你好,这是我的代码:
<table class="table table-hover table-bordered">
<tr>
<th class="well" style="text-align:center">ID</th>
<th class="well" style="text-align:center">Email</th>
<th class="well" style="text-align:center">User level</th>
</tr>';
<?
$code_sql = "SELECT user_id, user_email,user_level FROM users ORDER BY user_id ASC ";
$code_query = mysql_query($code_sql) or die(error_sql(mysql_error(),__LINE__,__FILE__));
$sql_rows = mysql_num_rows($code_query);
if($sql_rows > 0){
while($rows = mysql_fetch_object($code_query)){
$user_id = intval($rows->user_id);
$user_level= intval($rows->user_level);
$user_email = htmlspecialchars($rows->user_email);
echo ' <tr>
<td>'.$user_id.'</td>
<td>'.$user_email.'</td>
<td>'.$user_level.'</td>
</tr>';
}
mysql_free_result($code_query);
}else{
echo '<tr>
<td>
<font color="red">no data found</font>
</td>
</tr>';
}
echo '</table>';
?>
代码的输出就像。
<table class="table table-hover table-bordered">
<tr>
<th class="well" style="text-align:center">ID</th>
<th class="well" style="text-align:center">Email</th>
<th class="well" style="text-align:center">User level</th>
</tr>
<tr>
<td>1</td>
<td>test1@gmail.com</td>
<td>1</td>
</tr>
<tr>
<td>2</td>
<td>test2@gmail.com</td>
<td>2</td>
</tr>
<tr>
<td>3</td>
<td>test3@gmail.com</td>
<td>2</td>
</tr>
<tr>
<td>4</td>
<td>test4@gmail.com</td>
<td>1</td>
</tr>
</table>
(1 =普通用户,2 =管理员用户级别),但我想要的是。就像那样
<table class="table table-hover table-bordered">
<tr>
<th class="well" style="text-align:center">ID</th>
<th class="well" style="text-align:center">Email</th>
<th class="well" style="text-align:center">User level</th>
</tr>
<div id="users">
<tr>
<td>1</td>
<td>test1@gmail.com</td>
<td>1</td>
</tr>
<tr>
<td>4</td>
<td>test4@gmail.com</td>
<td>1</td>
</tr>
</div>
<div id="admins">
<tr>
<td>2</td>
<td>test2@gmail.com</td>
<td>2</td>
</tr>
<tr>
<td>3</td>
<td>test3@gmail.com</td>
<td>2</td>
</tr>
</div>
</table>
我想添加div id =&#34;用户&#34;将包含来自数据库的所有用户,其中user_level = 1,另一个div id =&#34; admins&#34; for user_level = 2。
答案 0 :(得分:0)
尝试以下代码,您需要根据条件创建div,例如,如果迭代编号1然后在内容之前和内容之后插入div(打开)以及在内容之后插入...(如果迭代编号为3)然后在内容之后再插入div(打开)内容(关闭)
<table class="table table-hover table-bordered">
<tr>
<th class="well" style="text-align:center">ID</th>
<th class="well" style="text-align:center">Email</th>
<th class="well" style="text-align:center">User level</th>
</tr>
<?php
$code_sql = "SELECT user_id, user_email,user_level FROM users ORDER BY user_id ASC ";
$code_query = mysql_query($code_sql) or die(error_sql(mysql_error(),__LINE__,__FILE__));
$sql_rows = mysql_num_rows($code_query);
if($sql_rows > 0){
$i=0;
while($rows = mysql_fetch_object($code_query)){
$user_id = intval($rows->user_id);
$user_level= intval($rows->user_level);
$user_email = htmlspecialchars($rows->user_email);
if($i==0){?><div id='users'><?php }elseif($i==2){?><div id='admins'><?php }
echo ' <tr>
<td>'.$user_id.'</td>
<td>'.$user_email.'</td>
<td>'.$user_level.'</td>
</tr>';
if($i==0){?></div><?php }elseif($i==2){?></div><?php }
$i++;
}
mysql_free_result($code_query);
}else{
echo '<tr>
<td>
<font color="red">no data found</font>
</td>
</tr>';
}
echo '</table>';
?>
答案 1 :(得分:0)
尝试使用tbody将以下代码分组表行
<?php
$code_sql = "SELECT user_id, user_email,user_level FROM users ORDER BY user_level ASC, user_id ASC ";
$code_query = mysql_query($code_sql) or die(error_sql(mysql_error(),__LINE__,__FILE__));
$sql_rows = mysql_num_rows($code_query);
if($sql_rows > 0){
$newgroup = false;
$groupid = array(1=>'users', 2=>'admins');
while($rows = mysql_fetch_object($code_query)){
$user_id = intval($rows->user_id);
$user_level= intval($rows->user_level);
$user_email = htmlspecialchars($rows->user_email);
if($newgroup != $user_level) {
if($newgroup != false) echo '</tbody>';
echo '<tbody id="'.$groupid[$user_level].'">';
$newgroup = $user_level;
}
echo ' <tr>
<td>'.$user_id.'</td>
<td>'.$user_email.'</td>
<td>'.$user_level.'</td>
</tr>';
}
echo '</tbody>';
mysql_free_result($code_query);
}else{
echo '<tr>
<td>
<font color="red">no data found</font>
</td>
</tr>';
}
echo '</table>';
?>
注意: 如果您只想对表行进行分组,它将起作用。但是如果你想添加像动画这样的附加功能,你可能需要添加更多的css / jQuery。
其他建议: 你应该开始使用mysqli或PDO 2.您应该开始使用css来替换字体标记。