Mysql查詢加入條件

這是我的數據庫表，我如何查詢，以便我可以得到如下所需的輸出。我可以查詢數據庫使用單個查詢語句來獲取JSON輸出？Mysql查詢加入條件

Table name: rowManager 
    +-------------+-----------------+----------------+ 
    | rowid  |  rowname | showid   | 
    +-------------+-----------------+----------------+ 
    |   1 | first   |    0 | 
    |   17 | second   |    2 | 
    |   18 | third   |    0 | 
    |   20 | forth   |    0 | 
    +-------------+-----------------+----------------+ 
    Table name: row_vid 
    +-------------+----------------+-----------------+-----------------+ 
    | rowid  | name   |  description | submission_date | 
    +-------------+----------------+-----------------+-----------------+ 
    |   1 | Learn PHP  | abavavaav  | 2007-05-24  | 
    |   17 | Learn MySQL | sdasdsa   | 2007-05-24  | 
    |   20 | JAVA Script | Sanjay   | 2007-05-06  | 
    |   1 | JAVA   | Sanjay   | 2007-05-06  | 
    |   18 | Android  | Sanjay   | 2007-05-06  | 
    |   17 | ios   | Sanjay   | 2007-05-06  | 
    |   1 | python   | Sanjay   | 2007-05-06  | 
    |   18 | c++   | Sanjay   | 2007-05-06  | 
    |   18 | c#    | Sanjay   | 2007-05-06  | 
    |   17 | ruby   | Sanjay   | 2007-05-06  | 
    |   18 | JQuery   | Sanjay   | 2007-05-06  | 
    |   17 | objective c | Sanjay   | 2007-05-06  | 
    |   1 | JAVA Tutorial | Sanjay   | 2007-05-06  | 
    +-------------+----------------+-----------------+-----------------+ 



$rowQuery='SELECT * from row_vid,rowManagr where rowManagr.showid=0';

我試圖查詢這些表，使得它給了我像輸出：

{ 
    "responses":{ 
     "First":[ 
     { 
      "name":"Learn PHP" 

     }, 
     { 
      "name":"JAVA" 

     }, 
     { 
      "name":"python" 

     }, 
     { 
      "name":"JAVA Tutorial" 
     } 
     ], 
     "Other Show":[ 
     { 
      "name":"Learn MySQL" 
     }, 
     { 
      "name":"ios" 
     }, 
     { 
      "name":"ruby", 

     }, 
     { 
      "name":"objective c" 
     } 
     ], 
     "Videos":[ 
     { 
      "name":"Android" 
     }, 
     { 
      "name":"c++ " 
     }, 
     { 
      "name":"c#", 

     }, 
     { 
      "name":"JQuery" 
     } 
     ] 
    } 
}

感謝

來源

2014-02-11 user3294288

我覺得你首先需要看看連接。 – Strawberry

我想但不能得到所需的輸出:( – user3294288

要開始，我相信這是你要找的查詢：

SELECT rowManager.rowname as category, row_vid.name as name 
FROM  rowManager, row_vid 
WHERE rowManager.rowid=row_vid.rowid 
ORDER BY rowManager.rowid, row_vid.name

在PHP中，你可以這樣做：

$sql = "SELECT rowManager.rowname as category, row_vid.name as name FROM rowManager, row_vid WHERE rowManager.rowid=row_vid.rowid ORDER BY rowManager.rowid, row_vid.name" 

$query_result = mysqli_query($sql); 

// Piece together the $responses array, which will have [category] 
// as key, while value is an array of 'name':[name]' pairs. 
$responses = array(); 
while ($row = mysqli_fetch_assoc($query_result)) { 
    if (!array_key_exists($row['category'], $responses)) { 
    $responses[$row['category']] = array(); 
    } 
    array_push($responses[$row['category']], array('name' => $row['name'])); 
} 

// Entire $response array is the value of 'responses' key in the 
// final $results array. 
$results = array('responses' => $responses); 

// Convert $results_array to JSON and print. 
print(json_encode($results));

來源

2014-02-11 09:45:11

謝謝............ :) – user3294288

你可能會想使用JOIN構建您收集的數據。但是，然後，你將不得不「手動」把我變成JSON。 mysql沒有內置函數返回JSON數據。您可以在php的響應中執行此操作，也可以使用CONCAT或類似命令在mysql中構建您的JSON響應。

你最好的選擇，以便能夠調試，跟蹤誤差和容易地通過遍歷您mysql結果，構建多維array具有相似的佈局，然後在運行json_encode處理您JSON佈局可能會在php做到這一點陣列。

來源

2014-02-11 09:36:36 inquam

如果你有rowManager與列名row_vid之間的關係ROWID，那麼你不能得到輸出，這些查詢 -

SELECT 
    * 
FROM 
    row_vid AS rvid 
     LEFT JOIN 
    rowManagr rm ON rvid.rowid = rm.rowid 
WHERE 
    rm.showid = 0

這個查詢後，你將不得不使用PHP來遍歷這個查詢結果和格式化您的預期數據並使用PHP將這些數據編碼爲json。

來源

2014-02-11 09:42:07

Mysql查詢加入條件

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