Javascript javascript中的整数？
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whole number in javascript?
提问by ktm
I get 28.6813276578 when i multiply 2 numbers a and b, how can i make it whole number with less digits
当我将 2 个数字 a 和 b 相乘时，我得到 28.6813276578，我怎样才能用更少的数字使它成为整数
and also, when i multiply again i get results after first reult like 28.681321405.4428.68 how to get only one result ?
而且，当我再次相乘时，我在第一次结果后得到结果，如 28.681321405.4428.68 如何只得到一个结果？
<script>
$(document).ready(function(){
$("#total").hide();
$("#form1").submit(function(){
var a = parseFloat($("#user_price").val());
var b = parseFloat($("#selling").val());
var total = a*b;
$("#total").append(total)
.show('slow')
.css({"background":"yellow","font-size":50})
;
return false;
});
});
</script>
回答by Jason
You can do several things:
你可以做几件事：
total = total.toFixed([number of decimals]);
total = Math.round(total);
total = parseInt(total);
toFixed()will round your number to the number of decimals indicated.Math.round()will round numbers to the nearest integer.parseInt()will take a string and attempt to parse an integer from it without rounding.parseInt()is a little trickier though, in that it will parse the first characters in a string that are numbers until they are not, meaningparseInt('123g32ksj')will return123, whereasparseInt('sdjgg123')will returnNaN.- For the sake of completeness,
parseInt()accepts a second parameter which can be used to express the base you're trying to extract with, meaning that, for instance,parseInt('A', 16) === 10if you were trying to parse a hexidecimal.
- For the sake of completeness,
toFixed()会将您的数字四舍五入到指定的小数位数。Math.round()将数字四舍五入到最接近的整数。parseInt()将采用一个字符串并尝试从中解析一个整数而不舍入。parseInt()不过有点棘手，因为它将解析字符串中的第一个字符，直到它们不是数字为止，这意味着parseInt('123g32ksj')将返回123，而parseInt('sdjgg123')将返回NaN。- 为了完整起见，
parseInt()接受第二个参数，该参数可用于表示您尝试提取的基数，这意味着，例如，parseInt('A', 16) === 10如果您试图解析一个十六进制。
- 为了完整起见，
回答by DuckMaestro
回答by SHug
In addition to the other answers about rounding, you are appending the answer to "total" by using
除了有关四舍五入的其他答案之外，您还使用以下方法将答案附加到“总计”
$("#total").append(total)
You need to replace the previous text rather than appending by using
您需要替换以前的文本而不是使用附加
$("#total").html(total)
