內部聯接在PHP中不起作用，但是在phpMyAdmin SQL中起作用

Question

我有一個MySql查詢，該查詢未在PHP中返回正確的值，但是如果我在phpMyAdmin中運行相同的MySql查詢，則它將返回一個值。 如果我在網絡瀏覽器中顯示選擇，則會在其結尾顯示“資源ID＃27”。

PHP代碼

$SQL_PhotoQueryList = "SELECT count(*) FROM `invoice_detail`".
                      " INNER JOIN `photos` ON invoice_detail.photo_id = photos.photo_id".
                      " INNER JOIN `invoice` ON invoice_detail.invoice_id = invoice.invoice_id".
                      " WHERE invoice.invoice_active = '$PassStatus' AND photos.user_id = '$SessionUserID'".
$SQL_PhotoResultList = mysql_query($SQL_PhotoQueryList);    
$ListPhotoCount      = mysql_result($SQL_PhotoResultList,0);
echo "SQL Query = $SQL_PhotoQueryList<br>";
echo "ListCount = $ListPhotoCount<br>";

屏幕輸出

SQL Query = SELECT count(*) FROM `invoice_detail` INNER JOIN `photos` ON invoice_detail.photo_id = photos.photo_id INNER JOIN `invoice` ON invoice_detail.invoice_id = invoice.invoice_id WHERE invoice.invoice_active = '2' AND photos.user_id = '2'Resource id #27
ListCount = 0

代碼調用例程（$ SessionUserID是$ _SESSION變量）

   $PassStatus = "2"; // Active
   require("get_invoice.php");
   $InfoTotalSales = $ListGalleryCount;

Answer 1

看來您有錯字。

$SQL_PhotoQueryList = "SELECT count(*) FROM `invoice_detail`".
                      " INNER JOIN `photos` ON invoice_detail.photo_id = photos.photo_id".
                      " INNER JOIN `invoice` ON invoice_detail.invoice_id = invoice.invoice_id".
                      " WHERE invoice.invoice_active = '$PassStatus' AND photos.user_id = '$SessionUserID'".
$SQL_PhotoResultList = mysql_query($SQL_PhotoQueryList);    
$ListPhotoCount      = mysql_result($SQL_PhotoResultList,0);
echo "SQL Query = $SQL_PhotoQueryList<br>";
echo "ListCount = $ListPhotoCount<br>";

注意查詢的最后一行的句號（實際上是串聯運算符）：

                      " WHERE invoice.invoice_active = '$PassStatus' AND photos.user_id = '$SessionUserID'".

這應該是一個分號。 是的，我也做到了。 有時最難發現的錯誤。

最后的資源ID來自mysql_query（）的結果。