i need a regular expression for decimal/float numbers like 12 12.2 1236.32 123.333 and +12.00 or -12.00 or ...123.123... for using in javascript and jQuery. Thank you.
选择性地配对+或——在开始时,先是一位或数位精干的数位数,然后是一位精干的数位数和一个或数位精度的数位数。
/^[+-]?d+(.d+)?$/
The right expression should be as followed:
[+-]?([0-9]*[.])?[0-9]+
申请:
+1
+1.
+.1
+0.1
1
1.
.1
0.1
这里的例子就是:
import re
#print if found
print(bool(re.search(r [+-]?([0-9]*[.])?[0-9]+ , 1.0 )))
#print result
print(re.search(r [+-]?([0-9]*[.])?[0-9]+ , 1.0 ).group(0))
产出:
True
1.0
如果你使用mac,你可以检查指挥线:
python -c "import re; print(bool(re.search(r [+-]?([0-9]*[.])?[0-9]+ , 1.0 )))"
python -c "import re; print(re.search(r [+-]?([0-9]*[.])?[0-9]+ , 1.0 ).group(0))"
您可以核对文本的确认,而且只有1个精准点的验证,即NaN
var val = $( #textbox ).val();
var floatValues = /[+-]?([0-9]*[.])?[0-9]+/;
if (val.match(floatValues) && !isNaN(val)) {
// your function
}
对任何人来说:
我为之做了一次科学论文的登记:(与全方位相匹配)。
const floatR = /^[+-]?(?:[0-9]+(?:.[0-9]+)?|.[0-9]+)(?:[eE][+-]?[0-9]+)?$/;
floatR.test("-2.3E23"); //=> true
floatR.test("-2.E"); //=> false
floatR.test(" 2 "); //=> false
If you prefer to use Unicode numbers, you could replace all [0-9] by d in the RegExp.
And possibly add the Unicode flag u at the end of the RegExp.
For a better understanding of the pattern see https://regexper.com/.
And for making RegExp, I can suggest https://regex101.com/.
EDIT:
EDIT2:虽然有人特别要求登记,但可以直接在联合材料中查证: Converting the string to a number (or EDIT3:我发现另一个冷却场:见。 EDIT 4: 如果你想要排除铅/tra0使用以下手段: 此外,在 Java,数字可以强调数字分离器,因此,为了完整起见,数字应当与任何有效的联合数字相匹配: 注:由于某些排气管重新作为未填入的字面处理(<0>代码>)。/**
* ## Parse {@linkcode str} to number
* @param {string} str - the string to parse
* @returns {number | null} the parsed number or `null` if not-a-number
*/
const parseNum = str => Number.isNaN(str = Number(str)) ? null : str;
parseNum("123.12345678E+3") ?? 1; //=> 123123.45678
parseNum("80F") ?? 1; //=> 1
NaN) with Number()
then checking if it is NOT NaN with !Number.isNaN()// Exclude leading but not trailing 0s
const floatR_L = /^[+-]?(?:[1-9][0-9]*(?:.[0-9]+)?|.[0-9]+)(?:[eE][+-]?[1-9][0-9]*)?$/;
// Exclude leading and trailing 0s
const floatR_LT = /^[+-]?(?:[1-9][0-9]*(?:.[0-9]*[1-9])?|.[0-9]*[1-9])(?:[eE][+-]?[1-9][0-9]*)?$/;
// Allow _ as separator, leading/trailing dot and 0s, and test for prefix 0b (binary), 0o (octal), and 0x (hexadecimal)
const numR = /^[+-]?(?:0[bB][01](?:_?[01]+)*|0[oO][0-7](?:_?[0-7]+)*|0[xX][0-9a-fA-F](?:_?[0-9a-fA-F]+)*|(?:[0-9](?:_?[0-9]+)*(?:.(?:[0-9](?:_?[0-9]+)*)?)?|.[0-9](?:_?[0-9]+)*)(?:[eE][+-]?[0-9](?:_?[0-9]+)*)?)$/;
numR.test("007.123e+0_1"); //=> true
numR.test("-0xAbc"); //=> true
numR.test("+0x00F"); //=> true
numR.test("0b0010_1010"); //=> true
numR.test("2."); //=> true
numR.test(".2"); //=> true
numR.test("00."); //=> true
numR.test(" 2 "); //=> false
numR.test("2_"); //=> false
numR.test("2._2"); //=> false
www.un.org/spanish/ecosoc 这是一个老的职位,但这是对“浮动点的经常表述”或像这种方式那样做的最主要搜索结果,没有回答问题。 页: 1 我在做这项工作时将分享我的结果,这样,下一位在座的人就不必为自己做工作。
All of the answers thus far accept a leading 0 on numbers with two (or more) digits on the left of the decimal point (e.g. 0123 instead of just 123) This isn t really valid and in some contexts is used to indicate the number is in octal (base-8) rather than the regular decimal (base-10) format.
这些表述也接受一种精度的缩略语(.14,而不是0.14,或无微小部分(3.,而不是3.0)。 在一些方案方面(包括 Java),这一点是有效的,但我想解一下(因为出于我的目的,这些错误可能比故意错误)。
“科学标志”如<条码>1.234E7。
/^(-)?
