2016-11-20 90 views
0

我约一个半月到PHP。我试图让函数获取结果从find_all_subjects()虽然循环表Mysqli结果

 function find_all_subjects() { 
     global $dbconnect; 

     $q = "SELECT * FROM subjects ORDER BY id ASC"; 
     $subject_set = mysqli_query($dbconnect, $q); 
     confirm_query($subject_set); 
     return $subject_set; 

     } 

然后又做了新的功能,这将循环结果在表行。问题是它循环只有一个结果......当我用<ul><li>做它时,它工作正常,但它似乎没有工作时,与表一起做。我的表核心标签在另一个文档中。所以它不是......

function navigacija() { 

       $s   = find_all_subjects(); 

      while($subjects = mysqli_fetch_assoc($s)) { 

       $out  = "<tr>"; 

       // Ime Teme 

       $out  .= "<td width='25%' height='40'>&nbsp;&nbsp;<a href=\"admin_content.php?subject=" . urlencode($subjects['id']) . "\">"; 
       $out  .= $subjects['menu_name']; 
       $out  .= "</a></td>"; 

       // Vidljiva 

      if($subjects['visible'] == 1) { 
        $subjects['visible'] = 'DA'; 
     } else { 
        $subjects['visible'] = 'NE'; 
     } 
       $out  .= "<td width='25%' height='40'><p align=\"center\">DA&nbsp;/&nbsp;NE"; 
       $out  .= "</p></td>"; 

       // Broj Strana 

       $out  .= "<td width='25%' height='40'>"; 

       $pages_set = find_sub_from_pages($subjects['id']); 

      while($pages = mysqli_fetch_assoc($pages_set)) { 

       $out  .= "<a href=admin_content.php?page=" . $pages['id'] . ">" . $pages['menu_name'] . "</a>"; 
       $out  .= "</td>"; 

      } 

       // Vidljiva 

       $out  .= "<td align=\"center\" width='25%' height='40'>"; 
       $out  .= "<img width=\"17px\" height=\"17px\" src=\"st/img/ic-arup.png\"> 
          </img>&nbsp;<img width=\"17px\" height=\"17px\" src=\"st/img/ic-ardown.png\"></img>&nbsp;"; 
       $out  .= "</td>"; 

       $out  .= "</tr>"; 
      } 

       return $out; 
} 

回答

4

的问题是,你重置

$out  = "<tr>"; 
在循环每次

。 此行更改为

$out  .= "<tr>"; 

只有放声明圈外

$out  = ""; 
+0

作品般的魅力的!非常感谢:) – KonB