2016-12-04 24 views
0

我有一個問題,我只是編程方面的新手而且在邏輯問題上有點低。我想從MySQL查詢結果中生成數組值,並且它必須嵌套。插入一個數組的值取決於具有條件的數據表(嵌套數組)

我實際的代碼,我一直在努力:

<?php 
    $mostSold_arr = array(); 
    $subData_array = array(); 
    $day_arr = array(); 
    $data = array(); 
    // Connect to MySQL 
    $link = new mysqli('localhost', 'root', '', 'db'); 
    if ($link->connect_errno) { 
     die("Failed to connect to MySQL: (" . $link->connect_errno . ") " . $link->connect_error); 
    } 
    $fetch_data = mysqli_query($link, "SELECT products.id as prodID, title, SUM(total_price) as value, orders.order_date as date FROM products LEFT JOIN order_details ON order_details.product_id=products.id LEFT JOIN orders ON orders.id=order_details.order_id GROUP BY order_details.product_id ORDER BY value DESC LIMIT 2 ") or die(mysqli_error($link)); 

    while ($row_mainData = mysqli_fetch_assoc($fetch_data)) { 
     $mostSold_arr['category'] = $row_mainData['title']; 
     $mostSold_arr['value'] = $row_mainData['value']; 
     $prod_id['prodID'] = $row_mainData['prodID']; 
     $mostSold_arr['url'] = "#"; 
     $mostSold_arr['description'] = "click to drill-down"; 
     $mostSold_arr['data'] = array(); 
     $fetch_subData = mysqli_query($link, "SELECT title, SUM(total_price) as value, MONTHNAME(orders.order_date) as date FROM products LEFT JOIN order_details ON order_details.product_id=products.id LEFT JOIN orders ON orders.id=order_details.order_id WHERE order_details.product_id=" . $prod_id['prodID'] ." GROUP BY title, date ORDER BY order_date LIMIT 3 ") or die(mysqli_error($link)); 
     while ($row_subData = mysqli_fetch_assoc($fetch_subData)) { 
      $subData_arr['category']=$row_subData['date']; 
      $subData_arr['value']=$row_subData['value']; 
      $day_arr['data'] = array(); 
      array_push($mostSold_arr['data'],$subData_arr); 

      $fetch_subDataDay = mysqli_query($link, "SELECT title, SUM(total_price) as value, DATE_FORMAT(orders.order_date, '%M-%d') as date FROM products LEFT JOIN order_details ON order_details.product_id=products.id LEFT JOIN orders ON orders.id=order_details.order_id WHERE order_details.product_id=" . $row_mainData['prodID'] ." AND DATE_FORMAT(orders.order_date, '%M-%d') LIKE '%March%' GROUP BY date ORDER BY order_date LIMIT 3 ") or die(mysqli_error($link)); 

      while ($row_subDataDay = mysqli_fetch_assoc($fetch_subDataDay)) { 
       $subDataDay_arr['category']=$row_subDataDay['date']; 
       $subDataDay_arr['value']=$row_subDataDay['value']; 
       //array_push($day_arr['data'],$subDataDay_arr); 

       if($subData_arr['category'] == 'March'){ 
        array_push($mostSold_arr['data'],$subDataDay_arr); 
       } 
       //array_push($day_arr['data'],$subDataDay_arr); 
       //var_dump($subData_arr); 
      } 
     } 
     array_push($data, $mostSold_arr); 
    } 

    $jsonData = json_encode($data, JSON_PRETTY_PRINT); 
    $fp = fopen('dataColMultiNested.json', 'w'); 
    fwrite($fp, $jsonData); 
    fclose($fp); 

    die('<pre>' . print_r($jsonData, 1)); 
    mysqli_close($link); 
    ?> 

上面的代碼的實際輸出:

