PHP數組不正確的HTML元素打印

Question

我一直想這個10分鐘，但無法找到一個方法。我把幾個名字到一個數據庫，並呼籲使用mysqli_fetch_array()這些名字在一個叫在功能sort_member_db()

我的輸出代碼功能是：echo $names['FirstName']. ' ' . $names['LastName']; echo '<br>';

它正確地打印出所有的名字，但是名字都沒有的div元素，我希望他們進來。他們出現在網站的最頂端。

這是我放在div元素：

<div class="myclass"> 
      <hgroup class="title"> 
       <h1>Text1</h1> 
       <h2> 
        text2 
       </h2> 
      </hgroup> 
      <p> 
       Array should print after this line:<br> 

      '. sort_member_db() .' 

      </p> 
     </div> 

編輯：這裏就是整個功能

function sort_member_db() 
{ 

    $openConn = mysqli_connect(
      "host", "user", "", ""); 

    //The sorting function for mysql goes here 
    $sortedMembers = mysqli_query($openConn, "SELECT * FROM Members ORDER BY LastName"); 

    //Get the names and insert them into a PHP array 
    while($names = mysqli_fetch_array($sortedMembers)) 
    { 
     //Escape sequence to find the possible error 
     if (!$sortedMembers) 
     { 
      printf("Error: %s\n", mysqli_error($openConn)); 
      exit(); 
     } 

     //Print out the names here   
     echo $names['FirstName']. ' ' . $names['LastName']; 
     echo '<br>'; 
    } 

    mysqli_close($openConn); 

} 

這裏的div元素我試圖讓數組：

page_body('  <div id="body"> 
      <!-- RenderSection("featured", required:=false) --> 
      <section class="content-wrapper main-content clear-fix"> 
       <!-- @RenderBody() --> 
        <section class="featured"> 
     <div class="content-wrapper"> 
      <hgroup class="title"> 
       <h1>Members:</h1> 
       <h2> 

       </h2> 
      </hgroup> 
      <p> 
       Our Team Members :<br> 

      '. sort_member_db() .' 

      </p> 
     </div> 
      </section> 
      </section>'); 

Answer 1

您需要在HTML返回並連接的名稱，而不是直接在所謂呼應名功能，否則這將首先完成。

function sort_member_db() 
{ 

    $openConn = mysqli_connect(
      "host", "user", "", ""); 

    //The sorting function for mysql goes here 
    $sortedMembers = mysqli_query($openConn, "SELECT * FROM Members ORDER BY LastName"); 

    //Get the names and insert them into a PHP array 
    $returnString = ''; 
    while($names = mysqli_fetch_array($sortedMembers)) 
    { 
     //Escape sequence to find the possible error 
     if (!$sortedMembers) 
     { 
      printf("Error: %s\n", mysqli_error($openConn)); 
      exit(); 
     } 

     //Print out the names here   
     $returnString .= $names['FirstName']. ' ' . $names['LastName']; 
     $returnString .='<br>'; 
    } 

    mysqli_close($openConn); 
    return $returnString; 
}