Question

我想问一下，

我有两个表，用户和帖子与列字段

用户：user_id，姓名，电子邮件

帖子：post_id，user_id，post_title

我想显示所有用户的所有帖子，

但我只希望 logged_in用户会话再增加两个按钮，而其他公开帖子只有两个按钮

p / s：我在users表中使用电子邮件列字段作为登录 $ _ SESSION 。

<?php 
    global $connect;
    global $user_id;

    $sql_post = "SELECT * FROM posts";
    $run_post = mysqli_query($connect, $sql_post);

    if($run_post && mysqli_num_rows($run_post) > 0 )
    {
        while($row_post = mysqli_fetch_array($run_post))
        {
            $post_id    = $row_post['post_id'];
            $user_id    = $row_post['user_id'];
            $post_title = $row_post['post_title'];

            $sql_user   =  "SELECT * FROM users WHERE user_id='$user_id'";
            $run_user   = mysqli_query($connect, $sql_user);
            $check_user = mysqli_fetch_array($run_user);

                    $user_id     = $check_user['user_id'];
                    $user_name   = $check_user['name'];
                    $user_email  = $check_user['email'];

                    $post_output = "<div id='posts_wrap'>
                                        <p>$user_name</p>
                                        <p>$user_email</p>
                                        <p>$post_title</p>
                                        <a href=''><button>Like</button></a>
                                        <a href=''><button>Comment</button></a> 

                                            // i want these two button (Edit and Delete) only available to logged in user
                                            <a href=''><button>Edit</button></a>
                                            <a href=''><button>Delete</button></a>  
                                    </div>
                                   ";
                    echo $post_output;
        }
        mysqli_free_result($run_post);          
    }
    else
    {
        echo "No post yet";
    }
?>

Answer 1

用户登录后，在会话中保留用户详细信息，如果用户登录则检查条件。例如，如果您尝试评论并且仅对登录用户感兴趣那么您可以做一些事情，如

<?php
session_start();
$_SESSION['email']='email@example.com';
$user_name='dd';
$user_email='ddd';
$post_title='gsdg';

$post_output = "<div id='posts_wrap'><p>$user_name</p><p>$user_email</p><p>$post_title</p>";
 if(isset($_SESSION['email'])){ 
        $post_output.="<a href=''><button>Like</button></a><a href=''><button>Comment</button></a> ";
    }  
    // i want these two button (Edit and Delete) only available to logged in user
    $post_output.= "<a href=''><button>Edit</button></a><a href=''><button>Delete</button></a> </div>";                                    
            print_r($post_output);      





?>

在上面的代码用户登录，所以所有按钮都是可见的。如果没有那么它对所有人都不可见。只是试图破坏session.i认为以前的会话电子邮件仍在那里

Answer 2

我找到了解决方案。事实证明，我需要创建另一个查询，结合user_id和电子邮件，以确保登录属于登录用户。这是代码

<?php
    global $connect;
    global $user_id;

        $get_post = "SELECT * FROM posts";
        $run_post = mysqli_query($connect, $get_post);

        if($run_post && mysqli_num_rows($run_post) > 0 )
        {
            while($row_post = mysqli_fetch_array($run_post))
            {
            $post_id         = $row_post['post_id'];
            $user_id         = $row_post['user_id'];
            $post_title      = $row_post['post_title'];

            $emailsql    = $_SESSION['email'];
            $get_email   =  "SELECT * FROM users WHERE user_id='$user_id' AND email='$emailsql'";
            $run_email   = mysqli_query($connect, $get_email);
            $check_email = mysqli_fetch_array($run_email);

                $d_email = $check_email['email'];

            $get_user   =  "SELECT * FROM users WHERE user_id='$user_id'";
            $run_user   = mysqli_query($connect, $get_user);
            $check_user = mysqli_fetch_array($run_user);

                    $user_id     = $check_user['user_id'];
                    $user_name   = $check_user['name'];

                    echo    "<div id='posts_wrap'>
                                <p><h3><a href='userprofile.php?user_id=$user_id'>$user_name</a></h3></p>

                                <div id='posts_title'>
                                <p><h3><a href='post.php?post_id=$post_id'>$post_title</a></h3></p>
                                </div>

                                <hr>
                                <a href=''><button>Like</button></a>
                                <a href=''><button>Comment</button></a>
                                ";

                                if($check_email){
                                    echo "
                                    <a href=''><button>Edit</button></a>
                                    <a href=''><button>Delete</button></a>";
                                }

                    echo    "</div>";
            }
            mysqli_free_result($run_post);          
        }
        else
        {
            echo "No post yet";
        }   
?>

如何在PHP MySQLi中创建if语句内部的echo

2 个答案: