SQL - SELECT計算多個字段

Question

我需要通過PHP生成條件SQL查詢，其中我總結並計算一些字段。 讓我們把它剪掉，直接進入我寫的查詢。

$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";
}

比計數和總和數據會像表格一樣

 $row_voucher1=mysql_fetch_row($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>";

比查詢中的*用於檢索另一個表中的更多常規數據

 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>";

在使用sum和count之前，查詢是一個簡單的SELECT * FROM voucher ，第二個表中的所有內容都正常工作。 現在我得到了一個非常常見的錯誤Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in /volume1/web/voucher/index6.php on line 285是查詢中出錯的標志（而不僅僅是那里）。

任何幫助表示贊賞。

Answer 1

在將結果傳遞給mysql_fetch_array之前檢查$ result。 您會發現它是錯誤的，因為查詢失敗了。 有關可能的返回值以及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'];
}