Question

如何通过使用LEFT JOIN方法来回显任何列值，在选择查询中使用的所有表中一个列名是公用的？在我的代码下面。我想同时获取所有列的“ app_id”值，但“ app_id”不显示任何内容。 app_id列在此查询使用的所有4个表中均可用。它会打印我要显示的所有行，但是app_id不返回任何值。

<?php

session_start();

include('../../connect.php');

$date = strtotime("+1 day", time());
$cdate = strtotime("-7 day", time());
$finaldate = date('Y-m-d', $date);
$scdate = date('Y-m-d', $cdate);

//$sql = "SELECT * FROM tblapps WHERE journeydate >= '".$finaldate."' AND appstatus = 'On-Hold' AND process = 'yes'";

$sql = "SELECT * from tblapps
LEFT JOIN tblapps2 ON tblapps.app_id =   tblapps2.app_id
LEFT JOIN tblapps3 ON tblapps.app_id =  tblapps3.app_id
LEFT JOIN payments ON tblapps.app_id = payments.app_id
WHERE
tblapps.journeydate >= '$finaldate' AND tblapps.appstatus = 'On-Hold' AND tblapps.process = 'yes'";

$result = $connect->query($sql);
if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "id: " . $row["app_id"]. " - Name: " . $row["firstname"]. " " . $row["journeydate"]. "<br>";
    }
} else {
    echo "0 results";
}
$conn->close();

?>

如果使用左联接的所有表中的一个列名相同，则打印列值

0 个答案: