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我有一个将数据插入数据库的表单。其中一个字段weDate是填写

$weDate=date('Y-m-d',strtotime('Friday'))

如果我使用以下内容:

$sql = mysql_query("SELECT ID as 'DB_ID', 
                         partnumber as 'Part_Number', 
                         pndesc as 'Part Number Description', 
                         name as 'Name', 
                         reason as 'Reason', 
                         comment as 'Comments', 
                         date as 'Date', 
                         time as 'Time', 
                         weDate as 'Weekend Date' 
                    FROM $table 
                    WHERE weDate = '2013-05-03'");

我得到结果。

当我尝试:

$sql = mysql_query("SELECT ID as 'DB_ID', 
                         partnumber as 'Part_Number', 
                         pndesc as 'Part Number Description', 
                         name as 'Name', 
                         reason as 'Reason', 
                         comment as 'Comments', 
                         date as 'Date', 
                         time as 'Time', 
                         weDate as 'Weekend Date'
                    FROM $table 
                    WHERE weDate = '$weDate2'");

我什么也得不到。为什么?

PHP 变量是:

$weDate2 = date('Y-m-d', strtotime('-1 weeks Friday')
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1 回答 1

3

strtotime("Friday")返回即将到来的星期五,即 5 月 10 日。这不是硬编码查询中的 5 月 3 日。

于 2013-05-06T19:13:39.597 回答