Php 如何计算预期的行数

当我有这样一个mysql查询时：

SELECT 
    i.`pk_faq_item`, 
    i.`title`, 
    i.`question`, 
    i.`featured`, 
    i.`published`, 
    DATE_FORMAT(i.`datecreated`, '%d-%m-%Y %H:%i') as `datecreated`, 
    DATE_FORMAT(i.`datechanged`, '%d-%m-%Y %H:%i') as `datechanged`, 
    i.`creator`, 
    i.`userchanged`, 
    if(ISNULL(s.`vote`), 0,SUM(s.`vote`)) as `votes`, 
    if(ISNULL(s.`vote`), 0,COUNT(*)) as `seen`
FROM `faq_item` i
LEFT JOIN `faq_item_stats` s ON i.`pk_faq_item` = s.`fk_faq_item`
GROUP BY i.`pk_faq_item`
LIMIT 0, 30

我怎么知道有多少结果没有限制0，30

伪查询：

这给了我一个这样的列表：

SELECT 
    i.`pk_faq_item`, 
    i.`title`, 
    i.`question`, 
    i.`featured`, 
    i.`published`, 
    DATE_FORMAT(i.`datecreated`, '%d-%m-%Y %H:%i') as `datecreated`, 
    DATE_FORMAT(i.`datechanged`, '%d-%m-%Y %H:%i') as `datechanged`, 
    i.`creator`, 
    i.`userchanged`, 
    if(ISNULL(s.`vote`), 0,SUM(s.`vote`)) as `votes`, 
    if(ISNULL(s.`vote`), 0,COUNT(*)) as `seen`
FROM `faq_item` i
LEFT JOIN `faq_item_stats` s ON i.`pk_faq_item` = s.`fk_faq_item`
GROUP BY i.`pk_faq_item`
LIMIT 0, 30

3. 1. 1. 1. 等等

但我期待着这样的事情：

SELECT 
    i.`pk_faq_item`, 
    i.`title`, 
    i.`question`, 
    i.`featured`, 
    i.`published`, 
    DATE_FORMAT(i.`datecreated`, '%d-%m-%Y %H:%i') as `datecreated`, 
    DATE_FORMAT(i.`datechanged`, '%d-%m-%Y %H:%i') as `datechanged`, 
    i.`creator`, 
    i.`userchanged`, 
    if(ISNULL(s.`vote`), 0,SUM(s.`vote`)) as `votes`, 
    if(ISNULL(s.`vote`), 0,COUNT(*)) as `seen`
FROM `faq_item` i
LEFT JOIN `faq_item_stats` s ON i.`pk_faq_item` = s.`fk_faq_item`
GROUP BY i.`pk_faq_item`
LIMIT 0, 30

137 我当前的解决方案 在我看来，这是一种糟糕的编程实践：

SELECT * FROM ints;
+---+
| i |
+---+
| 0 |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 8 |
| 9 |
+---+

SELECT SQL_CALC_FOUND_ROWS * FROM ints LIMIT 3;
+---+
| i |
+---+
| 0 |
| 1 |
| 2 |
+---+

SELECT FOUND_ROWS();
+--------------+
| FOUND_ROWS() |
+--------------+
|           10 |
+--------------+