Javascript javascript获取字符前的字符串
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javascript get string before a character
提问by Keith Power
I have a string that and I am trying to extract the characters before the quote.
我有一个字符串，我试图在引号之前提取字符。
Example is extract the 14from 14' - 14.99
示例是从14' - 14.99 中提取14
I am using the follwing code to acheive this.
我正在使用以下代码来实现这一点。
$menuItem.text().match(/[^']*/)[0]
My problem is that if the string is something like 0.88 I wish to get an empty string returned. However I get back the full string of 0.88.
我的问题是，如果字符串类似于 0.88，我希望返回一个空字符串。但是我得到了 0.88 的完整字符串。
What I am I doing wrong with the match?
我在比赛中做错了什么？
回答by 3on
This is the what you should use to split:
这是您应该用来拆分的内容：
string.slice(0, string.indexOf("'"));
And then to handle your non existant value edge case:
然后处理您不存在的价值边缘情况：
function split(str) {
var i = str.indexOf("'");
if(i > 0)
return str.slice(0, i);
else
return "";
}
回答by jfriend00
Nobody seems to have presented what seems to me as the safest and most obvious option that covers each of the cases the OP asked about so I thought I'd offer this:
似乎没有人提出在我看来是最安全和最明显的选项，涵盖了 OP 询问的每个案例，所以我想我会提供这个：
function getCharsBefore(str, chr) {
var index = str.indexOf(chr);
if (index != -1) {
return(str.substring(0, index));
}
return("");
}
回答by nandu
try this
尝试这个
str.substring(0,str.indexOf("'"));
回答by jhnstn
Here is an underscore mixin in coffescript
这是 coffescript 中的下划线 mixin
_.mixin
substrBefore : ->
[char, str] = arguments
return "" unless char?
fn = (s)-> s.substr(0,s.indexOf(char)+1)
return fn(str) if str?
fn
or if you prefer raw javascript : http://jsfiddle.net/snrobot/XsuQd/
或者，如果您更喜欢原始 javascript：http: //jsfiddle.net/snrobot/XsuQd/
You can use this to build a partial like:
您可以使用它来构建部分，如：
var beforeQuote = _.substrBefore("'");
var hasQuote = beforeQuote("14' - 0.88"); // hasQuote = "14'"
var noQoute = beforeQuote("14 0.88"); // noQuote = ""
Or just call it directly with your string
或者直接用你的字符串调用它
var beforeQuote = _.substrBefore("'", "14' - 0.88"); // beforeQuote = "14'"
I purposely chose to leave the search character in the results to match its complement mixin substrAfter(here is a demo: http://jsfiddle.net/snrobot/SEAZr/). The later mixin was written as a utility to parse url queries. In some cases I am just using location.searchwhich returns a string with the leading ?.
我故意选择在结果中保留搜索字符以匹配其补码 mixin substrAfter（这是一个演示：http: //jsfiddle.net/snrobot/SEAZr/）。后来的 mixin 是作为解析 url 查询的实用程序编写的。在某些情况下，我只是使用location.searchwhich 返回一个带前导的字符串?。
