在MYSQL查询中查找失败的条件（偏差摘要）

Question

如果任何mysql查询条件出现问题，我需要显示偏差摘要。 我需要显示由于没有得到任何结果而导致输入错误的数据。

更多的逻辑问题，所以试图谷歌搜索相同。

表字段：id，日期，日期，时间，程序，标题

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count。 如果任何“条件”值不满意，我需要显示“偏差摘要”。

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

同样，我需要显示由于发生了哪个计数0而导致哪个值是错误的。

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` 

它将根据条件满足返回1或0 。 希望它对您有任何帮助。

Answer 2

您可以使用switch语句来打印详细的摘要。 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";
}