prepared statement 결과에 broken code가 추가되었습니다.

Question

ID로 검색 할 때 prepare 준비를하기 전에 준비가 완료되었지만 prepared statement 결과를 추가 한 후 ID가 존재하지 않습니다!

이 내 코드입니다 : 아직 문을 준비 전혀 없기 때문에,

<form action="" method="post"> 
    <div class="col-lg-3"> 
     <label for=""><h4>Search by ID</h4></label> 
     <div class="input-group"> 
      <input name="search" type="number" class="form-control"> 
      <span class="input-group-btn"> 
       <button name="submit" class="btn btn-default" type="submit"> 
      <span class="glyphicon glyphicon-search"></span> 
      </button> 
      </span> 
     </div> 
<hr> 
    </div> 
    <div class="col-lg-12"> 
     <div class="table-responsive"> 
      <table class="table table-hover"> 
       <thead> 
        <tr> 
         <th>ID</th> 
         <th>Ime</th> 
         <th>Prezime</th> 
        </tr> 
       </thead> 
       <tbody> 
       </tbody> 

       <?php 

       if(isset($_POST['search'])) { 

       $search_id = $_POST['search']; 

       mysqli_set_charset($connection, "utf8"); 
       $stmt = mysqli_prepare($connection, "SELECT id, ime, prezime FROM anketa WHERE id = ? LIMIT 1"); 

       if(isset($stmt)) { 

        mysqli_stmt_bind_param($stmt, 'i', $search_id); 

        mysqli_stmt_bind_result($stmt, $ank_id, $ank_ime, $ank_prezime); 

        mysqli_stmt_execute($stmt); 

       } 

       if(!$stmt) { 
        die("QueryFailed" . mysqli_error($connection)); 
       } 

       if(mysqli_stmt_num_rows($stmt) > 0) { 

        while(mysqli_stmt_fetch($stmt)): 

          echo "<tr>"; 
          echo "<td>{$ank_id}</td>"; 
          echo "<td>{$ank_ime}</td>"; 
          echo "<td>{$ank_prezime}</td>";  
          echo "<td><a class='btn btn-info btn-xs' href='search.php?source=edit&edit={$ank_id}'><i class='fa fa-pencil-square-o'></i> Edit</a></td>"; 
          echo "<td><a class='btn btn-danger btn-xs' onClick=\"javascript: return confirm('Delete?'); \" href='search.php?delete={$ank_id}'><i class='fa fa-trash-o'></i> Delete</a></td>"; 

        endwhile; 

        } else { 

         echo "ID doesn't exist!"; 

        } 

       } 

       ?> 

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

<?php 

if(isset($_GET['delete'])) { 

$ank_id = $_GET['delete']; 

if(isset($_SESSION['user_role'])) { 

if($_SESSION['user_role'] == 'admin' || 'user_role'] == 'superadmin') { 

$the_anketa_id = mysqli_real_escape_string($connection, $_GET['delete']); 

$query = "DELETE FROM anketa WHERE id = {$ank_id} "; 
$delete_anketa = mysqli_query($connection, $query); 
header("Location: search.php"); 

} 

} 

} 

?> 

</form> 
</div> 

<?php 
if(isset($_GET['source'])) { 
$source = $_GET['source']; 
if($source = 'edit') { 
include "includes/edit.php"; 
} elseif($source = 'submit') { 
header("Location: search.php"); 
} 
} 
?> 

<?php 
} else { 
echo "<script>alert('No access!')</script>"; 
} 
?> 

나는 문제가 삭제를 위해 문 또는 코드를 준비하는 경우 몰라?

Answer 1

이것은 mysqli_stmt_num_rows ($ stmt)가 항상 0을 반환하기 때문일 수 있습니다.

PHP 설명서 : 올바른 값을 반환하는 기능을 얻을 수 있습니다 전에 http://php.net/manual/en/mysqli-stmt.num-rows.php

먼저) mysqli_stmt_store_result를 (호출 할 필요가 있습니다.