【发布时间】:2015-05-15 02:16:57
【问题描述】:
我正在编写一个wallpost 脚本,页面没有错误,我已经检查了数据库(查询在那里工作)并且我已经尝试了我认为可以的一切,但它就是不起作用。我不知道还能做什么,如果您对为什么它不起作用有任何想法,请告诉我。 PS:数据库中有一些虚拟数据。
<?php
include 'inc/dbc.php';
include 'inc/functions.php';
if(isset($_GET['user']) && !empty($_GET['user'])) {
$username = $_GET['user'];
} else {
$username = $_SESSION['username'];
}
$my_name = $_SESSION['username'];
$firstname = getuser($username, 'firstname');
$middlename = getuser($username, 'middlename');
$lastname = getuser($username, 'lastname');
$aboutme = getuser($username, 'aboutme');
$email = getuser($username, 'email');
$dob = getuser($username, 'dob');
$address = getuser($username, 'address');
$website = getuser($username, 'website');
$country = getuser($username, 'country');
$city = getuser($username, 'city');
$state = getuser($username, 'state');
$phone = getuser($username, 'phone');
$gender = getuser($username, 'gender');
$rank = getuser($username, 'rank');
$avatar = getuser($username, 'avatar');
$reg_date = getuser($username, 'reg_date');
?>
<?php
if (loggedIn() == true) {
$chech_posts = mysqli_query($mysqli, "SELECT * FROM wallposts WHERE posted_to = '$username' AND deleted = 0 ");
$check = mysqli_num_rows($chech_posts);
if ($check == false) {
echo '<li>Error Getting posts</li>';
} else {
while ($run = mysqli_fetch_array($check)) {
$post_id = $run['post_id'];
$postby = $run['posted_by'];
$postto = $run['posted_to'];
$post = $run['post'];
$post_date = $run['post_date'];
$post_time = $run['post_time'];
$p_avatar = getuser($postby, 'avatar');
$p_first = getuser($postby, 'firstname');
$p_last = getuser($postby, 'lastname');
?>
<li class='wall' id='<?php echo $post_id;?>'>
<div class='post'>
<div class='post-container'>
<div class='post-header'>
<div class='pull-left'><?php echo $postby;?></div>
<div class='post-img'><img src="images/users/<?php echo $p_avatar;?>" alt="<?php echo $p_first . ' ' . $p_last . '\'s Profile Picture' ;?>" class="img-circle" align="middle"></div>
<div class='pull-right'><?php echo $post_date?></div>
</div>
<div class='post-body'>
<p><?php echo $post;?></p>
</div>
<div class='post-footer'>
<div class='lk-cmt-shr'>
</div>
<span id='comments'>
</span>
</div>
</div>
</div>
</li>
<?php
}
}
} else {
echo '<li>You must be <a href="index.php">logged</a> in to view ' . $firstname . '\'s posts.</li>';
}
?>
【问题讨论】:
-
你希望它做什么?你是什么意思
"but it just will not work",你需要更好地描述它。怎么不行? -
它不起作用,因为什么也没发生,但在 phpmyadmin 中,脚本(选择查询)工作正常。我希望显示数据库中的数据。
-
那么,
$username有值吗?为什么不在运行查询之前echo告诉我们返回结果是什么。 -
是的,它有一个价值,我做了
echo 'test';并且它有效 -
我的意思是
echo $username....