检查null或0 [英] Check null or 0
问题描述
我正在使用以下代码创建总和:
预测=预测+ eval(f [i] .value);
这只适用于f [i] .value的情况。包含一个数字。
有人可以帮助我,以便它可以工作,如果f [i] .value
为空或包含文字。
问候,
S
I''m using the following code to create a sum:
forecast = forecast + eval(f[i].value);
This only works if "f[i].value" contains a number.
Can someone please help me so that it also works if "f[i].value" is
empty or contains text.
Regards,
S
推荐答案
sta ... @ gmail.com写道:
sta...@gmail.com wrote:
我正在使用以下代码创建一个总和:
预测=预测+ eval(f [i] .value);
这只适用于f [i] .value的情况。包含一个数字。
有人可以帮助我,以便它可以工作,如果f [i] .value是空的或包含文字。
问候,
S
I''m using the following code to create a sum:
forecast = forecast + eval(f[i].value);
This only works if "f[i].value" contains a number.
Can someone please help me so that it also works if "f[i].value" is
empty or contains text.
Regards,
S
试试这个:
预测=预测+ parseInt(f [i] .value);
http://www.w3schools.com/jsref/jsref_parseInt.asp
- JS >
st **** @ gmail.com 写道:
我正在使用以下代码创建一个总和:
forecast = forecast + eval(f [i] .value);
I''m using the following code to create a sum:
forecast = forecast + eval(f[i].value);
使用eval经常被滥用和不必要。
请尝试以下方法来创建总和:
>
预测+ = + f [i] .value;
The use of eval is most often misused and unnecessary.
Try the following to create your sum:
forecast += +f[i].value;
于2006年2月27日在comp.lang.javascript中写道 :
wrote on 27 feb 2006 in comp.lang.javascript:
我正在使用以下代码创建总和:
预测=预测+ eval(f [i] .value);
只有在f [i] .value时才有效。包含一个数字。
有人可以帮助我,以便它可以工作,如果f [i] .value是空的还是包含文本。
I''m using the following code to create a sum:
forecast = forecast + eval(f[i].value);
This only works if "f[i].value" contains a number.
Can someone please help me so that it also works if "f[i].value" is
empty or contains text.
eval()是邪恶的,我认为你没有理由在这里使用它。
===========
如果您的意思是:
var forecast = 7
var xxx = 8
var yyy =''xxx''
预测=预测+ eval(yyy);
提醒(预测)// 15
使用起来要好得多:
var forecast = 7
var xxx = 8
var yyy =''xxx''
预测=预测+窗口[yyy];
警报(预测)// 15
现在假如xxx没有初始化或NaN:
var forecast = 7
var xxx
var yyy =''xxx''
if(!window [yyy] || isNaN(window [yyy])){
var a = 0
yyy =''a''
}
预测=预测+窗口[yyy];
提醒(预测)// 7
这会有帮助吗?
-
Evertjan 。
荷兰。
(请在我的电子邮件地址中将x'变为点数)
eval() is evil, and I see no reason why you are using it here.
===========
If you mean the following:
var forecast = 7
var xxx = 8
var yyy =''xxx''
forecast = forecast + eval(yyy);
alert(forecast) //15
it is much better to use:
var forecast = 7
var xxx = 8
var yyy =''xxx''
forecast = forecast + window[yyy];
alert(forecast) //15
now what if xxx is not initialized or NaN:
var forecast = 7
var xxx
var yyy = ''xxx''
if (!window[yyy]||isNaN(window[yyy])) {
var a = 0
yyy = ''a''
}
forecast = forecast + window[yyy];
alert(forecast) //7
will this help?
--
Evertjan.
The Netherlands.
(Please change the x''es to dots in my emailaddress)
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