SQL - SELECT count multiple fields

Question

我需要通過PHP來生成條件SQL查詢，其中我總結和計算了一些字段。讓我們把它切出來，直接轉到我寫的查詢。

$query_voucher="SELECT *, 
count(case WHEN IDprestazione=0 then 1 else null end) as quantitaliberi, 
count(case WHEN IDprestazione != 0 AND pagato=0 then 1 else null end) as quantitaprenotati, 
count(case WHEN pagato=1 then 1 else null end) as quantitapagati, 
count(*) as quantita, 
sum(valorelordo case WHEN IDprestazione=0 then 1 else 0 end) as valoreliberi, 
sum(valorelordo case WHEN IDprestazione=1 AND pagato=0 then 1 else 0 end) as valoreprenotati, 
sum(valorelordo case WHEN pagato=1 then 1 else 0 end) as valorepagati, 
sum(*) as valore 
FROM voucher"; 
$conditions=array(); 

if (!empty($IDpersona_recuperato) && $IDpersona_recuperato!="nessunvalore") { 
     $conditions[]="IDprestazione IN (SELECT IDprestazione FROM prestazioni WHERE IDpersona=$IDpersona_recuperato)"; 
} 
if (!empty($cerca_idprestazione)) { 
    $conditions[]="IDprestazione='$cerca_idprestazione'"; 
} 
if (!empty($dataemissione)) { 
    $conditions[]="dataemissione <= '$dataemissione'"; 
} 
if (!empty($valore)) { 
    $conditions[]="valorelordo = '$valore'"; 
} 
if (!empty($codicecontrollo)) { 
    $conditions[]="codice = '$codicecontrollo'"; 
} 
if (!empty($numero)) { 
    $conditions[]="numero = '$numero'"; 
} 
if ($consegnato=="si"){ 
    $conditions[]="consegnato=1"; 
}elseif ($consegnato=="no") { 
    $conditions[]="consegnato=0"; 
} 

if (count($conditions) > 0) { 
    $query_voucher .=" WHERE " . implode(' AND ', $conditions); 
} 

if ($ordinaper=="numero") { 
    $query_voucher .=" ORDER BY numero"; 
}elseif ($ordinaper=="idprestazione") { 
    $query_voucher .=" ORDER BY IDprestazione"; 
}elseif ($ordinaper=="valore") { 
    $query_voucher .=" ORDER BY valorelordo DESC"; 
}elseif ($ordinaper=="consegnato") { 
    $query_voucher .=" ORDER BY consegnato"; 
} 

比起數和金額數據會在一個表像

$row_voucher1=mysql_fetch_row($query_voucher); 

echo "<form method='POST' name='elimina' id='elimina' action='php/elimina_voucher.php'> 
    <table class='table table-striped' style='margin-top:70px; width: 60%;'> 
    <caption>Riassunto Query Eseguita</caption> 
    <tr> 
     <th></th> 
     <th>liberi</th> 
     <th>prenotati</th> 
     <th>consegnati</th> 
     <th>Totale</th> 
    </tr>"; 
echo "<td> 
     <tr>" . $row_voucher1[quantitaliberi] . "</tr> 
     <tr>" . $row_voucher1[quantitaprenotati] . "</tr> 
     <tr>" . $row_voucher1[quantitapagati] . "</tr> 
     <tr>" . $row_voucher1[valoreliberi] . "</tr> 
     </td>"; 

    echo "</table>"; 

比在查詢中*使用另一個表來獲取更廣泛的數據這樣

while ($row_voucher=mysql_fetch_row($risultato_query_voucher)) { 
if ($row_voucher[1]!=0){ 
    $risultato_query_prestazioni=mysql_query("SELECT * FROM prestazioni WHERE IDprestazione='$row_voucher[1]'"); 
} 

    echo "<tr> 
      <th><div class='showdata'>+</div><div style='margin-top:-20px;margin-left:15px;'><input type='checkbox' name='IDvoucher' id='IDvoucher' value='" . $row_voucher[0] . "'></div></th> 
      <td>" . $row_voucher[2] . "</td> 
      <td>" . $row_voucher[3] . "</td> 
      <td>" . date_format(new DateTime($row_voucher[4]), 'd/m/Y') . "</td> 
      <td>" . $row_voucher[6] . "€</td> 
      <td>"; if ($row_voucher[1]==0){echo "<font color=green>Libero</font>";}else{echo "$row_voucher[1]";}echo "</td> 
      <td>";if ($row_voucher[7]==0) { 
       echo "No"; 
      }else{ 
       echo "Si"; 
      } echo "</td> 
      <td>";if ($row_voucher[7]==1){echo date_format(new DateTime($row_voucher[8]), 'd/m/Y');} echo "</td> 
     </tr>"; 

之前使用sum和count和查詢是一個簡單的SELECT * FROM voucher在第二個表中的一切工作正常。 現在我得到了一個很常見的錯誤Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in /volume1/web/voucher/index6.php on line 285在查詢中出現錯誤的跡象（而不僅僅是在那裏）。

任何幫助表示讚賞。

Answer 1

在將它傳遞給mysql_fetch_array之前檢查$結果。你會發現它是錯誤的，因爲查詢失敗。有關可能的返回值和如何處理它們的建議，請參閱mysql_query文檔。

$result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '$username'"); 

if($result === FALSE) { 
    die(mysql_error()); // TODO: better error handling 
} 

while($row = mysql_fetch_array($result)) 
{ 
    echo $row['FirstName']; 
}