2012-04-11 122 views
0

我寫了下面的查詢PHP MySQL查詢工作在phpMyAdmin但不能在網頁

SELECT COUNT(userID) From statistics WHERE userID = "" 

此查詢顯示的網站訪問unathunticated的數量。

的查詢工作在phpMyAdmin當我使用雙引號但是它不當我使用單引號像下面它只是給我存儲在表

$queryB = "SELECT COUNT(userID) From statistics WHERE userID = ''"; 
$resultB =mysql_query($queryA, $con) or die(mysql_error()); 
$authB = mysql_result($resultB, "COUNT(userID)"); 

echo "the number of authenticated visits were $authB<br />\n"; 

我不知道爲什麼它記錄的數量打破,任何想法?

回答

0

不知道這是否會工作......來考慮它只是先想到: 怎麼當u使用轉義雙引號呢?

$ queryB = 「SELECT COUNT(用戶ID)從統計其中userid = \」 \ 「」

4

您存儲在$queryB您的查詢,但您使用$queryA

+3

+1不錯catch.I敢打賭'$ queryA'是一個沒有'WHERE'子句的測試。 – 2012-04-11 14:03:07

0

試試這個:

$queryB = "SELECT COUNT(userID) AS total From statistics WHERE userID = ''"; 
$resultB =mysql_query($queryB, $con) or die(mysql_error()); 
$authB = mysql_fetch_assoc($resultB); 
echo "the number of authenticated visits were ".$authB['total']."<br />\n"; 
0

用戶ID是否有默認值?如果默認值爲NULL,然後改變你的查詢

$queryB = "SELECT COUNT(userID) From statistics WHERE userID IS NULL"; 
0

你應該有點改變了代碼

$queryB = "SELECT COUNT(userID) From statistics WHERE userID = ''"; 
$resultB =mysql_query($queryB, $con) or die(mysql_error()); 
$authB = mysql_result($resultB, 0, 0); 

echo "the number of authenticated visits were $authB<br />\n"; 
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