LED code?

The expression (c & 0x01) can only yield the value 0 or 1, and since the value of the ASCII character '0' is 0x30, neither 0 nor 1 will ever equal '0'.

this is my program code.im still unable to on only one led.

#include <putl.h>
char xdata OPORT _at_ 0x0000;
char xdata IPORT _at_ 0x0001;

void sInit(void)
{
        PCON=0x80;
        TMOD=0x20;
        SCON=0x50;
        TH1=243;
        TR1=1;
}

void sputCh(char c)
{
        while(T1==0);
        SBUF =c;
        TI=0;
}
unsigned char sgetCh(void)
{
        unsigned char c;
        while(RI==0);
        c=SBUF;
        RI=0;
        return c;
}

void main(void)
{

        unsigned char c;
        //unsigned char r;
        sInit();
        while(1)
        {
                c=sgetCh();
                OPORT=c;
                {
                        if(c=='1')
                        {
                                if((c&0x01)==0)
                                {
                                        OPORT=0x00;
                                }
                                else
                                {
                                        OPORT=0xff;
                                }
                        }

                        else if(c=='2')
                        {
                                if((c&0x02)
                                {
                                        OPORT=0x00;
                                }
                                else
                                {
                                        OPORT=0xff;
                                }
                        }


        }
}
}

I'm not going to rewrite your program, but I am going to tell you to think a bit more about what you are trying to do!

Exactly how do you think:

if (c=='1') {
    if ((c&0x01)==0) {
        OPORT=0x00;
    } else {
        OPORT=0xff;
    }
}

will work?

If you have already checked if c equals '1', how do you then think that a second if statement would be able to find c to be sometimes odd and sometimes even? Where would the variable c manage to change it's value between the outer and inner if?

Another thing: you want to change one single LED? How many bits should you then toggle so as not to change the value of the other 7 LED?

Turning on a single bit can be done by:
x |= (1 << bitpos);
Turning off a single bit can be done by:
x &= ~(1 << bitpos);
where bitpos is a value between 0 and 7 (if x is a byte).

"Turning on a single bit can be done by:
x |= (1 << bitpos);
Turning off a single bit can be done by:
x &= ~(1 << bitpos);"

This is standard 'C' - general and applicable to any compiler and any target; it is Portable

The 8051 has a particular feature that some addresses are bit-addressable; ie, individual bits can be addressed directly.

Keil C51 provides some specific, proprietary extensions to the 'C' standard to allow you to take advantage of this feature; of course, this is non-Portable

Ports P0-P3 are bit addressable...

Yes, I'm aware of the bit-addressing in the 8051.

Note the specific sentence in the first post: "the problem is when i ON the LED it light all the 7 leds.There is 7 output port."

Then note that the OP used the declaration:

char xdata OPORT _at_ 0x0000;

and then performed assigns:

OPORT = 0xff;

or

OPORT = 0x00;

To my eyes, that doesn't look like an attempt at using bit-addressable variables - hence my comment about the general method to set or clear a bit.

I think that the OP code will need a number of refining steps, and I don't think it is a good idea to just post a complete rewrite of the code. Each single change to the code must be followed my an understanding why.

I feel that switching to bdata, sbit, overloading of the ^ operator etc is a separate step.