查找MYSQL查询中哪个条件失败（偏差总结）答案

【问题标题】：Finding which condition Failed in MYSQL Query (Deviation Summary)查找MYSQL查询中哪个条件失败（偏差总结）
【发布时间】：2019-08-16 09:42:14
【问题描述】：

如果任何 mysql 查询条件失败，我需要显示偏差摘要。我需要显示哪个数据输入错误，因为我没有得到任何结果。

更多的是一个逻辑问题，所以试图用谷歌搜索同样的问题。

表格字段： id , day , date , time , program , title

查询

SELECT Count(id) FROM `table` 
WHERE day='Monday' and date = '2018-10-15' and time = '19:45:34' and title = 'NEW-BOTTLE'

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

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，我需要显示哪个值是错误的，因为发生了计数 0。

【问题讨论】：

标签： php mysql logic

【解决方案1】：

您可以使用if 条件选择值如下：

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。希望对您有所帮助。

【讨论】：

【解决方案2】：

您可以使用 switch 语句打印详细的摘要。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";
}

【讨论】：