如何在JavaScript中添加多個數字？

Question

我是JavaScript編程的新手。.我一直在研究解決方案，但是仍然..沒有運氣。 例如：我想添加用戶將輸入的6個（或更多）數字。 我使用此代碼，但僅計算前三個。 當我已經添加了四個數字時，就會出現“ NAN”。 Nan表示無效的計算。

<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>

僅計算B1，B5和B9。 這是工作代碼：

<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>

這是表單動作：

<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>

當我單擊“計算”按鈕時，結果將顯示在t1文本區域中。 這是該代碼。

<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>

請幫我。 :(

---

這是sysdocadd.php代碼：

<?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)
?> 

Answer 1

您可以稍微簡化一下自己的答案：

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;
}

Answer 2

問題解決了。

<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>

謝謝。 :)