我是JavaScript新手,在将多个值返回给表单时遇到问题。这是我的代码:
表格:
<form name="form1">
First option
<input type="text" size="20" name="OptA" rows="1">
Second option
<input type="text" size="20" name="OptB" rows="1">
<input type="button" name="calc" value="Calculate" onclick="Evaluate(document.form1.OptA, document.form1.OptB, document.form1.TheStr);">
<input type="text" size="5" name="TheStr" readonly><input type="text" size="5" name="TheSte" readonly>
</form>
剧本:
function Evaluate(OptA, OptB, TheStr)
{
var A_Count = 0;
var B_Count = 0;
var C_Count = 0;
var D_Count = 0;
var Str1 = 0;
var Str2 = "";
var Str3 = 0;
var Str4 = "";
var Char1 = "";
var Char2 = "";
var Char3 = "";
var Char4 = "";
var Total = 0;
var TheOptions = OptA.value + OptB.value;
var A1 = "";
TheStr.value = "";
TheOptions = TheOptions.toUpperCase();
for (var i = 0; i < TheOptions.length; i++)
{
if (TheOptions.charAt(i) == 'A')
{ A_Count++; }
else if (TheOptions.charAt(i) == 'B')
{ B_Count++; }
}
Str1 = "" + A_Count+B_Count;
Str2 = "";
while ((Str1.length > 2) & (Count < 20))
{
Str2 = "";
for (var i = 0; i < Str1.length - 1; i++)
{
Char2 = Str1.charAt(eval(i+1));
Str2 = Str2 + (parseInt(Str1.charAt(i)) + parseInt(Str1.charAt(eval(i+1))));
}
Count++;
Str1 = Str2;
}
if (Count > 19)
{
TheStr.value = '0%';
}
else
{
TheStr.value = parseInt(Str2,10) + '%';
}
for (var i = 0; i < TheOptions.length; i++)
{
if (TheOptions.charAt(i) == 'C')
{ C_Count++; }
else if (TheOptions.charAt(i) == 'D')
{ D_Count++; }
}
Str3 = "" + C_Count+D_Count;
Str4 = "";
while ((Str3.length > 2) & (Count < 20))
{
Str4 = "";
for (var i = 0; i < Str3.length - 1; i++)
{
Char4 = Str3.charAt(eval(i+1));
Str4 = Str4 + (parseInt(Str3.charAt(i)) + parseInt(Str3.charAt(eval(i+1))));
}
Count++;
Str3 = Str4;
}
if (Count > 19)
{
TheSte.value = '0%';
}
else
{
TheSte.value = parseInt(Str2,10) + '%';
}
TheStr=(TheStr.value, TheSte.value)
}
由于某种原因,它只返回第一个输入框的值,第二个输入框保持空白。我想我的问题是弄清楚如何返回多个值,然后如何将其解析为两个输入框。有什么建议吗?
答案 0 :(得分:1)
两个输入都命名为“TheStr”。如果我理解正确,你需要命名第二个输入“TheSte”,并在你的Evaluate函数中声明它。
编辑:此外,最后的指示
TheStr=(TheStr.value, TheSte.value)
无效。在您的函数中,TheStr是一个文本输入。您可以通过设置值属性
来更改其值