Php 两个表之间的SQL查询

我有一个包含两个表的数据库，我正在使用SugarCRM

我正在查询cases_audit表，以获取状态更改为closed的案例行计数。所有这些都很有效

我遇到的问题是如何从cases\u audit中获取id，并确保在表cases下相同的id具有type=support

// Query cases_aduit to find out how many cases were closed -0 days ago
$query_date_1_closed = "select * from cases_audit where after_value_string = 'Closed' and date_created LIKE '$date_1 %'";
$rs_date_1_closed = mysql_query($query_date_1_closed);
$num_rows_1_closed = mysql_num_rows($rs_date_1_closed);

假设cases\u audit中的列id引用不太可能出现的cases中的id，此查询将为您提供audit plus类型中的每一列：

SELECT
    A.*, C.type
FROM cases_audit A
    INNER JOIN cases C ON A.id=C.id
WHERE A.after_value_string = 'Closed' AND A.date_created LIKE '$date_1 %'

如果要计算已关闭的支持案例，请将C.type添加到WHERE条件：

cases\u audit.parent\u id是与cases.id相关的字段

SELECT
    COUNT(*)
FROM cases_audit A
    INNER JOIN cases C ON A.id=C.id
WHERE A.after_value_string = 'Closed' AND A.date_created LIKE '$date_1 %' AND C.type = 'support'