Question

我是JavaScript编程的新手。我一直在研究解决方案，但仍然没有运气。例如：我想添加用户将输入的6个数字（或更多）。我使用此代码，但只计算前三个。当我添加四个数字时，会出现“NAN”。 Nan意味着无效的计算。

<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>

以下是表单操作：

<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>

当我点击计算按钮时，结果将显示在t1文本区域。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>

请帮帮我。：（

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

问题已经解决了。

<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>

谢天谢地。：）

如何在JavaScript中添加多个数字？

2 个答案: