MySQLi中的计数[重复项]

Posted 2023-01-31

此问题已经在这里有了答案：

• mysqli_fetch_assoc() expects parameter / Call to a member function bind_param() errors. How to get the actual mysql error and fix it?1个答案

我有一个复杂的mysqli_query，它显示结果，但不计算在内。我应该两次运行相同的脚本？或其他建议？

$ sql应该给我结果，而$ count应该计算完全相同的条件。 $ count用于分页

我尝试了不同类型的计数，但仍然告诉我mysqli_fetch_array会获得布尔值而不是查询。

我有2个脚本，非常相似：

$sql = "(SELECT account_id FROM work 
    WHERE 
    (company LIKE '%".$includewords."%'
    OR function LIKE '%".$includewords."%' 
    OR duties LIKE '%".$includewords."%' 
    OR achievements LIKE '%".$includewords."%' 
    OR reason_for_leaving LIKE '%".$includewords."%')
    AND 
    (company NOT LIKE '%".$excludewords."%'
    OR function NOT LIKE '%".$excludewords."%' 
    OR duties NOT LIKE '%".$excludewords."%' 
    OR achievements NOT LIKE '%".$excludewords."%' 
    OR reason_for_leaving NOT LIKE '%".$excludewords."%'))
    UNION
    (SELECT account_id FROM personal 
    WHERE 
    actual_city LIKE '%".$includelocation."%'
    AND
    actual_city NOT LIKE '%".$excludelocation."%'
    )
    UNION
    (SELECT account_id FROM education 
    WHERE 
    institution LIKE '%".$school."%' OR
    grade_achieved LIKE '%".$school."%'
    )
";

$count="SELECT COUNT account_id FROM 
(SELECT account_id FROM work 
    WHERE 
    (company LIKE '%".$includewords."%'
    OR function LIKE '%".$includewords."%' 
    OR duties LIKE '%".$includewords."%' 
    OR achievements LIKE '%".$includewords."%' 
    OR reason_for_leaving LIKE '%".$includewords."%')
    AND 
    (company NOT LIKE '%".$excludewords."%'
    OR function NOT LIKE '%".$excludewords."%' 
    OR duties NOT LIKE '%".$excludewords."%' 
    OR achievements NOT LIKE '%".$excludewords."%' 
    OR reason_for_leaving NOT LIKE '%".$excludewords."%'))
    UNION
    (SELECT account_id FROM personal 
    WHERE 
    actual_city LIKE '%".$includelocation."%'
    AND
    actual_city NOT LIKE '%".$excludelocation."%'
    )
    UNION
    (SELECT account_id FROM education 
    WHERE 
    institution LIKE '%".$school."%' OR
    grade_achieved LIKE '%".$school."%'
    )";

第二脚本不起作用

mysqli_fetch_array()期望参数1为mysqli_result，布尔在C：\ xampp \ htdocs]中给出

我有一个复杂的mysqli_query，它显示结果，但不计算在内。我应该两次运行相同的脚本？或其他建议？ $ sql应该给我结果，而$ count应该精确计数...

答案

计算您想计算的领域周围的需求。请参阅：https://www.w3schools.com/sql/func_mysql_count.asp

另一答案

假设您有以下查询：