查找MYSQL查询失败的条件(偏差汇总)

Question

如果mysql查询条件失败，我需要显示偏差汇总。我需要显示哪个数据输入是错误的，因为我没有得到任何结果。

更多的是合乎逻辑的问题，所以试图用谷歌搜索同样的问题。

表字段: id、日期、日期、时间、节目、标题

查询

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all条件，上述查询将返回count。我需要显示偏差汇总，以防任何条件值未得到满足。

例如，如果新瓶子输入为NEW-BOTTLE1，那么我需要显示标题名称是不正确的。

以同样的方式，我需要显示哪个值是错误的，因为发生了哪个计数为0。

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` 

它将根据条件满足情况返回1或0。希望它能以任何方式帮助你。

Answer 2

您可以使用switch语句打印详细的摘要。下面是伪代码。

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";
}