php - PHP，MySQL - 如何参数化时间戳比较

Question

我有 $year 和 $month，我需要选择时间戳在一年/月之后和另一年/月之前的所有记录

WHERE m.id = :magazineId
      AND t.isChargedBack IS FALSE
      AND t.`timestamp` >= STR_TO_DATE(:year'-08-01', '%Y-%m-%d')
      AND t.`timestamp` < STR_TO_DATE('2013-09-01', '%Y-%m-%d')

刚开始，我只删除了一个年份条目，查询停止查找任何行。

应该如何参数化它？

Answer 1

本质上（为了示例而没有 PDO 参数）

PHP

$target = $year . '/' . $month . '/01';

SQL

" ...
WHERE m.id = :magazineId
AND t.isChargedBack IS FALSE
AND t.`timestamp` >= '$target'
AND t.`timestamp` < DATEADD('$target' + 1 MONTH);"

Answer 2

您是否尝试过使用：

`t`.`timestamp` BETWEEN "$DATE1" AND "$DATE2"?

MySQL 之间

您可能还想尝试而不是使用 STR_TO_DATE，将您的日期转换为时间戳

Answer 3

$mysqli = new mysqli("host", "user", "password", "world");
$mymagazineid = "Foo";
$year = "2013";

$parameter_clause = <<<EOD
   SELECT * FROM YourTable
   WHERE m.id = ?
   AND t.isChargedBack IS FALSE
   AND t.`timestamp` >= STR_TO_DATE(?, '%Y-%m-%d')
   AND t.`timestamp` < STR_TO_DATE('2013-09-01', '%Y-%m-%d')
EOD;

$stmt = mysqli_prepare($mysqli, $parameter_clause);
mysqli_stmt_bind_param($stmt, "ss", $mymagazineid, "$year-08-01");
mysqli_stmt_execute($stmt);

Answer 4

尝试这个：

$date1 = strtotime('01-08-$year');
$date2 = strtotime('01-09-2013');

$query = "SELECT * from your_table WHERE m.id = '{$magazineId}' AND t.isChargedBack IS FALSE
  AND t.`timestamp` BETWEEN  '{$date1}' AND '{$date2}'";