在MYSQL查询中查找哪个条件失败（偏差汇总）

Question

如果任何 mysql 查询条件失败，我需要显示偏差摘要。我需要显示哪些数据输入错误，因为我得到了 NONE 结果。

更多的逻辑问题所以试图用谷歌搜索同样的问题。

表字段：id、day、date、time、program、title

询问

SELECT Count(id) FROM `table` 

WHERE day='Monday' and date = '2018-10-15' and time = '19:45:34' and title = 'NEW-BOTTLE'

如果满足所有条件，上述查询将返回计数。如果任何条件值不满意，我需要显示偏差摘要。

例如，如果 NEW-BOTTLE 输入为 NEW-BOTTLE1，那么我需要显示标题名称不正确。

以同样的方式，我需要显示由于发生了哪个计数 0 而哪个值是错误的。

Answer 1

您可以使用if条件选择值，如下所示：SELECT IF(day = 'Monday', 'TRUE','FALSE') as dayResult, IF(date = '2018-10-15', 'TRUE','FALSE') as dateResult,IF(time = '19:45:34', 'TRUE','FALSE') as timeResult, IF(title = 'NEW-BOTTLE', 'TRUE','FALSE') as titleResult   FROM `table` 它将返回要么1或0根据条件满足。希望它以任何方式帮助你。

Answer 2

您可以使用 switch 语句打印详细摘要。这是伪代码。switch (true) {case 0:    SELECT Count(id) FROM `table`    WHERE day = 'Monday' and date = '2018-10-15' and time = '19:45:34' and title = 'NEW-BOTTLE'    // Get the count from query        if ($count != '0') {        // Criteria met successfully, not need to check other cases;        break;    }case 1:    SELECT Count(id) FROM `table`    WHERE day = 'Monday' and date = '2018-10-15' and time = '19:45:34'    //get the count    if ($count != '0')        echo "Title is incorrect";case 2:    SELECT Count(id) FROM `table`    WHERE day = 'Monday' and date = '2018-10-15' and title = 'NEW-BOTTLE'    //get the count    if ($count != '0')        echo "Time didn't match";case 3:    SELECT Count(id) FROM `table`    WHERE day = 'Monday' and time = '19:45:34' and title = 'NEW-BOTTLE'    if ($count != '0')        echo "Date didn't match";case 4:    SELECT Count(id) FROM `table`    WHERE date = '2018-10-15' and time = '19:45:34' and title = 'NEW-BOTTLE'    if ($count != '0')        echo "Day didn't match";}