데이터보기 표시 방법 PHP에서 mysql

Question

보기를 만들었습니다. mysql 및 php 필터를 통해 작업을 표시하려고하지만 결과를 표시 할 수 없습니다. 여기에 코드 이 기능은 데이터가이 문제를 해결하는 방법

<?php 
     //tampil berdasar filter 
     if($_POST['do']=="find"){ 
      $arrayBayarReg=$data->tampilDataRegFilter($_POST['q']); 
     } 
     if(count($arrayBayarReg)){ 
      foreach($arrayBayarReg as $data){ 
       ?> 
       <tr class="tabcont"> 
        <td class="tabtxt" align="center"><?php echo $c=$c+1; ?>.</td> 
        <td class="tabtxt" align="center"><?php echo $data['kode_bayar'];?></td> 
        <td class="tabtxt" align="center"><?php echo $data['nm_mhs'];?></td> 
        <td class="tabtxt" align="center"><?php echo $data['tgl_bayar']; ?></td> 
        <td class="tabtxt" align="center"><?php echo $data['jumlah']; ?></td> 
        <td class="tabtxt" align="center"><?php echo $data['keterangan']; ?></td> 
       </tr> 
       <?php 
      } 
     }else{ 
      echo 'Not Found !'; 
     } 
    ?> 

, 내가 PHP에서 초보자이야을 보여주기 위해 여기에 데이터

function tampilDataRegFilter($data){ 
     $query=mysql_query("SELECT * from dataPendaftaran WHERE nm_mhs LIKE '%$data%'"); 
     $no_rows=mysql_num_rows($query); 
     if($no_rows==1){ 
      while($row=mysql_fetch_array($query)){ 
       $data[]=$row; 
       return $data; 
      } 
     } 
    } 

을 보여주는 것입니다. 나는 또한 주변이 전환 것입니다 :)

Answer 1

function tampilDataRegFilter($data){ 
    $query=mysql_query("SELECT * from dataPendaftaran WHERE nm_mhs LIKE '%$data%'"); 
    $no_rows=mysql_num_rows($query); 
    if($no_rows >= 1){ // Does this if there was more than or 1 row. Not only if there was one row 
     $data = array(); 
     while($row=mysql_fetch_array($query)){ 
      $data[]=$row; // Does not return data here 
     } 
     return $data; // Return should only be done when all the data is in the array, which is after while loop is finished 
    } else { 
     return false; 
    } 
} 

을 도와 주셔서 너무 감사합니다 :

<?php 
    //tampil berdasar filter 
    if($_POST['do']=="find"){ 
     $arrayBayarReg=$data->tampilDataRegFilter($_POST['q']); 
    } 
    if($arrayBayarReg !== false){ // Returns false if no rows were found 
     foreach($arrayBayarReg as $data1){ // Wouldnt overwrite the $data variable which apparently holds your data object. 
      ?> 
      <tr class="tabcont"> 
       <td class="tabtxt" align="center"><?php echo $c=$c+1; ?>.</td> 
       <td class="tabtxt" align="center"><?php echo $data1['kode_bayar'];?></td> 
       <td class="tabtxt" align="center"><?php echo $data1['nm_mhs'];?></td> 
       <td class="tabtxt" align="center"><?php echo $data1['tgl_bayar']; ?></td> 
       <td class="tabtxt" align="center"><?php echo $data1['jumlah']; ?></td> 
       <td class="tabtxt" align="center"><?php echo $data1['keterangan']; ?></td> 
      </tr> 
      <?php 
     } 
    }else{ // Returned false, so no rows were found 
     echo 'Not Found !'; 
    } 
?> 

이 하나를 시도하십시오.