How to read the data of pin found and not pin found from the file?

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C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define STRING_LEN 200

char string[STRING_LEN], pinFind[STRING_LEN];
char * lineOne = NULL, *pinFound = NULL, *numbers[5];
int i;
char *read(FILE * fname, char * findPin){
    while(fgets(string, STRING_LEN, fname)){
        lineOne = strtok(string, "\n");
        pinFound = strstr(lineOne, pinFind);

        numbers[0] = strtok(lineOne, ",");
        for(i = 1; i < 5; i++){
            numbers[i] = strtok(NULL, ",");
        }
        if(pinFound)   return string;
        if(!pinFound)   return string;
    }
  return 0;
}

int main(){
  FILE * fp1 = fopen("file.csv", "r");
  printf("Enter your Pin Code: ");
    scanf("%s", pinFind);

  read(fp1, pinFind);
  for(i=0; i<5; i++){
    printf("Here is: %s\n", numbers[i]);
  }
  return 0;
}

What I have tried:

In this code, I am trying to make the `pinFound` and `!pinFound` work at the same time. But it is just giving me the result of `pinFound`

And when I use another if statement and enter the pin code from the first line in the file, it gives me the data of the first line, but I want the exceptional pin code data. And if enter the pin code from the second line of the file, it works and gives me the first line data.

Data in the file

    Bilal Khan,1111111111111,1122,1000,12255334
    Ali Ahmed,2222222222222,2233,2000,66778899

Expected result
    Both results have a pin code but one is found and another is not.

    Enter your Pin Code: 1122
    Pin found......
    Here is: Bilal Khan
    Here is: 1111111111111
    Here is: 1122
    Here is: 1000
    Here is: 12255334
    
    Pin not Found...
    Here is: Ali Ahmed
    Here is: 2222222222222
    Here is: 2233
    Here is: 2000
    Here is: 667788994

Posted 2-Nov-20 0:26am

ibilalkayy

Updated 2-Nov-20 1:54am

CPallini

v3

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3 solutions

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OriginalGriff · Answer 1 · 2020-11-02T00:52:00

Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.

So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.

Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:

Input   Expected output    Actual output
  1            2                 1
  2            4                 4
  3            6                 9
  4            8                16

Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:

C#

int Double(int value)
   {
   return value * value;
   }

Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on the first line of the method, and run your app. When it reaches the breakpoint, the debugger will stop, and hand control over to you. You can now run your code line-by-line (called "single stepping") and look at (or even change) variable contents as necessary (heck, you can even change the code and try again if you need to).
Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
Hopefully, that should help you locate which part of that code has a problem, and what the problem is.
This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!

Richard MacCutchan · Answer 2 · 2020-11-02T01:10:00

C++

if(pinFound)   return string;
if(!pinFound)   return string;

So you return the string if the pin is found, or if it is not found. And you do that on the first line of the file which is not logical. If the pin is not found you should not do anything until the end of the program when you report "pin not found".

Your logic should be something like:

Set PinFound = FALSE
While TRUE DO
  Read next line of text
  If end of file break
    Search for pin in current line
    If found then report the details and set PinFound = TRUE
DONE
If PinFound == FALSE then print "pin not found"

CPallini · Answer 3 · 2020-11-02T01:55:00

Try (note: the code is not robust):

C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

#define PIN_FOUND 1
#define FIELDS  5
#define PIN_NOT_FOUND 0
#define STRING_LEN 200


int find(char * line, char * pin, char * field[])
{
  int i;
  line = strtok(line, "\n");

  field[0] = strtok(line, ",");
  for(i = 1; i < FIELDS; i++)
  {
    field[i] = strtok(NULL, ",");
  }
  return strstr(field[2], pin) == NULL ? PIN_NOT_FOUND : PIN_FOUND;
}

int main()
{
  char line[STRING_LEN], pin[STRING_LEN], * field[FIELDS];

  printf("Enter your Pin Code: ");
    scanf("%s", pin);

  FILE * fp = fopen("file.csv", "r");
  while ( fgets(line, STRING_LEN, fp))
  {
    int i;
    if ( find(line, pin, field) == PIN_FOUND)
      printf("\npin found.\n");
    else
      printf("\npin not found.\n");

    for(i=0; i<FIELDS; i++)
      printf("Here is: %s\n", field[i]);
  }
  fclose(fp);
  return 0;
}

How to read the data of pin found and not pin found from the file?

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