Mysql查詢似乎混淆了

Question

$q=2013 

$sql="SELECT * FROM vip_sales WHERE type = 'vecka' and date >= '"$q ."-01-01' and date <= '"$q ."-01-31' "; 

$result = mysql_query($sql, $con); 
$janw = mysql_num_rows($result); 

$sql="SELECT * FROM vip_sales WHERE type = 'manad' and date >= '"$q ."-01-01' and date <= '"$q ."-01-31' "; 
$result = mysql_query($sql, $con); 
$janm = mysql_num_rows($result); 

$sql="SELECT * FROM vip_sales WHERE type = 'ar' and date >= '"$q ."-01-01' and date <= '"$q ."-01-31' "; 

這行''$ q。「 - 01-01'is it =」2013-01-01「？

我的，我用來建立代碼程序與說，有一個錯誤，但我不能找到我的PHPadmin

$sql="SELECT * FROM vip_sales WHERE type = 'ar' and date >= '2013-01-01' and date <= '2013-01-31' "; 

我測試了這個任何錯誤，它工作正常。我只是想確保這是正確的使用線。 Cuse當我測試

$sql="SELECT * FROM vip_sales WHERE type = 'ar' and date >= '"2013 ."-01-01' and date <= '"2013 ."-01-31' "; 

它dident工作。我可能會混淆「和」。

感謝所有幫助

Answer 1

試試這個

$sql="SELECT * FROM vip_sales WHERE type = 'vecka' and date >= '".$q."-01-01' and date <= '".$q."-01-31'"; 

它會給你下面的SQL查詢：

SELECT * FROM vip_sales WHERE type = 'vecka' and date >= '2013-01-01' and date <= '2013-01-31' 

Answer 2

我覺得雅需要另一個點。當您連接，如果您有任何字符串，任何變量之間的時間段。在這裏閱讀更多http://php.net/manual/en/language.operators.string.php

"SELECT * FROM vip_sales WHERE type = 'manad' and date >= '" . $q ."-01-01' and date <= '" . $q ."-01-31' ";