php
  • count
  • 2011-04-07 46 views 0 likes 
    0

    我要計數的行總數,樣品:PHP計數進行計數的行總計

    row1 = 1 
    row2 = 1 
    row3 = -1 
    row4 = 1 
    row5 = -1 
    

    總共= 1,becouse 1 + 1 +( - 1)1 +( - 1)= 1

    我想從數據庫指望它

    我的代碼是:

    <? $result_count_like = mysql_query("SELECT * FROM like_mood WHERE img_id='$imageid'"); 
    while($row_like_co = mysql_fetch_array($result_count_like)) 
        { 
        $total_num=$row_like_co['mood_num']; // THIS IS ROWS NAME 
        } 
    ?> 
    

    謝謝:)

    +0

    在這種情況下會重新分配在每個循環'$ total_num',從而使其-1到底。 – arma 2011-04-07 23:42:37

    回答

    0
    <? 
    $toCount = array(); 
    $result_count_like = mysql_query("SELECT * FROM like_mood WHERE img_id='$imageid'"); 
    while($row_like_co = mysql_fetch_array($result_count_like)) 
    { 
        $toCount[]=$row_like_co['mood_num']; // THIS IS ROWS NAME 
    } 
    
    echo array_sum($toCount); 
    
    ?> 
    
    +1

    你可以''total_num + = $ row_like_co ['mood_num'];' – 2011-04-07 23:42:18

    +0

    謝謝你verey很多,我嘗試它,它的工作很好:) – user674152 2011-04-07 23:43:08

    2

    嘗試...

    SELECT SUM(mood_num) FROM like_mood WHERE img_id= '$imageid' 
    

    例子..

    $sql = mysql_query("SELECT SUM(mood_num) FROM like_mood WHERE img_id= '$imageid'"); 
    $sql = mysql_fetch_array($sql); 
    $total = $sql[0]; 
    
    +0

    謝謝,但我嘗試你的代碼。但它不起作用。 – user674152 2011-04-07 23:37:52

    +0

    你可以轉儲你的表列和類型,它應該工作。 – 2011-04-07 23:41:20

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