2012-07-03 23 views
0

我試圖把數據放入from按某個鍵/值到現有陣列

[comments] => Array 
    (
     [count] => 2 
     [data] => Array 
      (
       [0] => Array 
        (
         [idcomments] => 1 
         [from] => Array 
           (
            [idusers] => 1 
            [username] => 
            [full_name] => AndrewLiu 
           ) 
         [text] => testing this comment out 
        ) 
       [1] => Array 
        (
         [idcomments] => 2 
         [from] => Array 
          (
           [idusers] => 1 
           [username] => 
           [full_name] => AndrewLiu 
          ) 
         [text] => more comments yeah 
        ) 
      ) 
    ) 

我有這樣的事情:

while($row = $SQL_products -> fetch(PDO::FETCH_ASSOC)){ 
    $json['data'][] = array(
     "comments" =>array(
      "count"=>$SQL_ccount[0], 
      "data" => array($SQL_comments->fetchAll(PDO::FETCH_ASSOC), 
      "from" => array($SQL_comments_from->fetchAll(PDO::FETCH_ASSOC))) 
     ), 
    ); 
} 

不過,這並不落入from 。不知道正確的語法是讓$SQL_comments_from進入data這樣像上面的例子是什麼。

這就是我得到

[comments] => Array 
    (
     [count] => 2 
     [data] => Array 
      (
       [0] => Array 
        (
         [0] => Array 
          (
           [idcomments] => 1 
           [from] => 1 
           [text] => testing this comment out 
          ) 
         [1] => Array 
          (
           [idcomments] => 2 
           [from] => 2 
           [text] => more comments yeah 
          ) 
        ) 
       [from] => Array 
        (
         [0] => Array 
          (
           [idusers] => 1 
           [username] => 
           [full_name] => AndrewLiu 
          ) 
        ) 
      ) 
    ) 

我試圖讓"from"進入"data"。我提供的例子不「從」進入其他的「從」做(這就是正確的下idcomments)

提前感謝!

回答

1

您的數據字段使用來自SQL_comments_from$SQL_comments_from的數據混合在一起,這應該會生成您想要的數據。

while($row = $SQL_products -> fetch(PDO::FETCH_ASSOC)){ 
    $from = $SQL_comments_from->fetchAll(PDO::FETCH_ASSOC); 
    $data= array(); 
    foreach ($SQL_comments->fetchAll(PDO::FETCH_ASSOC as $items){ 
     $data[] = array('idcomments' => $items['idcomments'], 'from' => $from[$items['from']], 'text' => $items['text']); 
    } 
    $json['comments'][] = array(
      "count"=>$SQL_ccount[0], 
      "data" => $data; 
      ) 
     ), 
    ); 

}       
+0

你是什麼意思它混在一起? – hellomello

+0

感謝穆薩,但有另一種方式,而不是正從[0],如果我有什麼多個陣列只需$?就像我匹配id的$是從[0]改爲其他的?謝謝! – hellomello

+0

我認爲,但由於$只有一個項目,我使用'$ from [0]' – Musa