Преобразование mysql в mysqli в php?

Question

Мне нравится преобразовывать mysql в mysqli, потому что mysql устарел.

Следующий код, Не могли бы вы мне помочь?

 <?php 
     error_reporting(0); 
     if ($_REQUEST['submit']) { 
      $search_file = $_POST['search_file']; 
      $sql = mysql_query("select * from information where firstname like '%$search_file%' or lastname like '%$search_file%' or ID like '%$search_file%' Order by lastname ASC") or die('Error in query : $sql. ' .mysql_error()); 

      if (empty($search_file)) { 
      echo '<script language="javascript">'; 
      echo 'alert("Text field cannot be empty. Please Try it again.")'; 
      echo '</script>'; 
      header("refresh:2; url=index.php"); 
      } 
      else if (mysql_num_rows($sql) > 0) 
      {    
      $i = 1; 
      while ($row = mysql_fetch_array($sql)) { 
       // Print out the contents of the entry 
       echo '<tr>'; 
       echo '<td class="text-center">' . $i . '</td>'; 
       echo '<td class="text-center">' . $row['firstname'] . '</td>'; 
       echo '<td class="text-center">' . $row['lastname'] . '</td>'; 
       echo '<td class="text-center">' . $row['phonenumber'] . '</td>'; 
       echo '<td class="text-center">' . $row['email'] . '</td>'; 
       $i++; 
      } 
      } 
      else 
      { 
      echo '<div class="alert alert-danger" style="width:130px; float:right; margin-top:-142px;">No Results Found!!!</div>'; 
      } 
     } 
     ?> 

Я только вставить "я" в MySQL вот что я придумал, но он показывает ошибку:

"Error in query : $sql."

 <?php 
     error_reporting(0); 
     if ($_REQUEST['submit']) { 
      $search_file = $_POST['search_file']; 
      $sql = mysqli_query($conn, "select * from information where firstname like '%$search_file%' or lastname like '%$search_file%' or ID like '%$search_file%' Order by lastname ASC") or die('Error in query : $sql. ' .mysqli_error()); 

      if (empty($search_file)) { 
      echo '<script language="javascript">'; 
      echo 'alert("Text field cannot be empty. Please Try it again.")'; 
      echo '</script>'; 
      header("refresh:2; url=index.php"); 
      } 
      else if (mysqli_num_rows($sql) > 0) 
      {    
      $i = 1; 
      while ($row = mysqli_fetch_array($sql)) { 
       // Print out the contents of the entry 
       echo '<tr>'; 
       echo '<td class="text-center">' . $i . '</td>'; 
       echo '<td class="text-center">' . $row['firstname'] . '</td>'; 
       echo '<td class="text-center">' . $row['lastname'] . '</td>'; 
       echo '<td class="text-center">' . $row['phonenumber'] . '</td>'; 
       echo '<td class="text-center">' . $row['email'] . '</td>'; 
       $i++; 
      } 
      } 
      else 
      { 
      echo '<div class="alert alert-danger" style="width:130px; float:right; margin-top:-142px;">No Results Found!!!</div>'; 
      } 
     } 
     ?>  

Просьба помочь! Спасибо!

Answer 1

ли эти изменения

<?php 
    error_reporting(0); 
    if ($_REQUEST['submit']) { 
    $search_file = $_POST['search_file']; 
    // Create connection 
    $conn = new mysqli($servername, $username, $password, $dbname); 

    $query = "select * from information where firstname like '%$search_file%' or lastname like '%$search_file%' or ID like '%$search_file%' Order by lastname ASC"; 

    if ($conn->query($query) === TRUE) { 
     echo "New record created successfully"; 
    } else { 
     echo "Error: " . $sql . "<br>" . $conn->error; 
    } 

    //add this where connection ends 
    //$conn->close(); 
?> 

Найти отдых другой код на PHP сайте.