Pascal, 210 B

In Standard Pascal a string has necessarily a minimum length of two char values.^† Therefore it is not possible to find the count of occurrences of a “string” of length one in another string.

type c=char;z=integer;function f(s:array[m..n:z]of c;t:array[p..q:z]of c):z;var
i,j,a,r:z;begin r:=0;i:=n-q+p;for i:=m to i do begin a:=1;for j:=0 to q-p do
begin a:=a*ord(s[i+j]=t[p+j])end;r:=r+a end;f:=r end;

Antighoulfed:

    function occurrencesCount(
            { In Pascal a string is a `packed array[1‥N] of char`, N ≥ 2. }
            sample: packed array[sampleFirst‥sampleLast: integer] of char;
            needle: packed array[needleFirst‥needleLast: integer] of char
        ): integer;
        var
            sampleIndex, needleOffset, intermediateResult, result: integer;
        begin
            result ≔ 0;
            
            sampleIndex ≔ sampleLast − needleLast + needleFirst;
            for sampleIndex ≔ sampleFirst to sampleIndex do
            begin
                intermediateResult ≔ 1;
                
                for needleOffset ≔ 0 to needleLast − needleFirst do
                begin
                    { The product is zero if any factor is zero. }
                    intermediateResult ≔ intermediateResult *
                        ord(
                            sample[sampleIndex + needleOffset]
                            =
                            needle[needleFirst + needleOffset]
                        )
                end;
                
                result ≔ result + intermediateResult
            end;
            { In Pascal the function’s result is defined by assigning a value
              to the implicitly declared variable bearing the function’s name. }
            occurrencesCount ≔ result
        end;

_{^† Extended Pascal allows an empty string (''), too.}

K/Kona 6

+/y~'x

where x is the string and y the substring. ~ is the negate operator, with ', it is applied to every element in x; it will return 0 if it does not match and 1 if it does match. Since it is applied element-wise, the result of y~'x is a vector, the +/ then sums the result giving the total number of occurrences.

Unfortunately, this method requires that y is only one character, otherwise we will be comparing a multi-character string to a single character string, resulting in a length error.

Cobra - 25

print a.split(b).length-1

Python 2.x - 49 23 22 bytes

This is assuming variable input is okay. Both strings can be of any length.

Shortened @avall.

a='s'
b='aaaabbbbsssffhd'
print~-len(b.split(a))

49 bytes version, counts every instance of the substring ('aba' is in 'ababa' twice).

a='s'
b='aaaabbbbsssffhd'
print sum(a==b[i:i+len(a)]for i in range(len(b)))

Java (38)

System.out.print(a.split(b).length-1);

(The question did not require a complete program or function.)

C++ 225

int n,k,m;
int main()
{
string s1,s2;
cin>>s1;
cin>>s2;
int x=s1.size(),y=s2.size();
if(x>=y)
{
for(int i=0;i<x;i++)
{
k=0,m=0;
for(int j=0;j<y;j++)
{
if(s2[j]==s1[i+m])
{
    k++,m++;
}
else break;
}
if(k==y)
{
n++;
i+=(y-1);
}
}
}
cout<<n<<endl;
return 0;
}

Mathematica 26 23

Works like Dennis' algorithm, but wordier:

Length@StringCases[a,b]

Three chars shaved off by Szabolics.

Fortran 90: 101

Standard abuse of implicit typing, works for any length arrays a and b, though one should expect that len(a) < len(b).

function i();i=0;k=len(trim(b))-1;do j=1,len(trim(a))-k;if(a(j:j+k)==b(1:1+k))i=i+1;enddo;endfunction

This function must be contained within a full program to work. a and b are received from stdin and can be entered either on the same line (either comma or space separated) or on different lines. Compile via gfortran -o main main.f90 and execute as you would any other compiled program.

program main
   character(len=256)::a,b
   read*,a,b
   print*,i()
 contains
   function i()
     i=0
     k=len(trim(b))-1
     do j=1,len(trim(a))-k
        if(a(j:j+k)==b(1:1+k))i=i+1
     end do
   end function
end program main

Tests:

>ababa aba
2

I could make the above return 1 if I add 4 characters (,k+1) for the do loop

> aaaabbbbbsssffhd s
3

Lua (48)

So i thought I could just submit another answer, this time in lua. Its very possible this could be improved a lot, im very new to this.

print((a.len(a)-a.len(a.gsub(a,b,"")))/b.len(b))

Delphi XE3 (113)

Takes 2 strings, removes substring from string and substracts new length from old length followed by a division of the substring length.

function c(a,b:string):integer;begin c:=(Length(a)-Length(StringReplace(a,b,'',[rfReplaceAll])))div Length(b)end;

Testing:

c('aaaabbbbsssffhd','s') = 3
c('aaaabbbbsssffhd','a') = 4
c('ababa','aba') = 1
c('ababa','c') = 0

C 130 120

Note: will probably crash if called with incorrect arguments.

r;main(int c,char**a){char*p=*++a,*q,*t;while(*p){for(q=a[1],t=p;*q&&*q==*t;q++)t++;*q?p++:(p=t,r++);}printf("%d\n",r);}

Ungolfed (kinda):

int main(int argc, char *argv[]) {
    int result = 0;
    char *ptr = argv[1];
    while (*ptr) {
        char *tmp, *tmp2 = ptr;
        // str(n)cmp
        for (tmp = argv[2]; *tmp; tmp++, tmp2++)
            if (*tmp != *tmp2)
                break;
        if (*tmp) {
            ptr++;
        } else {
            result++;
            ptr += tmp;
        }
    }
    printf("%d\n", result);
}

Old version with strstr and strlen: 103

l;main(int c,char**a){char*p=a[1];l=strlen(a[2]);while(c++,p>l)p=strstr(p,a[2])+l;printf("%d\n",c-5);}

C# - 66 Bytes

//s = "aba"
//t = "ababa"

Console.Write(t.Split(new[]{s},StringSplitOptions.None).Length-1);

//Output: 1

C# - 73

//a = "aba";
//b = "ababa";

Console.Write(b.Split(new string[]{a},StringSplitOptions.None).Length-1);

// output = "1"

Applescript, 106 bytes

Applescript is a fun, but silly language to golf in.

on run a
set AppleScript's text item delimiters to (a's item 1)
(count of (a's item 2)'s text items)-1
end

Run with osascript:

$ osascript instr.scpt s aaaabbbbsssffhd
3
$ 

J (7)

No use of external libraries Check! , or your language's API. Check...? I don't know what an language API is. You have to implement it manually Check! No file I/O Check! No connecting with a server, website, et cetera Check!

+/a E.b

How it works:

E. is WindowedMatch: the J Refsheet gives 're' E. 'reread' as example. This gives 1 0 1 0 0 0. Then, the only thing left to do is simply adding this with +/ (basically sum).

I don't think this counts as using your language's built-in function or method for counting occurences, but that's disputable.

EDIT: Just to be clear:

   +/'aba'E.'ababa'
2

Powershell 32

($args[0]-split$args[1]).count-1

Works like this:

PS C:\MyFolder> .\ocurrences.ps1 ababa aba
1

Explanation: Uses -split to separate the first argument by the second one, returns the size of the array resulting from the split (minus 1)

JavaScript 32

Nothing really interesting here...

(p=prompt)().split(p()).length-1

split main purpose is to create an array from a string using the delimiter in argument.

GolfScript, 3 bytes

/,(

Assumes string and substring are on the stack.

Try it online.

How it works

/  # Split the string around occurrences of the substring.
,  # Get the length of the split array.
(  # Subtract 1.