我不能显示第二个表，当我用不同的数据更改第二个表但没有显示任何错误？ [英] i cant show the second table, when im changing the second table with different data but doesn't show any error?

问题描述

$dep_re = $_POST['dpt_from'];
    $arive_re = $_POST['arive_to'];
    $day_re = $_POST['day_re'];
    $month_re = $_POST['month_re'];
    $year_re = $_POST['year_re'];
    $time_re = $_POST['time_re'];
    $adult_re = $_POST['no_of_pax_adult'];
    $child_re = $_POST['no_of_pax_child'];
   $date_re = "$year_re-$month_re-$day_re";

    $connectSQL = mysql_connect("localhost","root","");
     if (!$connectSQL)
        die("Database Not Found".mysql_error());

    $connectSelectdb = mysql_select_db("bobdatabse");
    if (!$connectSelectdb)
        die("Error connect to database".mysql_error());

    else{

        $query = "SELECT * FROM trip_info WHERE date='$date' AND time='$time' AND depart='$dep' AND arive='$arive'";
        //$query_re = "SELECT * FROM round_trip_info WHERE date_re='$date_re' AND time_re='$time_re' AND depart_re='$dep_re' AND arive_re='$arive_re'";
        $result = mysql_query($query);
        //$result_re = mysql_query($query_re);

        if(!$result)
            die("invalid query!".mysql_error());
        else
            echo "<form name=listtrip method=post action=Reservation.php>";
            echo "<b>$dep</b> to <b>$arive</b>";
            echo "<table border=1px>";
            echo "<tr><td>Date</td><td>Time</td><td>Departure</td><td>Arive</td><td>Price Adult</td><td>Price Child</td><td></td></tr>";
            while($row=mysql_fetch_array($result))
            {
                echo "<tr><td>".$row['date']."</td><td>".$row['time']."</td><td>".$row['depart']."</td><td>".$row['arive']."</td><td>".$row['adult']." x ".$adult."=".$row['adult'] * $adult."</td><td>".$row['child']." x ".$child."=".$row['child'] * $child."</td><td><input type=submit value=Booking Now></td></tr>";
            }
            echo "</table><br>";




            $connectSQL = mysql_connect("localhost","root","");
            if (!$connectSQL)
                die("Database Not Found".mysql_error());

            $connectSelectdb = mysql_select_db("bobdatabse");
                if (!$connectSelectdb)
                    die("Error connect to database".mysql_error());

            else{

            $query_re = "SELECT * FROM round_trip_info WHERE date_re='$date_re' AND time_re='$time_re' AND depart_re='$dep_re' AND arive_re='$arive_re'";
            $result_re = mysql_query($query_re);

            if(!$query_re)
                die("invalid query!".mysql_error());
            echo "<form name=listtrip method=post action=Reservation.php>";
            echo "<b>$arive</b> to <b>$dep</b>";
            echo "<table border=1px>";
            echo "<tr><td>Date</td><td>Time</td><td>Departure</td><td>Arive</td><td>Price Adult</td><td>Price Child</td><td></td></tr>";
            while($row_re = mysql_fetch_array($result_re,MYSQL_BOTH))
            {
                echo "<tr><td>".$row_re['date_re']."</td><td>".$row_re['time_re']."</td><td>".$row_re['depart_re']."</td><td>".$row_re['arive_re']."</td><td>".$row_re['adult_re']." x ".$adult_re."=".$row_re['adult_re'] * $adult_re."</td><td>".$row_re['child_re']." x ".$child_re."=".$row_re['child_re'] * $child_re."</td><td><input type=submit value=Booking Now></td></tr>";
            }
            echo "</table>";





}
}

推荐答案

dep_re =

dep_re =

_POST [' dpt_from' ];

_POST['dpt_from'];

arive_re =

arive_re =

这篇关于我不能显示第二个表，当我用不同的数据更改第二个表但没有显示任何错误？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！