将日期变量phpMyAdmin到SELECT命令-致命错误:未捕获ArgumentCountError

Question

我是新来的，我学习PHP和SQL连接。我想从SQL数据库中列出未预订的房间号。我创建了两张桌子(房间和预订)，填充了一些示例。

第一步

如果我把直接日期作为“预订结束”的日期(“2019-03-11”)，然后列出房间号就行了。

function reserv3()
{
    include "config.php";
    $html = "";
    $sql = 'SELECT roomnr , reserv_start FROM rooms LEFT OUTER JOIN reservation ON reservation.room_id = rooms.room_id AND DATE(reservation.reserv_start)  >= DATE("2019-03-11")';
    $result = mysqli_query($conn, $sql);
    while( $sor = mysqli_fetch_array($result) )
        $html .= "<DIV>".$sor['roomnr'].", ".$sor['reserv_start']."</DIV>";
    return $html;
}

第二步

我想将固定日期(“2019-03-11”)替换为变量。

$today = DATE("2019-03-11");
function reserv3($today)
{
    include "config.php";
    $html = "";
    $sql = 'SELECT roomnr , reserv_start FROM rooms LEFT OUTER JOIN reservation ON reservation.room_id = rooms.room_id AND DATE(reservation.reserv_start)  >= '.$today;     
    $result = mysqli_query($conn, $sql);
    while( $sor = mysqli_fetch_array($result) )
        $html .= "<DIV>".$sor['roomnr'].", ".$sor['reserv_start']."</DIV>";
    return $html;
}

在第2步中，传入了致命错误Fatal error: Uncaught :函数reserv3()的参数太少，0

你能告诉我如何把变量放到这个命令中吗？谢谢

Answer 1

您的代码中存在错误。PHP date()生成日期。如果您想要当前日期，请将日期写入日期('Y-m-d')，将日期和时间写入日期('Y-m-d H: i: s)‘。在mysql中，如果字段是日期类型，则不必使用mysql date()函数进行比较。只有当它是datetime类型时，才使用date()函数来提取日期。

<?php
include "config.php";

$today = DATE('Y-m-d');

function reserv3($today)
{
    global $conn;

    $sql = 'SELECT t1.roomnr , t2.reserv_end FROM reservation as t2 LEFT JOIN rooms as t1 ON t2.room_id = 
               t1.room_id where t2.reserv_end <= '.$today;     
    $result = mysqli_query($conn, $sql);

    if(mysqli_num_rows($result)>0){
     // do something
    }

}