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Php 如何在html中呈现多级mysql查询(codeIgniter)_Php_Mysql_Html_Codeigniter - Fatal编程技术网

Php 如何在html中呈现多级mysql查询(codeIgniter)

Php 如何在html中呈现多级mysql查询(codeIgniter),php,mysql,html,codeigniter,Php,Mysql,Html,Codeigniter,这是mysql查询 SELECT * FROM users AS up JOIN users AS u ON up.id = u.parent_id JOIN users AS uc ON u.id = uc.parent_id WHERE up.id =16 这是视图文件中的输出 <?php foreach($query->result() as $q): ?> <?php echo $q->id; ?> <?php end

这是mysql查询

SELECT * 
FROM users AS up
JOIN users AS u ON up.id = u.parent_id
JOIN users AS uc ON u.id = uc.parent_id
WHERE up.id =16
这是视图文件中的输出

   <?php foreach($query->result() as $q): ?>
    <?php echo $q->id; ?>

    <?php endforeach; ?>

mysql结果给出了3个级别的结果,但在html中它只输出1个级别,是最后一个级别,但我需要在层次结构中呈现它

将查询更改为:

选择up.id作为id1,u.id作为id2,uc.id作为id3
从用户到用户
以u ON up.id=u.parent\u id的身份加入用户
以uc ON u.id=uc.parent\u id的身份加入用户
其中up.id=16
然后您可以访问三个级别,即
id1
id2
id3

<?php
  foreach($query->result() as $q) {
    echo $q->id1;
    echo $q->id2;
    echo $q->id3;
  }
?>

将查询更改为:

选择up.id作为id1,u.id作为id2,uc.id作为id3
从用户到用户
以u ON up.id=u.parent\u id的身份加入用户
以uc ON u.id=uc.parent\u id的身份加入用户
其中up.id=16
然后您可以访问三个级别,即
id1
id2
id3

<?php
  foreach($query->result() as $q) {
    echo $q->id1;
    echo $q->id2;
    echo $q->id3;
  }
?>