Question

  $from = "2017-7-1";
  $to = "2017-8-25";

我不知道为什么我似乎无法正确选择 HTML5日期选择器中的从日期，例如当我选择2017-7-1时，结果页面不会在2017-7-1上显示结果数据，但是当我选择2017-6-30 一天之前时，它可以显示。知道为什么吗？我尝试了BETWEEN CLause和＆gt; = AND＆lt; =，他们都不工作知道为什么吗？感谢

  $result = mysql_query("SELECT * FROM treatmentdetail WHERE nextdate BETWEEN 
  '$from' AND '$to'");

  $result = mysql_query("SELECT * FROM treatmentdetail WHERE nextdate >= 
  '$from' AND nextdate <= '$to'");

Answer 1

You should convert properly to date

   SELECT * 
   FROM treatmentdetail 
   WHERE str_to_date(nextdate, '%Y-%m-%d') 
      BETWEEN str_to_date('$from','%Y-%m-%d' ) AND  str_to_date('$to','%Y-%m-%d' )

and you should use most recent mysql driver .. with param binding too

Answer 2

您缺少连接变量我想你将不得不使用连接变量。

$result = mysqli_query($conn,"SELECT * FROM treatmentdetail WHERE nextdate BETWEEN 
  '$from' AND '$to'") or die(mysqli_error($conn));

Answer 3

将您的字段更改为DATE和将您的日期保存为YYYY-MM-DD，如您所示

  $from = "2017-07-01";
  $to = "2017-08-25";

您可以使用mysql（i）正确搜索

SELECT * FROM treatmentdetail 
    WHERE 
        nextdate >= '$from' AND nextdate <= '$to';

查看文档

MySQL DATATYPES DATE

无法在MYSQL语句中正确选择FROM日期

3 个答案: