Php 基于用户显示mysql数据库中的特定数据
我试图让它,如果一个用户看他们的页面,他们可以看到他们已经工作或已经完成的任务。然后我想这样做,如果他们看另一个用户页面,他们可以看到他们创建的项目Php 基于用户显示mysql数据库中的特定数据,php,mysql,database,Php,Mysql,Database,我试图让它,如果一个用户看他们的页面,他们可以看到他们已经工作或已经完成的任务。然后我想这样做,如果他们看另一个用户页面,他们可以看到他们创建的项目 $my_id = $_SESSION['user_id']; $frnd_query = mysql_query("SELECT user_one, user_two FROM frnds WHERE user_one = '$my_id' OR user_two = '$my_id'"); while($r
$my_id = $_SESSION['user_id'];
$frnd_query = mysql_query("SELECT user_one, user_two FROM frnds WHERE user_one = '$my_id' OR user_two = '$my_id'");
while($run_frnd = mysql_fetch_array($frnd_query)) {
$user_one = $run_frnd['user_one'];
$user_two = $run_frnd['user_two'];
if($user_one == $my_id) {
$user = $user_one;
} else {
$user = $user_two;
}
$username = getuser($user, 'username');
echo "<a href = 'my_ideas.php?user=$user' class = 'list' style=display:block>Ideas</a>";
?>
<ul>
<li><a href = '#my_links'>Linked Ideas</a></li>
<li><a href = '#idea_chemistry'>My Idea Chemistry</a></li>
<li><a href = 'settings.php'>Profile Settings</a></li>>
</ul>
</div>
<div id = 'my_projects'>
<?php
if($user != $_SESSION['user_id']) {
$ideas_sql = "SELECT * FROM projects WHERE user_id = $username";
$query = mysql_query($ideas_sql) or die(mysql_error());
$rsIdeas = mysql_fetch_assoc($query);
do {
?>
<h2><a href = 'edit_post.php?id=<?php echo $rsIdeas['id']; ?>'><?php echo $rsIdeas['name']; ?></a></h2>
<?php echo $rsIdeas['keywords']; ?></p>
<p><?php echo $rsIdeas['description']; ?></p>
<?php } while ($rsIdeas = mysql_fetch_assoc($query));
} else {
$ideas_sql = "SELECT * FROM projects WHERE user_id = $my_id";
$query = mysql_query($ideas_sql) or die(mysql_error());
$rsIdeas = mysql_fetch_assoc($query);
do {
?>
<h2><a href = 'edit_post.php?id=<?php echo $rsIdeas['id']; ?>'><?php echo $rsIdeas['name']; ?></a></h2>
<p><?php echo $rsIdeas['keywords']; ?></p>
<p><?php echo $rsIdeas['description']; ?></p>
<?php } while ($rsIdeas = mysql_fetch_assoc($query));
}
?>
<?php
}
?>
$my\u id=$\u会话['user\u id'];
$frnd_query=mysql_query(“从frnd中选择user_one、user_two,其中user_one='$my_id'或user_two='$my_id');
while($run\u frnd=mysql\u fetch\u数组($frnd\u query)){
$user\u one=$run\u frnd['user\u one'];
$user\u two=$run\u frnd['user\u two'];
如果($user\u one==$my\u id){
$user=$user\u one;
}否则{
$user=$user\u二;
}
$username=getuser($user,'username');
回声“;
?>
>
我在这里搜索了相关问题,但似乎找不到任何有助于解决问题的方法。这基本上是一个代码审查问题。SO网络上还有另一个网站,您可以在这里提出此类问题。尽管您应该表明您对编码有一定的了解 我不确定为什么没有起始PHP标记,但这里是第一部分
<?php
$my_id = $_SESSION['user_id'];
$frnd_query = mysql_query("SELECT user_one, user_two FROM frnds WHERE user_one = '".$my_id."' OR user_two = '".$my_id."'");
while($run_frnd = mysql_fetch_array($frnd_query)) {
$user_one = $run_frnd['user_one'];
$user_two = $run_frnd['user_two'];
if($user_one == $my_id) {
$user = $user_one;
} else {
$user = $user_two;
}
$username = getuser($user, 'username');
echo "<a href = 'my_ideas.php?user='".$user."' class = 'list' style=display:block>Ideas</a>";
?>
<ul>
<li><a href = '#my_links'>Linked Ideas</a></li>
<li><a href = '#idea_chemistry'>My Idea Chemistry</a></li>
<li><a href = 'settings.php'>Profile Settings</a></li>>
</ul>
<div id = 'my_projects'>
<?php
if($user != $_SESSION['user_id']) {
$ideas_sql = "SELECT * FROM projects WHERE user_id = '".$username."'";
$query = mysql_query($ideas_sql) or die(mysql_error());
$rsIdeas = mysql_fetch_assoc($query);
do {
?>
<h2><a href = 'edit_post.php?id=<?php echo $rsIdeas['id']; ?>'><?php echo $rsIdeas['name']; ?></a></h2>
<?php echo $rsIdeas['keywords']; ?></p>
<p><?php echo $rsIdeas['description']; ?></p>
<?php } while ($rsIdeas = mysql_fetch_assoc($query));
} else {
$ideas_sql = "SELECT * FROM projects WHERE user_id ='".$my_id."'";
$query = mysql_query($ideas_sql) or die(mysql_error());
$rsIdeas = mysql_fetch_assoc($query);
do {
?>
<h2><a href = 'edit_post.php?id=<?php echo $rsIdeas['id']; ?>'><?php echo $rsIdeas['name']; ?></a></h2>
<p><?php echo $rsIdeas['keywords']; ?></p>
<p><?php echo $rsIdeas['description']; ?></p>
<?php } while ($rsIdeas = mysql_fetch_assoc($query));
}
?>
<?php
}
?>
这是一个示例$frnd\u query=mysql\u query(“从frnd中选择user\u one,user\u two,其中user\u one='$my\u id'或user\u two='$my\u id'”)
在$frnd\u query=mysql\u query中更改此示例(“从frnd中选择user\u one,user\u two,其中user\u one='''.$my\u id''或user\u two='.$my\u id'.”)
注意变量周围的引号和圆点。您需要将它们添加到代码中。这样可以完全解决问题,还是我需要做更多的工作。不,我不这么认为。我在您的代码中看到了更多的问题,但您需要从何处开始。谢谢。我询问的唯一原因是,这会显示数据,但只显示当前lo显示的数据登录用户已输入。我想知道我应该怎么做才能在其配置文件页上获取其他用户的数据。是的,这是一个自定义函数。如果这看起来像代码审阅问题,我很抱歉,但我真正的问题是,当我运行此操作时,无论我在哪个用户配置文件页上,它只显示登录用户已输入的数据。此外,我不是su这两个部分发生了什么。我想问题编辑就是这样编排的。