或者如果你真想接受一个主要<代码>+,那么:
/^((+-)? (0)([1-9][0-9]*)([0-9]+)$/
我认为,定期表达将严格检验典型的惯性激化或mal式浮动点。
相应:
$1 contains the full number that matched$2 contains the (possibly empty) leading sign (+/-)$3 contains the value to the left of the decimal point$5 contains the value to the right of the decimal point, including the leading .我通过“限制”表示,数字必须是你正在检测的唯一东西。
如果你想从包含其他内容的插图中仅提取浮动值:
/((<><>>>0>[1-9][0-9]*)([0-9]+)/
将在<代码>中找到<代码>-3.14> ,“否定式复制件大约为-3.14>>>> /code>或(-3.14)“ 等。
编号组别的意思与上文相同(但<代码>2现为无主要标志的空档(>>)。
但是,人们知道,它也将努力提取它能够找到的任何数字。 例如,它将从<代码>127.0.0.1中提取<>127.0。
如果你想要比那时更精练,我认为你可能希望研究的是灵活的分析,而不是经常表达。 I m guessing one can make a look-ahead-based expression that will recognized that Pi is 3.14 contained aroting Point number but 家庭是127.0.0.1。 不是,但最复杂。 如果您的格局取决于其之后的特性,以非维系的方式,你就开始在固定的言辞中冒险。
保罗普惠制和酒吧每周的答复使我得以这样做:
re=/^[+-]?(?:d*.)?d+$/;
>> /^[+-]?(?:d*.)?d+$/
re.exec("1")
>> Array [ "1" ]
re.exec("1.5")
>> Array [ "1.5" ]
re.exec("-1")
>> Array [ "-1" ]
re.exec("-1.5")
>> Array [ "-1.5" ]
re.exec(".5")
>> Array [ ".5" ]
re.exec("")
>> null
re.exec("qsdq")
>> null
www.un.org/spanish/ecosoc 这里是我的js 方法,处理0s , 载体
1- ^0[0-9]+.?[0-9]*$ : will find numbers starting with 0 and followed by numbers bigger than zero before the decimal seperator , mainly ".". I put this to distinguish strings containing numbers , for example, "0.111" from "01.111".
2-([1-9]{1}[0-9]? 如果从0开始就进行扼杀,则将考虑到大于0的部分。 此处使用的是母体,因为我只想捕捉符合规范的部分。
3-([0-9]<>>>? [0-9]:仅能抓住扼杀的精髓部分。
在Javascript ,pin.match (regex)中,首先包含部分内容的返回阵列。 我将这一方法用于“死板”的输入要素,即,如果用户进入某种违反舱面而不是违反部分内容的物品,则其价值完全没有显示,但如果部分内容符合规定,则其价值不变。
const floatRegexCheck = (st) => {
const regx1 = new RegExp("^0[0-9]+\.?[0-9]*$"); // for finding numbers starting with 0
let regx2 = new RegExp("([1-9]{1}[0-9]*\.?[0-9]*)"); //if regx1 matches then this will remove 0s at the head.
if (!st.match(regx1)) {
regx2 = new RegExp("([0-9]*\.?[0-9]*)"); //if number does not contain 0 at the head of string then standard decimal formatting takes place
}
st = st.match(regx2);
if (st?.length > 0) {
st = st[0];
}
return st;
}
如果您希望通过<代码>e开展工作,则使用这一表述:
[+-]?[0-9]+([.][0-9]+)?([eE][+-]?[0-9]+)?
Here的一个范例:
var re = /^[+-]?[0-9]+([.][0-9]+)?([eE][+-]?[0-9]+)?$/;
console.log(re.test( 1 ));
console.log(re.test( 1.5 ));
console.log(re.test( -1 ));
console.log(re.test( -1.5 ));
console.log(re.test( 1E-100 ));
console.log(re.test( 1E+100 ));
console.log(re.test( .5 ));
console.log(re.test( foo ));这里的答案更为严格。
^[+-]?0(?![0-9]).[0-9]*(?![.])$|^[+-]?[1-9]{1}[0-9]*.[0-9]*$|^[+-]?.[0-9]+$
下面的数值将相应(+-标号也可行)
.11234
0.1143424
11.21
1.
以下数值不匹配:
00.1
1.0.00
12.2350.0.0.0.0.
.
....
www.un.org/spanish/ecosoc 如何运作
?
不要让与逻辑OR运营商一样的运营商 break碎 。
^[+-]?0(?![0-9]).[0-9]*(?![.])$
这一节制将从0美元开始检查价值。
第一张支票+和——以0或1个时间签署<代码>[+-]
然后检查一下它是否领导了零0。
如果它有的话,那么,由于我们不希望看到00.123(?![0-9]),那么它的下一个价值决不能是零。
然后,只检查一分钟,用无限制的数码时间检查部分。 [0-9]∗
最后,如果它有一部分没有遵循,我们就抛弃了它。
Now see the second part
^[+-]?[1-9]{1}[0-9]*.[0-9]*$
www.un.org/french/ecosoc
如果从非零开始,则与第一个数字完全吻合一个时间,并随附无限制的时间。
只好一次,无限制的数位数沿用<代码>。 [0-9]*
www.un.org/spanish/ecosoc 现在见第三部分。
^[+-]?.{1}[0-9]+$
这将从12、34565美元等开始检查价值。
www.un.org/french/ecosoc
取而代之的是<代码>。 [0-9]+
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