下面是一个可重用的函数，它允许你使返回的子字符串包含或排除，然后可选地修剪它:

function get_substring(full_string, substring_1, substring_2, inclusive, trim)
{
    if (full_string === null) { return null; };
    let substring_1_start = full_string.indexOf(substring_1);
    if (substring_1_start === -1 ) { return null; }
    let substring_2_start = full_string.indexOf(substring_2, substring_1_start);
    if (substring_2_start === -1 ) { return null; }
    let substring_1_end = substring_1_start + substring_1.length;
    let substring_2_end = substring_2_start + substring_2.length;
    let return_string = inclusive ? (full_string.substring(substring_1_start, substring_2_end)) : (full_string.substring(substring_1_end, substring_2_start));
    return trim ? return_string.trim() : return_string;
}

使用例子:

//Returns 'cake and ice cream'
get_substring('I like cake and ice cream', 'cake', 'cream', true, true);

//Returns ' and ice '
get_substring('I like cake and ice cream', 'cake', 'cream', false, false);

//Returns 'and ice'
get_substring('I like cake and ice cream', 'cake', 'cream', false, true);

//Returns null
get_substring('I like cake and ice cream', 'cake', 'cookies', false, false);

//Returns null
get_substring('I like cake and ice cream', null, 'cream', false, false);

这可能是可行的解决方案

var str = 'RACK NO:Stock;PRODUCT TYPE:Stock Sale;PART N0:0035719061;INDEX NO:21A627 042;PART NAME:SPRING;';  
var newstr = str.split(':')[1].split(';')[0]; // return value as 'Stock'

console.log('stringvalue',newstr)

下面是一个可重用的函数，它允许你使返回的子字符串包含或排除，然后可选地修剪它:

function get_substring(full_string, substring_1, substring_2, inclusive, trim)
{
    if (full_string === null) { return null; };
    let substring_1_start = full_string.indexOf(substring_1);
    if (substring_1_start === -1 ) { return null; }
    let substring_2_start = full_string.indexOf(substring_2, substring_1_start);
    if (substring_2_start === -1 ) { return null; }
    let substring_1_end = substring_1_start + substring_1.length;
    let substring_2_end = substring_2_start + substring_2.length;
    let return_string = inclusive ? (full_string.substring(substring_1_start, substring_2_end)) : (full_string.substring(substring_1_end, substring_2_start));
    return trim ? return_string.trim() : return_string;
}

使用例子:

//Returns 'cake and ice cream'
get_substring('I like cake and ice cream', 'cake', 'cream', true, true);

//Returns ' and ice '
get_substring('I like cake and ice cream', 'cake', 'cream', false, false);

//Returns 'and ice'
get_substring('I like cake and ice cream', 'cake', 'cream', false, true);

//Returns null
get_substring('I like cake and ice cream', 'cake', 'cookies', false, false);

//Returns null
get_substring('I like cake and ice cream', null, 'cream', false, false);

你可以试试这个

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

我使用@tsds的方式，但只使用分裂函数。

var str = 'one:two;three';    
str.split(':')[1].split(';')[0] // returns 'two'

警告:如果字符串中没有“:”，访问数组的“1”索引将抛出错误!str.split(“:”)[1]

因此，如果存在不确定性，@tsds的方式更安全

str.split(':').pop().split(';')[0]

一般的和简单的:

function betweenMarkers(text, begin, end) { var firstChar = text.indexOf(begin) + begin.length; var lastChar = text.indexOf(end); var newText =文本。substring (firstChar lastChar); 返回newText; ｝ console.log (betweenMarkers(“MyLongString: StringIWant ;",":",";"));