Question

我有一个活动表格供我的会员填写。表格正在运行并转到数据库。

但是，我怎样才能显示哪个成员发布了该活动，以便其他人知道？因为我尝试在sql字符串中使用user_id并且只显示数据库中的最后一篇帖子而不是所有用户的所有帖子。

这是表格php部分

<?php

include("config/db_connect.php");
include("config/ckh_session.php");


// Inserting user Details code
if (isset($_POST['btnSave'])) {
    $event_title = mysqli_real_escape_string($conn, $_POST['event_title']);
    $myDate = mysqli_real_escape_string($conn, $_POST['myDate']);
    $cboStartTime = mysqli_real_escape_string($conn, $_POST['cboStartTime']);
    $dteEndDate = mysqli_real_escape_string($conn, $_POST['dteEndDate']);
    $dteEndTime = mysqli_real_escape_string($conn, $_POST['dteEndTime']);
    $event_type = mysqli_real_escape_string($conn, $_POST['event_type']);
    $country = mysqli_real_escape_string($conn, $_POST['country']);
    $event_region = mysqli_real_escape_string($conn, $_POST['event_region']);
    $union_territory = mysqli_real_escape_string($conn, $_POST['union_territory']);
    $event_town = mysqli_real_escape_string($conn, $_POST['event_town']);
    $event_postalcode = mysqli_real_escape_string($conn, $_POST['event_postalcode']);
    $event_title = mysqli_real_escape_string($conn, $_POST['event_title']);
    $event_description = mysqli_real_escape_string($conn, $_POST['event_description']);
    $event_ltm = mysqli_real_escape_string($conn, $_POST['event_ltm']);


    $insert = mysqli_query($conn, "update events set event_title = '" . $event_title . "' , 
myDate            = '" . $myDate . "' ,
cboStartTime      = '" . $cboStartTime . "' , 
dteEndDate        = '" . $dteEndDate . "' , 
dteEndTime        = '" . $dteEndTime . "' , 
event_type        = '" . $event_type . "' , 
country           = '" . $country . "' , 
event_region      = '" . $event_region . "' , 
union_territory   = '" . $union_territory . "' , 
event_town        = '" . $event_town . "' , 
event_postalcode  = '" . $event_postalcode . "' , 
event_title       = '" . $event_title . "' , 
event_description = '" . $event_description . "' , 
event_ltm         = '" . $event_ltm . "' ") or die(mysqli_error($conn));


    if ($_POST['event_ltm'] == 'MF Cpl') {
        header("location: CoupleffEvent.php");
    } else if ($_POST['event_ltm'] == 'MM Cpl') {
        header("location: CouplemmEvent.php");
    } else if ($_POST['event_ltm'] == 'FF Cpl') {
        header("location: CoupleffEvent.php");
    }
}

// Fetch user details
$query = mysqli_query($conn, "select * from user where user_id =  '" . $_SESSION['last_id'] . "' ");
$fetch_user = mysqli_fetch_array($query);
$user_country = $fetch_user['user_country'];
$user_gender = $fetch_user['user_gender'];
?>

任何帮助都会非常感激，所以我可以进入下一步

Answer 1

我不确定你要了解的是什么。如果您正在获取user_id，则必须选择* FROM events然后加入用户表，例如INNER JOIN用户ON events.user_id = user.id

此类请求将显示有关user.id

的所有事件和联接信息

资源：https://www.w3resource.com/mysql/advance-query-in-mysql/inner-join-with-multiple-tables.php

希望它会有所帮助，

填写表格后列出所有用户

1 个答案: