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' ";

此行'“$ q。” - 01-01'是=“2013-01-01”？

我用来构建代码的程序说有错误但我找不到任何错误我在PHPadmin中测试了这个

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

它工作正常。我只是想确保这是正确的使用方法。当我测试时，Cuse

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

它正在工作。我可能会混淆'和'。

感谢您的帮助

Answer 1

我想你需要另一个点。连接时，您应该在任何字符串和任何变量之间有一个句点。在这里阅读更多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' ";

Answer 2

试试这个

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

它将为您提供以下SQL查询：

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

Mysql查询似乎搞砸了

2 个答案: