我有2個while循環，但我只想調用一個。可能嗎？

Question

我有一個表格顯示來自數據庫中2個不同表格的一些數據。我的腳本正在工作，但我已爲每個表呼叫2 while while循環。我的問題是，我可以使用兩個表只有一個循環？謝謝所有!!!!!

<?php 
    if(isset($_POST['check'])) { 
     $kontata_gas = "select id, supervizori, crt_gas, operatori, data, status, nome, cognome, tel, pdr, codice_fiscale from kontrata_gas where (data between '$data_e_inserimit' and '$data_e_inserimit_2' or tel='$tel' or codice_fiscale='$fiscale' or pdr='$POD_PDR')"; 
     $result_kontata_gas = mysqli_query($dbCon, $kontata_gas); 
     if(!$result_kontata_gas) { 
      die("Error1"); 
     } 

     $kontata_luce = "select id, supervizori, crt_luce, operatori, data, status, nome, cognome, tel, pod, codice_fiscale from kontrata_luce where (data between '$data_e_inserimit' and '$data_e_inserimit_2' or tel='$tel' or codice_fiscale='$fiscale' or pod='$POD_PDR')"; 
     $result_kontata_luce = mysqli_query($dbCon, $kontata_luce); 
     if(!$result_kontata_luce) { 
      die("Error2"); 
     } 

    echo "<table style='width:auto;'> 
     <tr> 
     <td>ID</td> 
     <td>TIPO CONTRATTO</td> 
     <td>OPERATOR</td> 
     <td>DATA</td> 
     <td>STATUS</td> 
     <td>NOME</td> 
     <td>COGNOME</td> 
     <td>TEL</td> 
     <td>PDR/POD</td> 
     </tr>"; 
    while ($row = mysqli_fetch_assoc($result_kontata_gas)) { //first loop 
     $id = $row['id']; 
     echo "<tr> 
      <td>"."<a href='main.php?backoffice=&mod_contratto=gas_$id'><img id='img' src='images/modify.png'></a>".$row['id']."</td> 
      <td>".$row['crt_gas']."</td> 
      <td>".$row['operatori']."</td> 
      <td>".$row['data']."</td> 
      <td>".$row['status']."</td> 
      <td>".$row['nome']."</td> 
      <td>".$row['cognome']."</td> 
      <td>".$row['tel']."</td> 
      <td>".$row['pdr']."</td> 
      </tr>"; 
      $id++; 
    } 
     "<tr> 
     <td>ID</td> 
     <td>TIPO CONTRATTO</td> 
     <td>OPERATOR</td> 
     <td>DATA</td> 
     <td>STATUS</td> 
     <td>NOME</td> 
     <td>COGNOME</td> 
     <td>TEL</td> 
     <td>PDR/POD</td> 
     </tr>"; 
    while ($row = mysqli_fetch_assoc($result_kontata_luce)) { // second loop 
     $id = $row['id']; 
     $luce = $row['crt_luce']; 
     echo "<tr> 
      <td>"."<a href='main.php?backoffice=&mod_contratto=luce_$id'><img id='img' src='images/modify.png'></a>".$row['id']."</td> 
      <td>".$row['crt_luce']."</td> 
      <td>".$row['operatori']."</td> 
      <td>".$row['data']."</td> 
      <td>".$row['status']."</td> 
      <td>".$row['nome']."</td> 
      <td>".$row['cognome']."</td> 
      <td>".$row['tel']."</td> 
      <td>".$row['pod']."</td> 
      </tr>"; 
      $id++; 
    } 

     echo "</table>"; 
     $_SESSION['$id'] = @$id; 
     $_SESSION['$gas'] = @$gas; 
     $_SESSION['$luce'] = @$luce; 
} 
?> 

Answer 1

不知道，因爲你只顯示一個SQL查詢，但你很可能使用union（http://dev.mysql.com/doc/refman/5.7/en/union.html）同時SQL querys結合起來。

Answer 2

如果兩個數組都具有相同的字段，則可以將一個函數傳遞給數組並返回（或打印）該表。

printTable($result_kontata_gas); 
printTable($result_kontata_luce); 

function printTable($array) { 
    echo "<table style='width:auto;'> 
     <tr> 
     <td>ID</td> 
     <td>TIPO CONTRATTO</td> 
     <td>OPERATOR</td> 
     <td>DATA</td> 
     <td>STATUS</td> 
     <td>NOME</td> 
     <td>COGNOME</td> 
     <td>TEL</td> 
     <td>PDR/POD</td> 
     </tr>"; 
    while ($row = mysqli_fetch_assoc($array)) { //first loop 
     $id = $row['id']; 
     echo "<tr> 
      <td>"."<a href='main.php?backoffice=&mod_contratto=gas_$id'><img id='img' src='images/modify.png'></a>".$row['id']."</td> 
      <td>".$row['crt_gas']."</td> 
      <td>".$row['operatori']."</td> 
      <td>".$row['data']."</td> 
      <td>".$row['status']."</td> 
      <td>".$row['nome']."</td> 
      <td>".$row['cognome']."</td> 
      <td>".$row['tel']."</td> 
      <td>".$row['pdr']."</td> 
      </tr>"; 
      $id++; 
    } 

    echo "</table>"; 

或者你可以做一個函數，返回數組中取出已經做出2個不同的變量爲每個表和它們串聯各自爲陣的信息。