I'm new to JavaScript Programming.. I've been researching for the solution but still.. no luck. E.g: I want to add like 6 numbers (or more) that the user will input. I use this code but only the first three are calculated. When I add like four numbers already, 'NAN' appears. Nan means invalid putation.

<script type="text/javascript">
function show() {
 var a = document.calc.B1.value*1; 
 var b = document.calc.B5.value*1;
 var c = document.calc.B9.value*1;
 var d = document.calc.B12.value*1;
 var e = document.calc.B17.value*1;
 var f = document.calc.B21.value*1;

 document.calc.t1.value = a + b + c + d + e + f;
}
</script>

Only B1, B5 and B9 are calculated. Here's the working code:

<script type="text/javascript">
function show() {
 var a = document.calc.B1.value*1; 
 var b = document.calc.B5.value*1;
 var c = document.calc.B9.value*1;
 document.calc.t1.value = a + b + c;
}
</script>

Here's the form action:

<form action="sysdocadd.php" method="post" name="calc">
t1= TOTAL (text type
<td><div align="center" class="style66"><input name="t1" type="text" size="18" id="t1" value="0.00"/></div></td>

When I click calculate button, the result will be shown on t1 text area. here's the code for that..

<tr>
<td><span class="style77">Click to add</span></td>
<td><div align="center" class="style66"><input type=button onClick='show()'value=Calculate /></div></td>
</tr>

Please help me. :(

---

Here's the sysdocadd.php code:

<?php
$con = mysql_connect("localhost","user","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("dbconnect", $con);

$sql="INSERT INTO contents (reportnum, postedby, sysdate, userdateinp, B1, B2, B3, B4, B5, B6, B7, B8, B9, B10, B11, B12,
                            B13, B14, B15, B16, B17, B18, B19, B20, B21, B22, B23, B24, B25, B26, B27, B28, B29, B30,
                            B31, B32, B33, B34, B35, B36, B37, B38, B39, B40, B41, B42, B43, B44, B45, B46, B47, B48, B49, B50, B51, B52,
                            B53, B54, B55, B56, t1, t2, t3, t4)
                            VALUES
('$_POST[reportnum]','$_POST[postedby]','$_POST[sysdate]','$_POST[userdateinp]','$_POST[B1]','$_POST[B2]','$_POST[B3]',
'$_POST[B4]','$_POST[B5]','$_POST[B6]','$_POST[B7]','$_POST[B8]','$_POST[B9]','$_POST[B10]','$_POST[B11]','$_POST[B12]',
'$_POST[B13]','$_POST[B14]','$_POST[B15]','$_POST[B16]','$_POST[B17]','$_POST[B18]','$_POST[B19]','$_POST[B20]','$_POST[B21]',
'$_POST[B22]','$_POST[B23]','$_POST[B24]','$_POST[B25]','$_POST[B26]','$_POST[B27]','$_POST[B28]','$_POST[B29]','$_POST[B30]',
'$_POST[B31]','$_POST[B32]','$_POST[B33]','$_POST[B34]','$_POST[B35]','$_POST[B36]','$_POST[B37]','$_POST[B38]','$_POST[B39]','$_POST[B40]',
'$_POST[B41]','$_POST[B42]','$_POST[B43]','$_POST[B44]','$_POST[B45]','$_POST[B46]','$_POST[B47]','$_POST[B48]','$_POST[B49]','$_POST[B50]',
'$_POST[B51]','$_POST[B52]','$_POST[B53]','$_POST[B54]','$_POST[B55]','$_POST[B56]','$_POST[t1]','$_POST[t2]','$_POST[t3]','$_POST[t4]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

mysql_close($con)
?> 

I'm new to JavaScript Programming.. I've been researching for the solution but still.. no luck. E.g: I want to add like 6 numbers (or more) that the user will input. I use this code but only the first three are calculated. When I add like four numbers already, 'NAN' appears. Nan means invalid putation.

<script type="text/javascript">
function show() {
 var a = document.calc.B1.value*1; 
 var b = document.calc.B5.value*1;
 var c = document.calc.B9.value*1;
 var d = document.calc.B12.value*1;
 var e = document.calc.B17.value*1;
 var f = document.calc.B21.value*1;

 document.calc.t1.value = a + b + c + d + e + f;
}
</script>

Only B1, B5 and B9 are calculated. Here's the working code:

<script type="text/javascript">
function show() {
 var a = document.calc.B1.value*1; 
 var b = document.calc.B5.value*1;
 var c = document.calc.B9.value*1;
 document.calc.t1.value = a + b + c;
}
</script>

Here's the form action:

<form action="sysdocadd.php" method="post" name="calc">
t1= TOTAL (text type
<td><div align="center" class="style66"><input name="t1" type="text" size="18" id="t1" value="0.00"/></div></td>

When I click calculate button, the result will be shown on t1 text area. here's the code for that..

