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Found 4 results

  1. id | mid | pid | owgh | nwgh |1 3 12 1.5 0.62 3 12 1.5 0.33 3 14 0.6 0.44 3 15 1.2 1.15 4 16 1.5 1.06 4 17 2.4 1.27 3 19 3.0 1.4 Select mid , COUNT(distinct pid) as cpid , SUM(nwgh) as totalnwgh from test GROUP BY mid sqlfiddle : link of below result with above query mid cpid totalnwgh3 4 3.84 2 2.2 But above i need one more column that's as below : **totowgh** mid cpid totalnwgh totowgh3 4 3.8 6.3 (DISTINCT value as per pid column)4 2 2.2 3.9 where totowgh = 6.3 come by DISTINCT value as per pid columnthat's mid = 3 has count 5 but distinct pid = 4 for mid=3 same way "distinct" owgh = 6.3 for mid=3 and distinct pid.As pid=12 is count 1 time hence,1.5 + 0.6 + 1.2 + 3 = 6.3 ( please not this is as per DISTINCT value of pid )Please note : i need owgh value as per distinct pid or group by pid .. because if i replace value of owgh 0.6 with 1.5 then it will be 5.7 instead of 7.2 but value of owgh 0.6 belong to pid = 14 and not pid = 12 hence totalcount of owgh change ...but i need is 7.2SEE WHAT I MEANS : sqlfiddle.com/#!9/2a53c/6
  2. Hi! My problem is that I've made a function with "var kgTotales = 0". But when I sum "kgTotales = kgTotales + 2" the result is "02", it returns a string and not a number. I use this var in a bucle so the result is "kgTotales = 02...." and next to it the next numbers I want to sum. This is the code: <script type="text/javascript"> function multiplicar() { var contador1; var contador2 = 2; var contador3 = 1; var n1; var n2; var id1; var id2; var id3; var kgTotales = 0; var precioTotal = 0; for(contador1=1; contador1 < 60; (contador1)+2){ id1="n" +(contador1); id2="n" +(contador2); id3="result" +(contador3); contador1=(contador1)+2; contador2=(contador2)+2; contador3++; n1=document.getElementById(id1).value; n2=document.getElementById(id2).value; document.getElementById(id3).value = (n1*n2); precioTotal = precioTotal + (n1*n2); kgTotales = kgTotales+n2; document.getElementById("precioTotal").value = precioTotal + " Euros"; document.getElementById("kgTotales").value = kgTotales + " Kg"; } } </script> And this is the result: I don't understand why it happens because precioTotal is a similar var and also shows a sum and works well. I hope someone could helps me. Regards!
  3. i want to know how to sum the numbers returned from a php query of mysql data. please guide through...
  4. i tried this $TotalSales = mysql_fetch_array($sales);$TotalReceipts = mysql_fetch_array($receipt);$TotalBalance = $TotalSales - $TotalReceipts;echo $TotalBalance; what is the problem????
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