELECT * FROM governorate_table;"; 
        $run_query = mysqli_query($conn, $sql); 

        while ($row = mysqli_fetch_array($run_query)) 
        { 
         $gId = $row['g_id']; 
         echo "<li data-value='{$gId}'><a href='area_page.php' class='ui-btn'>$gId - $row[name]</a></li>"; 
        } 
       ?> 
       </ul> 
      </form> 
    </div> 
    <!-- FOOTER INCLUDE --> 
    <?php include "partials/footer.php"; ?> 
</div> 

================================ =======================================

這是我第二頁的代碼:

<?php 
session_start(); 
include "partials/connectDb.php"; 

$goverChoice = $_SESSION["goverVal"]; 
?> 

<div data-role="page" id="areaPage"> 
    <!-- HEADER INCLUDE --> 
    <?php include "partials/header.php"; ?> 
    <div data-role="main" class="ui-content" align="center"> 
     <h2>Select an Area</h2> 
     <?php 
      include "partials/connectDb.php"; 

      $q = "SELECT * FROM area_table WHERE governorate = '$goverChoice';"; 

      $run_sql = mysqli_query($conn, $q); 

      echo "<ul data-role='listview'>"; 
      while ($row = mysqli_fetch_array($run_sql)) 
      { 
       echo "<li><a href='#'>".$row['area_name']."</a></li>"; 
      } 
      echo "</ul>"; 
     ?> 
    </div> 
    <!-- FOOTER INCLUDE --> 
    <?php include "partials/footer.php"; ?> 
</div> 

回答

0

沒關係。我終於找到了一個合適的解決方案。 這是我用,如果有人遇到同樣的問題:

在第一頁:

$(document).ready(function(){ 
    $('#goverVal').on("click", "li", function(){ 
     $.ajax("update_session.php",{method:"post", data:{val:$(this).attr("data-value")}}); 
    }); 
}); 

而這在一個單獨的文件:

<?php 
//Another File 
session_start(); 
if (isset($_POST["val"])) 
{ 
    $_SESSION["val"] = $_POST["val"]; 
} 
?> 

而此行的第二頁面在哪裏顯示數據,頂部:

$goverChoice = $_SESSION["val"]; 
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