我试图从一个更大的字符串中提取一个字符串,它得到了a:和a之间的所有东西;

当前的

Str = 'MyLongString:StringIWant;'

期望输出值

newStr = 'StringIWant'

当前回答

获取两个子字符串之间的字符串(包含多于1个字符)

function substrInBetween(whole_str, str1, str2){
   if (whole_str.indexOf(str1) === -1 || whole_str.indexOf(str2) === -1) {
       return undefined; // or ""
  }
  var strlength1 = str1.length;
  return whole_str.substring(
                whole_str.indexOf(str1) + strlength1, 
                whole_str.indexOf(str2)
               );

   }

注意,我使用indexOf()而不是lastIndexOf(),因此它将检查这些字符串的第一次出现

其他回答

你也可以试试这个:

var str = 'one:two;three';    
str.split(':').pop().split(';')[0]; // returns 'two'

使用分割()

var s = 'MyLongString:StringIWant;';
var arrStr = s.split(/[:;]/);
alert(arrStr);

arrStr将包含所有由:或; 通过for循环访问每个字符串

for(var i=0; i<arrStr.length; i++)
    alert(arrStr[i]);
var str = '[basic_salary]+100/[basic_salary]';
var arr = str.split('');
var myArr = [];
for(var i=0;i<arr.length;i++){
    if(arr[i] == '['){
        var a = '';
        for(var j=i+1;j<arr.length;j++){
            if(arr[j] == ']'){
                var i = j-1;
                break;
            }else{
                a += arr[j];
            }
        }
        myArr.push(a);
    }
    var operatorsArr = ['+','-','*','/','%'];
    if(operatorsArr.includes(arr[i])){
        myArr.push(arr[i]);
    }
    var numbArr = ['0','1','2','3','4','5','6','7','8','9'];
    if(numbArr.includes(arr[i])){
        var a = '';
        for(var j=i;j<arr.length;j++){
            if(numbArr.includes(arr[j])){
                a += arr[j];
            }else{
                var i = j-1;
                break;
            }
        }
        myArr.push(a);
    }
}
myArr = ["basic_salary", "+", "100", "/", "basic_salary"]

你可以试试这个

var mySubString = str.substring(
    str.indexOf(":") + 1, 
    str.lastIndexOf(";")
);

使用' get_between '实用函数:

get_between <- function(str, first_character, last_character) {
    new_str = str.match(first_character + "(.*)" + last_character)[1].trim()
    return(new_str)
    }

字符串

my_string = 'and the thing that ! on the @ with the ^^ goes now' 

用法:

get_between(my_string, 'that', 'now')

结果:

"! on the @ with the ^^ goes