Question

我有两张桌子，

1 - 类别

2 - 子类别

 Category (table 1)
---------------
Cid    |   Name

1      |   Vegetable

2      |   Fruit


SubCategory  (table 2)
----------------
Cid  |  Sid | S_name

 1   |   1  | Carrot

 1   |   2  | Beans

 2   |  3   | Mango

我想在Json中得到这样的结果： -

  -> Category found

       ->  Vegetable

            -> carrot

             -> beans

       ->  Fruit

             -> mango

我正在使用下面的代码！任何人都可以帮我纠正我的代码和得到像我上面提到的输出！我多次尝试过。但没有得到解决方案！请帮我获取目录表单的输出。感谢

`$srtResult = "SELECT * FROM `category` `c` LEFT JOIN `subcategory` `s` ON 
(`c`.`cid` = `s`.`cid`)";

 //Execute Qu**strong text**ery
    $result=mysql_query($srtResult);
   //Iterate Throught The Results

  while ($row = mysql_fetch_assoc($result, MYSQL_ASSOC)) 
{ $count = $row['cid'];
 while($row['cid'] = $count)
 { $subcatitem[] = Array( "cid" => $row['cid'], "sid" =>$row['sid'],          
"sname" => $row['sc_name'] );
  }
  $json['category'][] = Array("category" => Array( "cid" => $row['cid'], 
"name" => $row['name'], "subcategory" => Array( $subcatitem))); 
  } 

  header('Content-type: application/json');
echo json_encode($json);`

Answer 1

这不够吗？

$srtResult = "SELECT * FROM `category` `c` LEFT JOIN `subcategory` `s` ON 
(`c`.`cid` = `s`.`cid`)";
$result = mysql_query($srtResult);

while ($row = mysql_fetch_assoc($result, MYSQL_ASSOC)) {
  $json['category'][$row['name']] = $row['sc_name']; 
} 

header('Content-type: application/json');
echo json_encode($json);

修改

在注释中进行了更多解释后，所需的PHP数组结构（给出预期的JSON字符串）应如下所示：

Array ( [success] => 1 [message] => Category found in Database [Category] => Array ( [0] => Array ( [id] => 1 [name] => vegetable [Subcategory] => Array ( [0] => Array ( [cid] => 1 [sid] => 1 [sc_name] => carrot ) [1] => Array ( [cid] => 1 [sid] => 2 [sc_name] => beans ) ) ) [1] => Array ( [id] => 2 [name] => Fruit [Subcategory] => Array ( [0] => Array ( [cid] => 1 [sid] => 1 [sc_name] => Mango ) ) ) ) )

PHP中两个mysql表的Json需要在层次结构中输出

1 个答案: