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Web Infos
Asked: 2018-01-13 02:12:46 +0800 CST

Mysql 更新查询未提供所需的结果

  • 772

在我的Php 代码中，我有查询作为

SELECT COUNT(id) AS todayleads 
FROM `lead_data` 
where `date` = '$todate' AND lead_status = 1 
AND uid = '$userid' OR assign_to = '$userid'

当我回应查询时，结果是我

SELECT COUNT(id) AS todayleads
FROM `lead_data` 
where `date` = '2018-01-12' AND lead_status = 1 
AND uid = '3' OR assign_to = '3'

输出为我提供了从头到尾的所有数据。但是预期的 O/p只是具有 lead_status = 1 和 uid = 3 或 assign_to = 3 的数据。

mysql php

1 个回答

  1. Best Answer

    据我了解，根据您的评论，您应该在 WHERE 子句中添加一个括号：

    SELECT COUNT(id) AS todayleads
FROM   `lead_data` 
WHERE  `date` = '2018-01-12' 
AND    lead_status = 1 
AND    (uid = '3' OR assign_to = '3')
    • 0

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