如何在 JavaScript 中获取指定字符之前的子字符串？

Question

I am trying to extract everything before the ',' comma. How do I do this in JavaScript or jQuery? I tried this and not working..

1345 albany street, Bellevue WA 42344

I just want to grab the street address.

var streetaddress= substr(addy, 0, index(addy, '.')); 

Answer 1

var streetaddress = addy.substr(0, addy.indexOf(',')); 

虽然它不是获取有关每种方法作用的确切信息的最佳位置（ mozilla 开发人员网络对此更好）但w3schools.com非常适合向您介绍语法。

Answer 2

var streetaddress = addy.split(',')[0];

Answer 3

尝试这个：

streetaddress.substring(0, streetaddress.indexOf(','));

Answer 4

//split string into an array and grab the first item

var streetaddress = addy.split(',')[0];

Also, I'd recommend naming your variables with camel-case(streetAddress) for better readability.

Answer 5

如果您喜欢简短，只需使用RegExp ：

var streetAddress = /[^,]*/.exec(addy)[0];

Answer 6

almost the same thing as David G's answer but without the anonymous function, if you don't feel like including one.

s = s.substr(0, s.indexOf(',') === -1 ? s.length : s.indexOf(','));

in this case we make use of the fact that the second argument of substr is a length, and that we know our substring is starting at 0.

the top answer is not a generic solution because of the undesirable behavior if the string doesn't contain the character you are looking for.

if you want correct behavior in a generic case, use this method or David G's method, not the top answer

regex and split methods will also work, but may be somewhat slower / overkill for this specific problem.

Answer 7

You can also use shift() .

var streetaddress = addy.split(',').shift();

According to MDN Web Docs:

The shift() method removes the first element from an array and returns that removed element. This method changes the length of the array.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/shift

Answer 8

var newString = string.substr(0,string.indexOf(','));

Answer 9

var streetaddress = addy.substr(0, addy.indexOf('.')); 

（您应该阅读javascript 教程，尤其是关于字符串函数的部分）

Answer 10

If you want to return the original string untouched if it does not contain the search character then you can use an anonymous function (a closure):

var streetaddress=(function(s){var i=s.indexOf(',');
   return i==-1 ? s : s.substr(0,i);})(addy);

This can be made more generic:

var streetaddress=(function(s,c){var i=s.indexOf(c);
   return i==-1 ? s : s.substr(0,i);})(addy,',');

Answer 11

You could use regex as this will give you the string if it matches the requirements. The code would be something like:

const address = "1345 albany street, Bellevue WA 42344";
const regex = /[1-9][0-9]* [a-zA-Z]+ [a-zA-Z]+/;
const matchedResult = address.match(regex);

console.log(matchedResult[0]); // This will give you 1345 albany street.

So to break the code down. [1-9][0-9]* basically means the first number cannot be a zero and has to be a number between 1-9 and the next number can be any number from 0-9 and can occur zero or more times as sometimes the number is just one digit and then it matches a space. [a-zA-Z] basically matches all capital letters to small letters and has to occur one or more times and this is repeated.

Answer 12

You can use Azle to get substrings before :

str = 'This is how we go to the place!'

az.get_everything_before(str, 'place')

Result : This is how we go to the

after

str = 'This is how we go to the place!'

az.get_everything_after(str, 'go')

Result : to the place!

and in between :

str = 'This is how we go to the place!'

az.get_everything_between(str, 'how', 'place')

Result : we go to the

Answer 13

If you are worried about catching the case where no comma is present, you could just do this:

var end = Math.min(addy.indexOf(","), addy.length)
var streetaddress = addy.substr(0, end);

Nobody said it had to go on one line.