<tr>
<td><span class="style77">Click to add</span></td>
<td><div align="center" class="style66"><input type=button onClick='show()'value=Calculate /></div></td>
</tr>

Please help me. :(

---

Here's the sysdocadd.php code:

<?php
$con = mysql_connect("localhost","user","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("dbconnect", $con);

$sql="INSERT INTO contents (reportnum, postedby, sysdate, userdateinp, B1, B2, B3, B4, B5, B6, B7, B8, B9, B10, B11, B12,
                            B13, B14, B15, B16, B17, B18, B19, B20, B21, B22, B23, B24, B25, B26, B27, B28, B29, B30,
                            B31, B32, B33, B34, B35, B36, B37, B38, B39, B40, B41, B42, B43, B44, B45, B46, B47, B48, B49, B50, B51, B52,
                            B53, B54, B55, B56, t1, t2, t3, t4)
                            VALUES
('$_POST[reportnum]','$_POST[postedby]','$_POST[sysdate]','$_POST[userdateinp]','$_POST[B1]','$_POST[B2]','$_POST[B3]',
'$_POST[B4]','$_POST[B5]','$_POST[B6]','$_POST[B7]','$_POST[B8]','$_POST[B9]','$_POST[B10]','$_POST[B11]','$_POST[B12]',
'$_POST[B13]','$_POST[B14]','$_POST[B15]','$_POST[B16]','$_POST[B17]','$_POST[B18]','$_POST[B19]','$_POST[B20]','$_POST[B21]',
'$_POST[B22]','$_POST[B23]','$_POST[B24]','$_POST[B25]','$_POST[B26]','$_POST[B27]','$_POST[B28]','$_POST[B29]','$_POST[B30]',
'$_POST[B31]','$_POST[B32]','$_POST[B33]','$_POST[B34]','$_POST[B35]','$_POST[B36]','$_POST[B37]','$_POST[B38]','$_POST[B39]','$_POST[B40]',
'$_POST[B41]','$_POST[B42]','$_POST[B43]','$_POST[B44]','$_POST[B45]','$_POST[B46]','$_POST[B47]','$_POST[B48]','$_POST[B49]','$_POST[B50]',
'$_POST[B51]','$_POST[B52]','$_POST[B53]','$_POST[B54]','$_POST[B55]','$_POST[B56]','$_POST[t1]','$_POST[t2]','$_POST[t3]','$_POST[t4]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

mysql_close($con)
?> 

• javascript
• numbers

Share Improve this question Follow edited Dec 8, 2013 at 4:47 Josh Crozier 242k5656 gold badges400400 silver badges313313 bronze badges asked Mar 21, 2011 at 1:43 catsgirl008catsgirl008 69122 gold badges1212 silver badges1515 bronze badges 5

• where is your PHP code - place where you want to add mulitiple numbers ? – bensiu Commented Mar 21, 2011 at 1:48
• This is JavaScript, not PHP.... – Yahel Commented Mar 21, 2011 at 1:55
• The code you're showing is Javascript. Is that what you meant? – Cfreak Commented Mar 21, 2011 at 1:57
• @Bensiu: Here is the sysdocadd.php – catsgirl008 Commented Mar 21, 2011 at 2:03
• @yc and cfreak: yes, sorry. Javascript. :) – catsgirl008 Commented Mar 21, 2011 at 2:07

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 1

You can simplify your own answer a little bit:

function addNums() {
    var sum = 0;

    for(i=0; i<14; i++)
        sum += parseFloat(document.forms["addition"]["B" + (4*i+1)].value);

    document.forms["addition"].t1.value = sum;
}

Problem's solved.

<script language="javascript" type="text/javascript">

function addNums(){
  num_1=Number(document.addition.B1.value);
  num_2=Number(document.addition.B5.value);
  num_3=Number(document.addition.B9.value);
  num_4=Number(document.addition.B13.value);
  num_5=Number(document.addition.B17.value);
  num_6=Number(document.addition.B21.value);
  num_7=Number(document.addition.B25.value);
  num_8=Number(document.addition.B29.value);
  num_9=Number(document.addition.B33.value);
  num_10=Number(document.addition.B37.value);
  num_11=Number(document.addition.B41.value);
  num_12=Number(document.addition.B45.value);
  num_13=Number(document.addition.B49.value);
  num_14=Number(document.addition.B53.value);
  valNum=num_1+num_2+num_3+num_4+num_5+num_6+num_7+num_8+num_9+num_10+num_11+num_12+num_13+num_14;
  document.addition.t1.value=valNum;
}
</script>

THANKS Y'ALL. :)