[ 
    { 
     "category": "ACER E5-473-53C0", 
     "value": "2455200.00", 
     "url": "#", 
     "description": "click to drill-down", 
     "data": [ 
      { 
       "category": "March", 
       "value": "809100.00" 
      }, 
      { 
       "category": "March-02", 
       "value": "27900.00" 
      }, 
      { 
       "category": "March-03", 
       "value": "27900.00" 
      }, 
      { 
       "category": "March-04", 
       "value": "55800.00" 
      }, 
      { 
       "category": "April", 
       "value": "781200.00" 
      }, 
      { 
       "category": "May", 
       "value": "418500.00" 
      } 
     ] 
    }, 
    { 
     "category": "ACER ASPIRE E5-551G-F57K LAPTOP", 
     "value": "2339220.00", 
     "url": "#", 
     "description": "click to drill-down", 
     "data": [ 
      { 
       "category": "March", 
       "value": "449850.00" 
      }, 
      { 
       "category": "March-02", 
       "value": "29990.00" 
      }, 
      { 
       "category": "March-03", 
       "value": "59980.00" 
      }, 
      { 
       "category": "March-07", 
       "value": "59980.00" 
      }, 
      { 
       "category": "April", 
       "value": "689770.00" 
      }, 
      { 
       "category": "May", 
       "value": "779740.00" 
      } 
     ] 
    } 

預期輸出(它應該是這樣的)

[ 
     { 
      "category": "ACER E5-473-53C0", 
      "value": "2455200.00", 
      "url": "#", 
      "description": "click to drill-down", 
      "data": [ 
       { 
        "category": "March", 
        "value": "809100.00", 
        "data": [ 
         { 
         "category": "March-02", 
         "value": "27900.00" 
        }, 
        { 
         "category": "March-03", 
         "value": "27900.00" 
        }, 
        { 
         "category": "March-04", 
         "value": "55800.00" 
        } 
        ] 
       }, 
       { 
        "category": "April", 
        "value": "781200.00" 
       }, 
       { 
        "category": "May", 
        "value": "418500.00" 
       } 
      ] 
     }, 
     { 
      "category": "ACER ASPIRE E5-551G-F57K LAPTOP", 
      "value": "2339220.00", 
      "url": "#", 
      "description": "click to drill-down", 
      "data": [ 
       { 
        "category": "March", 
        "value": "449850.00", 
        "data": [ 
         { 
         "category": "March-02", 
         "value": "29990.00" 
        }, 
        { 
         "category": "March-03", 
         "value": "59980.00" 
        }, 
        { 
         "category": "March-07", 
         "value": "59980.00" 
        } 
        ] 
       }, 
       { 
        "category": "April", 
        "value": "689770.00" 
       }, 
       { 
        "category": "May", 
        "value": "779740.00" 
       } 
      ] 
     } 
] 

我正在爲此工作約2晚或更長時間,但可以爲我的問題找到正確的解決方案。 我希望你們能幫助我。謝謝。

回答

0

這是一個完全退還的答案,基於你的問題和它的標籤中缺乏的信息:你想要的結果是JSON格式(但我必須承認你的陳述符合它)。

在PHP中,你必須先創建一個等價陣列,這將是這樣的:

[ 
    0 => [ 
     "year" => "2016", 
     "data" => [ 
      0 => [ 
       "month" => "Dec", 
       "data" => [ 
        0 => [ 
         "day": "Dec 04", 
        ], 
       ], 
      ], 
     ], 
    ], 
] 

可以用的東西不遠處您最初的代碼生成:

$fetch_data1 = "Select * From tbl1"; 
$year = []; 

while($row_year = mysqli_fetch_assoc($fetch_data1)){ 
    $year['year'] = $row_year['year']; 

    $fetch_data2 = "Select * From tbl2"; 
    /* my query needs to loop inside because 
    it will contain condition depends on the retrieved data row */ 
    $data = []; 
    while($row_month = mysqli_fetch_assoc($fetch_data2)){ 
     $data['month'] = $row_month['month']; 

     $fetch_data3 = "Select * From tbl3"; 
     /* my query needs to loop inside because 
     it will contain condition depends on the retrieved data row */ 
     $day = []; 
     while($row_day = mysqli_fetch_assoc($fetch_data3)){ 
      $day['day'] = $row_month['month'] . ' ' . $row_day['day']; 
     } 
     $data['data'] = $day; 
    } 
    $year[] = $data; 
} 
+0

我需要它json格式,所以輸出應該有'{}'。順便謝謝你的回覆。我會試試.. – Marksmanship

+0

@Marksmanship無論如何,在PHP中,你會像我展示的那樣得到它。如果你想用JSON格式,你必須實際做'$ json_year = json_encode($ year);'。它會導致只使用'{}'而不使用'[]'的東西。 – cFreed

+0

沒有'[]'它仍然算作JSON格式?對不起,我只是一個新手。 – Marksmanship