如何在 JavaScript 中添加多个数字？

Question

我是 JavaScript 编程新手..我一直在研究解决方案，但仍然..没有运气。例如：我想添加用户输入的 6 个（或更多）数字。我使用这段代码，但只计算前三个。当我添加四个数字时，会出现“NAN”。 Nan的意思是无效计算。

<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>

仅计算B1、B5和B9。这是工作代码：

<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>

这是表单操作：

<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>

当我单击计算按钮时，结果将显示在 t1 文本区域上。这是代码......

<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>

请帮助我。 :(

---

这是 sysdocadd.php 代码：

<?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

您可以稍微简化一下自己的答案：

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

问题解决了。

<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>

谢谢你们。 :)