Question

我有一个邻接模型列表，这是查询：

SELECT t1.FIO AS lev1, t2.FIO AS lev2, t3.FIO AS lev3, t4.FIO AS lev4, t5.FIO AS lev5, t6.FIO AS lev6, t7.FIO AS lev7, t8.FIO AS lev8, t9.FIO AS lev9, t10.FIO AS lev10, t11.FIO AS lev11, t12.FIO AS lev12, t13.FIO AS lev13, t14.FIO AS lev14, t15.FIO AS lev15, t16.FIO AS lev16, t17.FIO AS lev17, t18.FIO AS lev18, t19.FIO AS lev19, t20.FIO AS lev20, t21.FIO AS lev21, t22.FIO AS lev22, t23.FIO AS lev23, t24.FIO AS lev24 FROM users AS t1 LEFT JOIN users AS t2 ON t2.parent_id = t1.id LEFT JOIN users AS t3 ON t3.parent_id = t2.id LEFT JOIN users AS t4 ON t4.parent_id = t3.id LEFT JOIN users AS t5 ON t5.parent_id = t4.id LEFT JOIN users AS t6 ON t6.parent_id = t5.id LEFT JOIN users AS t7 ON t7.parent_id = t6.id LEFT JOIN users AS t8 ON t8.parent_id = t7.id LEFT JOIN users AS t9 ON t9.parent_id = t8.id LEFT JOIN users AS t10 ON t10.parent_id = t9.id LEFT JOIN users AS t11 ON t11.parent_id = t10.id LEFT JOIN users AS t12 ON t12.parent_id = t11.id LEFT JOIN users AS t13 ON t13.parent_id = t12.id LEFT JOIN users AS t14 ON t14.parent_id = t13.id LEFT JOIN users AS t15 ON t15.parent_id = t14.id LEFT JOIN users AS t16 ON t16.parent_id = t15.id LEFT JOIN users AS t17 ON t17.parent_id = t16.id LEFT JOIN users AS t18 ON t18.parent_id = t17.id LEFT JOIN users AS t19 ON t19.parent_id = t18.id LEFT JOIN users AS t20 ON t20.parent_id = t19.id LEFT JOIN users AS t21 ON t21.parent_id = t20.id LEFT JOIN users AS t22 ON t22.parent_id = t21.id LEFT JOIN users AS t23 ON t23.parent_id = t22.id LEFT JOIN users AS t24 ON t24.parent_id = t23.id LEFT JOIN users AS t25 ON t25.parent_id = t24.id WHERE t1.id = 16

这是使用邻接模型列表进行24级深度的查询

然后我做了这个：

<? for($i = 0; $i < $query->num_rows(); $i++): ?>
<? $row = $query->row($i); ?>
    <? for($n = 1; $n < 25; $n++): ?>
    <? $lev = "lev$n"; ?>
    <?= $row->$lev; ?>
    <? endfor; ?>
<? endfor; ?>

它只渲染每一行的字段，我真的不知道热点使它成为hieararchical，我使用的是codeigniter，在这里使用行或对象更好？

我需要做这样的事情：

root_parent {
   parent_1 {
      child_1.name
      child_2.name
      child_3.name
   }

   parent_2 {
      child_1.name
      child_2.name
      child_3.name
   }

   parent_3 {
      child_1.name
      child_2.name
      child_3.name
   }
}

没有重复可能吗？

Answer 1

看起来有点复杂和不灵活，你在这里做的吉尔斯。创建数据库表不是更好，每行都有自己的category_id和parent_id吗？即给予顶级父母一个parent_id为0，孩子们取其父级别category_id的parent_id。这将为您提供无限的深度，并且在渲染树时更容易编码。例如：

第一级层次结构的SQL -

SELECT * FROM (your_table) WHERE parent_id=0

用于第二级层次结构的SQL -

SELECT * FROM (your_table) WHERE parent_id=(category_id of first level)

第三级层次结构的SQL -

SELECT * FROM (your_table) WHERE parent_id=(category_id of second level)

依旧......

如何在html + php中呈现一个树（codeIgniter）

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