用PHP解析mysql數據到json用PHP

Question

嗨，對不起我的英文。

我想使用PHP解析從MySQL表中的數據到JSON。 我的表格有2列：'reg'，'base'代表該地區的地區和基地。 例子：

reg base 
KIE KIE1 
KIE KIE2 
KIE KIE3 
ZYT ZYT1 
ZYT ZYT2 
CHK CHK1 

等

我想這個值解析成JSON，看起來像：

{"text":"Bases","leaf":false,"children":{"text":"KIE","leaf":false,"children":[{"text":"KIE1","leaf":true},{"text":"KIE2","leaf":true},{"text":"KIE3","leaf":true}]},{"text":"ZYT","leaf":false,"children":[{"text":"ZYT1","leaf":true},{"text":"ZYT2","leaf":true}]},{"text":"CHK","leaf":false,"children":[{"text":"CHK1","leaf":true}]}} 

我的PHP代碼：

<?php 
      header("Content-type: application/json; charset=utf-8"); 
      $conn = mysqli_connect('localhost','test', '123','test'); 
      $bases = array('text'=>'Bases','leaf'=>false); 
      $regions = "select distinct reg from bases"; 
      $resreg = mysqli_query($conn,$regions); 
      while($reg=mysqli_fetch_assoc($resreg)){ 
       $bases['children'] = array('text'=>$reg['reg'],'leaf'=>false, 'children'=>array()); 
       $regi = $reg['reg']; 
       $query = "select base from bases where reg = '$regi' order by 1"; 
       $result = mysqli_query($conn,$query); 
       while($row = mysqli_fetch_assoc($result)){ 
        $c = array('text'=>$row['base'],'leaf'=>true); 
        $bases['children']['children'][]=$c; 
       } 
      } 
      $d = json_encode($bases); 
      echo $d; 
      mysqli_close($conn); 
?>  

這個問題我在JSON中只能從一個reg和don牛逼得到別人的

例子：

{"text":"Bases","leaf":false,"children":{"text":"KIE","leaf":false,"children":[{"text":"KIE1","leaf":true},{"text":"KIE2","leaf":true},{"text":"KIE3","leaf":true}]}  

我不知道，這是爲了得到我想要的，但我是新來的編程最好的梅託德。

任何人都可以幫助或者告訴我更好的方法來做到這一點？ 我需要它爲我的ExtJs應用程序樹。謝謝。

Answer 1

找到了一些解決方案：

list.php的

 <?php 
      header("Content-type: application/json; charset=utf-8"); 
      $conn = mysqli_connect('localhost','test', '123','test'); 
      $bases = array('text'=>'Bases','leaf'=>false,'children'=>''); 
      $regions = "select distinct reg from bases"; 
      $resreg = mysqli_query($conn,$regions); 
      while($reg=mysqli_fetch_assoc($resreg)){ 
       $bases['children'][] = array('text'=>$reg['reg'],'leaf'=>false,'children'=>''); 
      } 
      foreach($bases['children'] as $k=>$v){ 
       $query = "select base from bases where reg = '".$bases['children'][$k]['text']."'"; 
       $result = mysqli_query($conn,$query); 
       while($row=mysqli_fetch_assoc($result)){ 
        $bases['children'][$k]['children'][] = array('text'=>$row['base'],'leaf'=>true); 
       } 
      } 
      $d = json_encode($bases); 
      echo $d; 
      mysqli_close($conn); 
      ?> 

它是否類同this

我仍然認爲有更簡單更智能的解決方案。

無論如何，謝謝大家。

Answer 2

while($reg=mysqli_fetch_assoc($resreg)){ 
       $bases['children'] = array('text'=>$reg['reg'],'leaf'=>false, 'children'=>array()); 

應該是：

while($reg=mysqli_fetch_assoc($resreg)){ 
       $bases['children'][] = array('text'=>$reg['reg'],'leaf'=>false, 'children'=>array()); 

因爲現在您要添加的array('text'=>$reg['reg'],'leaf'=>false, 'children'=>array());到的$基地[ '孩子']數組的索引。

在哪裏 - 和以前每次while循環運行時重寫$bases['children']一樣，因此它只會等於分配給它的最後一個數據（最後一次迭代）的值。

這就回答瞭如何讓$bases['children']不止一個價值的問題，因爲'我不知道這是否是最好的方法'將它帶到代碼審查。