我只创建了一个简单的博客，但我无法从帖子中获得我想要的输出，我只想通过使用第一个表中的userID从第二个表中获得名字和姓氏。
controllers/posts.hp:
<?php
class Posts extends CI_Controller {
function __construct() {
parent::__construct();
$this->load->model('post');
}
function index() {
$data['posts'] = $this->post->get_posts();
$data['users'] = $this->post->get_users();
$this->load->view('post_index', $data);
}
function post($postID) {
$data['post'] = $this->post->get_post($postID);
$this->load->view('post', $data);
}
function correct_permissions($required) {
$user_type = $this->session->userdata('user_type');
if ($required == "User") {
if ($user_type) {
return true;
}
} elseif ($required == "Blogger") {
if ($user_type == "Blogger") {
return true;
}
}
}
function deletepost($postID) {
$user_type = $this->session->userdata('user_type');
if ($user_type != 'Blogger') {
echo "<script>alert:('Please log in to continue.');</script>";
redirect(base_url());
}
$this->post->delete_post($postID);
redirect(base_url() . 'posts');
}
}
型号/邮政.php
<?php
class Post extends CI_Model{
function get_posts($num=50,$start=0){
$this->db->select()->from('posts')->where('active',1)->order_by('date_added','desc')->limit($num,$start);
$query = $this->db->get();
return $query->result_array();
}
function get_post($postID){
$this->db->select()->from('posts')->where(array('active'=>1,'postID'=>$postID))->order_by('date_added','desc');
$query=$this->db->get();
return $query->first_row('array');
}
function get_user($userID){
$this->db->select()->from('users')->where(array('userID'=>$userID));
$query=$this->db->get();
return $query->first_row('array');
}
function get_users(){
$userID = $this->session->userdata('userID');
$this->db->select('firstname','lastname')->from('users')->where('userID',$userID);
$query = $this->db->get();
return $query->result_array();
}
function insert_post($data){
$this->db->insert('posts',$data);
return $this->db->insert_id();
}
function update_post($postID,$data){
$this->db->where('postID',$postID);
$this->db->update('posts',$data);
}
function delete_post($postID){
$this->db->where('postID',$postID);
$this->db->delete('posts');
}
}
我的代码来自views/post_index.php
<div class="panel">
<?php if (!isset($posts)) { ?>
<p>There are currently no posts on my blog.</p>
<?php
} else {
foreach ($posts as $row) {
?>
<div class="panel-heading">
<h2 href="<?= base_url() ?>posts/post/<?= $row['postID'] ?>"><?= $row['title'] ?></h2>
</div>
<div class="panel-body">
<p><?= substr(strip_tags($row['post']), 0, 200) . "..." ?></p>
<br><br>
<div class="panelver">
<h6 href="<?= base_url() ?>posts/post/<?= $row['userID'] ?>">Posted by:
<?php
if($row['userID'] == 0) {
echo "Someone";
} else {
echo $this->db->select('firstname','lastname')->from('users')->where('userID',$row['userID']);
}
?></h6>
<p href="<?= base_url() ?>posts/post/<?= $row['postID'] ?>">Added last: <?= $row['date_added'] ?></p>
</div>
<p><a href="<?= base_url() ?>posts/post/<?= $row['postID'] ?>">Read More</a> - - - <a href="<?= base_url() ?>edit/editpost/<?= $row['postID'] ?>">Edit</a> | <a href="<?= base_url() ?>posts/deletepost/<?= $row['postID'] ?>">Delete</a></p>
<hr/>
</div>
<?php
}
}
?>
</div>
输出：
博客文章
这是一个帖子，废话，废话。。。。。"发布者："必须输出author-name使用第一个表中的userID作为下面的输出，帮帮我！T_T。。。
发布者：2
上一篇：2014-09-20 11:30:00
阅读更多--编辑|删除
但当我做这个代码时：
<div class="panel">
<?php if (!isset($posts)) { ?>
<p>There are currently no posts on my blog.</p>
<?php
} else {
foreach ($posts as $row) {
?>
<div class="panel-heading">
<h2 href="<?= base_url() ?>posts/post/<?= $row['postID'] ?>"><?= $row['title'] ?></h2>
</div>
<div class="panel-body">
<p><?= substr(strip_tags($row['post']), 0, 200) . "..." ?></p>
<br><br>
<div class="panelver">
<h6 href="<?= base_url() ?>posts/post/<?= $row['userID'] ?>">Posted by:
<?php
if($row['userID'] == 0) {
echo "Someone";
} else {
echo $this->db->select('firstname','lastname')->from('users')->where('userID',$row['userID']);
}
?></h6>
<p href="<?= base_url() ?>posts/post/<?= $row['postID'] ?>">Added last: <?= $row['date_added'] ?></p>
</div>
<p><a href="<?= base_url() ?>posts/post/<?= $row['postID'] ?>">Read More</a> - - - <a href="<?= base_url() ?>edit/editpost/<?= $row['postID'] ?>">Edit</a> | <a href="<?= base_url() ?>posts/deletepost/<?= $row['postID'] ?>">Delete</a></p>
<hr/>
</div>
<?php
}
}
?>
</div>
这将发生：
博客文章
这是一个帖子，废话，废话。。。。。"postedby:"必须使用第一个表中的userID输出第二个表值中的作者名称getinguserID，请帮帮我！T_T。。。
"postedby:"必须使用第一个表中的userID作为下面的输出，输出从第二个表值获取userID的作者名称。
Posted by:
A PHP Error was encountered
Severity: 4096
Message: Object of class CI_DB_mysql_driver could not be converted to string
Filename: views/post_index.php
Line Number: 42
Added last: 2014-09-20 11:30:00
Read More - - - Edit | Delete
p.S：我是一名初级程序员，也是CodeIgniter的新手。
好的。所以我从你的问题中得到的是，你有两个表，其中一个表存储了userId，另一个表中存储了用户详细信息。所以尝试这个，因为你需要在回显之前得到行。
echo $this->db->select('firstname','lastname')->from('users')->where('userID',$row['userID'])->get()->row();
上面的代码也可以这样写：
$this->db->select('firstname','lastname')->where('userID',$row['userID'])->get('users')->row();
如果你有任何疑问，请告诉我。