mysql總和輸出在flot圖表

我試圖輸出兩個表項與flot/jquery圖表上的日期的運行總數。mysql總和輸出在flot圖表

到目前爲止，代碼工作正常，可以在給定的日期輸出兩個值的折線圖，但我試圖顯示同一圖上每列的運行總數。不知道是否可以使用單一的mysql查詢使用SUM（值）來實現這個?? - 只是不能得到它的工作。

<?php 

$query=("SELECT hours, credits, entryDate FROM entry WHERE uid='1'"); 

$result = mysql_query($query);   

while($row = mysql_fetch_assoc($result)) 
{ 
    $credits[] = array (strtotime ($row['entryDate'])*1000,($row['credits'])); 
    $hours[] = array (strtotime ($row['entryDate'])*1000,($row['hours'])); 
} ?> 

</script> 
var credits = <?php echo json_encode($credits);?> 
var hours = <?php echo json_encode($hours);?>

來源

2013-11-26 Zeb99

你是什麼意思的「總爲電子商務做ach專欄'？如果你使用SUM，你將得到一個結果（總和），並且很難有任何圖表。 – Krzysztof

_Running_總爲每列，不知道這是否可以幫助，但是這是我想要做的事 - 信用小時日期 C1 H1 D1 C2 H2 D2 給兩個圖： GRAPH 1 [X1 = C1 Y1 = d1] [x2 = c1 + c2 y2 = d2] GRAPH 2 [x1 = h1 y1 = d1] [x2 = h1 + h2 y2 = d2] – Zeb99

，最簡單的是做這將是剛在PHP運行總和，演出了對於每一日期，以及： -

<?php 

$query=("SELECT hours, credits, entryDate FROM entry WHERE uid='1' ORDER BY entryDate"); 

$result = mysql_query($query);   

while($row = mysql_fetch_assoc($result)) 
{ 
    $runningtotalcredits += $row['credits']; 
    $runningtotalhours += $row['hours']; 
    $credits[] = array (strtotime ($row['entryDate'])*1000,$row['credits']); 
    $creditssofar[] = array (strtotime ($row['entryDate'])*1000,$runningtotalcredits); 
    $hours[] = array (strtotime ($row['entryDate'])*1000,$row['hours']); 
    $hourssofar[] = array (strtotime ($row['entryDate'])*1000,$runningtotalhours); 
} ?> 

</script> 
var credits = <?php echo json_encode($credits);?> 
var hours = <?php echo json_encode($hours);?> 
var creditssofar = <?php echo json_encode($creditssofar);?> 
var hourssofar = <?php echo json_encode($hourssofar);?>

它也應該能夠做到這一點的SQL使用相關子查詢，但效率可能不會很大： -

<?php 

$query=("SELECT hours, 
       credits, 
       entryDate, 
       (SELECT SUM(hours) AS hourssofar FROM entry b WHERE uid='1' AND b.entryDate <= a.entryDate), 
       (SELECT SUM(credits) AS creditssofar FROM entry c WHERE uid='1' AND c.entryDate <= a.entryDate) 
     FROM entry a 
     WHERE uid='1' 
     ORDER BY entryDate"); 

$result = mysql_query($query);   

while($row = mysql_fetch_assoc($result)) 
{ 
    $credits[] = array (strtotime ($row['entryDate'])*1000,$row['credits']); 
    $creditssofar[] = array (strtotime ($row['entryDate'])*1000,$row['creditssofar']); 
    $hours[] = array (strtotime ($row['entryDate'])*1000,$row['hours']); 
    $hourssofar[] = array (strtotime ($row['entryDate'])*1000,$row['hourssofar']); 
} ?> 

</script> 
var credits = <?php echo json_encode($credits);?> 
var hours = <?php echo json_encode($hours);?> 
var creditssofar = <?php echo json_encode($creditssofar);?> 
var hourssofar = <?php echo json_encode($hourssofar);?>

（沒有測試）

來源

2013-11-26 11:55:27 Kickstart

謝謝！完美的作品。 – Zeb99

mysql總和輸出在flot圖表

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