由先前的選擇

Question

我有時所選擇的第一個，我想3個選擇下拉菜單確定下一選擇菜單，所述第二顯示出來，並且當選擇第二時，第三個顯示了，通過使用if(isset($_post[first_one]))和第三一個使用if(isset($_post[second_one]))

SQL：

$conn = new mysqli($servername, $username, $password, $dbname); 
$sql = "SELECT * FROM sp_meeting_log "; 
$result1 = $conn->query($sql); 

PHP/HTML：

<div style="position:absolute; top:300px; left:500px;"> 
    <form method="post"> 
     <p>Choose:</p> 
     <!--Get all orgs ,forums ,Users data available for admin--> 
     //select the first 
     <select style="display:inline-block;" name="org"> 
      <option value="All Orgs">All Orgs</option> 
      //first drop down info release 
      <?php 
      if ($result1->num_rows > 0) { 
       echo "<table><tr><th>orgID</th><th>orgName</th></tr>"; 
       // output data of each row 
       while($row = $result1->fetch_assoc()) { 
        echo "<option>" .$row["orgID"]." /".$row["orgName"]."</option>"; 
       } 

       echo "</table>"; 
      } else { 
       echo "0 results"; 
      } 
      ?> 
     </select> 
     <select style="display:inline-block;" name="forum"> 
      <option value="forum1"><h5>All Forums</h5></option> 
      <?php 
      // if the first dropdown post set 
      if(isset($_POST['org'])){ 
       $result2 = $conn->query($sql); 
       if ($result2->num_rows > 0) { 
        echo "<table><tr><th>forumID</th><th>forumName</th></tr>"; 
        // output data of each row 
        while($row = $result2->fetch_assoc()) { 
         echo "<option>".$row["forumID"]."/".$row["forumName"]."</option>"; 
        } 

        echo "</table>"; 
       } else { 
        echo "0 results"; 
       } 
      } 
      ?> 
     </select> 
     //select the second 
     <select style="display:inline-block;" name="user"> 
      <option><h5>All Users</h5></option> 
      <?php 
      // if the second drop down is set 
      if(isset($_POST['forum'])){ 
       $result3 = $conn->query($sql); 
       if ($result3->num_rows > 0) { 
        echo "<table><tr><th>userID</th><th>username</th></tr>"; 
        // output data of each row 
        while($row = $result3->fetch_assoc()) { 
         echo "<option>".$row["userID"]."/".$row["username"]. "</option>"; 
        } 

        echo "</table>"; 
       } else { 

       echo "0 results"; 
      } 
     } 
     ?> 

Answer 1

基本上這就是這個想法。你想要一個頁面，只是從兩個頁面獲取和結果回發到一個頁面到正確的地點：

page1.php中

<script type="text/javascript" src="http://code.jquery.com/jquery-1.9.1.js"></script> 
<form> 
    <label>Org 
     <select id="org" name="org"> 
      <option value="1">One</option> 
      <option value="2">Two</option> 
     </select> 
    </label> 
    <!-- This is where the forum html will drop into after ajax runs --> 
    <div id="forum"></div> 
    <!-- This is where the user html will drop into after ajax runs --> 
    <div id="user"></div> 
</form> 
<script type="text/javascript"> 
    $(document).ready(function() { 
     // On change of a select menu 
     $(this).on('change','select',function(e){ 
      // Assign selection 
      var thisSelect = $(this); 
      // Get the id name, this will tell page two 
      // what it's receiving 
      var sendType = thisSelect.attr('id'); 
      // Get the actual value of the selection 
      var sendVal  = thisSelect.val(); 
      // Create essentially a POST 
      var sendData = { field: sendType, value: sendVal }; 
      $.ajax({ 
       // Send to page 2 
       url : '/page2.php', 
       // Use post method 
       type: "POST", 
       // Use post data from above 
       data : sendData, 
       // This is what will run on success 
       success:function(response){ 
        // Parse the json coming back for placement 
        var jSon = JSON.parse(response); 
        // Save the correct html into the correct drop spot 
        $('#'+jSon.type).html(jSon.html); 
       }, 
       error: function(response){ 
        console.log(response); 
       } 
      }); 
     }); 
    }); 
</script> 

使page2.php

if(!empty($_POST)) { 
    $data = ''; 
    ob_start(); 
    if(isset($_POST['field'])) { 
     if($_POST['field'] == 'org') { 
      $type = 'forum'; 
?> 
    <label>Forum 
     <select id="forum" name="forum"> 
      <option value="1">One</option> 
      <option value="2">Two</option> 
     </select> 
    </label> 
<?php 
    } 
    elseif($_POST['field'] == 'forum') { 
     $type = 'user'; 
?> 
    <label>user 
     <select id="user" name="user"> 
      <option value="1">One</option> 
      <option value="2">Two</option> 
     </select> 
    </label> 
<?php } 
     $data = ob_get_contents(); 
     ob_end_clean(); 

     die(json_encode(array('type'=>$type,'html'=>$data))); 
    } 

    die(json_encode(array('type'=>'error','html'=>false))); 
}