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
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"]
其他回答
var s = 'MyLongString:StringIWant;';
/:([^;]+);/.exec(s)[1]; // StringIWant
我喜欢这个方法:
var str = 'MyLongString:StringIWant;';
var tmpStr = str.match(":(.*);");
var newStr = tmpStr[1];
//newStr now contains 'StringIWant'
获取所有子字符串。
var out = []; 'MyLongString:StringIWant;'
.replace(/(:)\w+(;)+/g, (e) => {
out.push(e.replace(':', '').replace(';', ''))
return e;
});
console.log(out[0])
尝试使用javascript在两个字符之间获取子字符串。
$("button").click(function(){
var myStr = "MyLongString:StringIWant;";
var subStr = myStr.match(":(.*);");
alert(subStr[1]);
});
从@ Find子字符串之间的两个字符与jQuery
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"]
