2012-04-11 126 views
0

我写了下面的查询PHP MySQL查询工作在phpMyAdmin但不能在网页

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

此查询显示的网站访问unathunticated的数量。

的查询工作在phpMyAdmin当我使用双引号但是它不当我使用单引号像下面它只是给我存储在表

$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"; 

我不知道为什么它记录的数量打破,任何想法?

回答

0

不知道这是否会工作......来考虑它只是先想到: 怎么当u使用转义双引号呢?

$ queryB = “SELECT COUNT(用户ID)从统计其中userid = \” \ “”

4

您存储在$queryB您的查询,但您使用$queryA

+3

+1不错catch.I敢打赌'$ queryA'是一个没有'WHERE'子句的测试。 – 2012-04-11 14:03:07

0

试试这个:

$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

用户ID是否有默认值?如果默认值为NULL,然后改变你的查询

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

你应该有点改变了代码

$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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