delete all consonents from the string "Hello, have a good day" - c++

Question

I have seen many good code to do this problem. I am new to coding. My question is where my logic went wrong. I think that problem is with second string str1. I din't initialize it. even when I am printing element by element withing if , it is working. but it is not working, when I am trying to print whole string str1.
#include<iostream>
#include<string>
using namespace std;
int main()
{
string str = "Hello, have a good day", str1;
for (int i = 0, j =0; i < str.length(); ++i)
{
if((str[i]>='a'&& str[i]<='z') || (str[i]>='A'&& str[i]<='Z'))
if (str[i] == 'I' || str[i] == 'i' || str[i] == 'U' || str[i] == 'u' || str[i] == 'O' || str[i] == 'o' ||
str[i] == 'A' || str[i] == 'a' || str[i] == 'E' || str[i] == 'e' )
{
str1[j] = str[i];
//std::cout << str1[j] ;
j++;
}
else
{
str1[j] = str[i];
j++;
}
}
cout << str1 <<'\n';
}
output is just blank.

Answer 1

The first thing to do is to write a function that determines whether a character is a consonant:
bool is_not_consonant(char ch) {
static char consonants[] = "bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ";
return std::find(std::begin(consonants), std::end(consonants), ch) == std::end(consonants);
}
Then use that function as a predicate to std::copy_if:
std::string result;
std::string input = whatever;
std::copy_if(std::begin(input), std::end(input),
std::back_inserter(result),
is_not_consonant);

Answer 2

Explanation
The problem is that you don't need the else condition. All you need to do is check for a vowel, and print if found which is rightly covered in your if condition.
Code
Try this:
#include<string>
using namespace std;
int main()
{
string str = "Hello, have a good day", str1;
for (int i = 0; i < str.length(); ++i)
{
if((str[i]>='a'&& str[i]<='z') || (str[i]>='A'&& str[i]<='Z'))
if (str[i] == 'I' || str[i] == 'i' || str[i] == 'U' || str[i] == 'u' || str[i] == 'O' || str[i] == 'o' || str[i] == 'A' || str[i] == 'a' || str[i] == 'E' || str[i] == 'e' )
{
str1 += str[i];
}
}
cout << str1 <<'\n';
}