Question

我正在尝试根据日期在表格中显示数据，目前正在使用以下代码。

if($_GET){
$datefrom = $_GET['datefrom'];
$dateto = $_GET['dateto'];
$date0 = date('Y-m-d', strtotime($datefrom . " +0 days"));
$date1 = date('Y-m-d', strtotime($datefrom . " +1 days"));
$date2 = date('Y-m-d', strtotime($datefrom . " +2 days"));
$date3 = date('Y-m-d', strtotime($datefrom . " +3 days"));
$date4 = date('Y-m-d', strtotime($datefrom . " +4 days"));
}

然后我使用以下语句重复查询我的数据库

        $stmtHR0 = $db->query("SELECT * FROM table WHERE Form_Date = '$date0' AND State = 'HR'");
        $numHR0 = $stmtHR0->rowCount();
        $stmtHR1 = $db->query("SELECT * FROM table WHERE Form_Date = '$date1' AND State = 'HR'");
        $numHR1 = $stmtHR1->rowCount();
        $stmtHR2 = $db->query("SELECT * FROM table WHERE Form_Date = '$date2' AND State = 'HR'");
        $numHR2 = $stmtHR2->rowCount();
        $stmtHR3 = $db->query("SELECT * FROM table WHERE Form_Date = '$date3' AND State = 'HR'");
        $numHR3 = $stmtHR3->rowCount();
        $stmtHR4 = $db->query("SELECT * FROM table WHERE Form_Date = '$date4' AND State = 'HR'");
        $numHR4 = $stmtHR4->rowCount();

“HR”只是其中一个州，我有30个州可以执行这些查询。代码有效，但我想知道是否有更好的方法来做到这一点，如果专家可以帮助我，我将不胜感激。请原谅我，因为我还在学习PHP。

Answer 1

这可以通过一个查询完成。假设From_Date字段为DATE字段类型：

if ($_GET) {
    $dateF = date('Y-m-d', strtotime($_GET['datefrom']));
    $dateT = date('Y-m-d', strtotime($_GET['dateto']));
    $sql = "
        SELECT   `From_Date`, COUNT(*) AS `count_num`
        FROM     `table`
        WHERE    `From_Date` BETWEEN '$dateF' AND '$dateT' AND `State` = 'HR'
        GROUP BY `From_Date`
        ORDER BY `From_Date` ASC
    ";
    // db fetch all
}

Answer 2

试试这段代码：实际上你可以使用数组作为日期

$date = array();
$date[0] = date('Y-m-d', strtotime($datefrom . " +0 days"));
$date[1] = date('Y-m-d', strtotime($datefrom . " +1 days"));
$date[2] = date('Y-m-d', strtotime($datefrom . " +2 days"));
$date[3] = date('Y-m-d', strtotime($datefrom . " +3 days"));
$date[4] = date('Y-m-d', strtotime($datefrom . " +4 days"));

for($i = 0; $i<count($date); $i++{
    $query_temp[] = "(
        SELECT COUNT(*) FROM table WHERE Form_Date = '" . $date[$i]. "' AND State = 'HR'
        ) AS count" . $i;

}

$query = "SELECT " . implode(", ",$query_temp);
$stmtHR4 = $db->query($query);

希望这段代码可以解决你的问题

使用PHP和Mysql根据表中的日期显示数据

2 个答案: