Anyone help me write this program better?

Can anyone help me to improve on my coding techniques by helping me to rewrite this program in a better way?

An example of the input file is attached below!

Code:

I100140100060002
I100060100060001
I100220128070008
I100490121060099
I100650544020100
C10081COVER SYSTEMS LTD   ALBOY RD;KINVER;SOUTH STAFFS; ;                             0002345000010000
I101030201090507
I100060493010009
I130050303090007
I140010134040044
I200010224030202
I222250160040032
I117030354080090
I117030544020300
R1986X0121060200
I199840101030085
I1518X0125050002
I1518X0140010002
D31003
I191860133070090
I193210215040088
I100060350090009
I1932103530X0092
I193210414080077
I1332305200X0250
I1454X0110020066
I1454X0142069999
R1454X0214070099
I197800522050100
R197800500080075
I197800415050055
I100220301040040
I100060302010030
I100060141090100
C10367MAXIMES PAINT STORESWATERLOO TERRACE;BRIDGNORTH;STAFFS; ;                       0086575000099999
I1036702040X0333
I107400211050332
I1074001180X0003
I133400203030003
I133400211050012
I133400140010012
C19925CHUCKERS DIY        19 HIGH STREET;STOURBRIDGE;WEST MIDS; ;                     0000000000000000
I1258X01180X0102
I1258X0116061234
I112820210080065
I112820224030055
I134040116060009
I134040117030009
D13404
I1116903480X0008
I111690404010007
R100060542160552
I103830307080045
I103830313050035
I103670505040200
R103670534060100
I103240410090015
I103240310030015
I135440526040020
I135440121060010
I136250148000005
I124670114010045
I124670110020675
I198440130050456
I198440302010790
D13358
I1308X0117200122
I1308X0354080033
I100060400020067
I100060401000056
I100060522050555
C10219SPLASH AND DRIBBLE  33 THE POLES;ALAWAY HILL;BIRMINGHAM;WEST MIDS;              0000100000008000
I183090342070044
I183090505040023
I183090309020010
I1812X0351060222
I1812X0133070123
R1000601390X0067
R100060134040099
I100060132000088
I145240117620006
I138890114010004
I138890100490003
R192240212020032
I192240304060100
I192240346060067
I100060342230002
I100060344010002
I189290526040099
I189290532010062
I120090342070062
I151050349080044
I151050350090004
I112820141090004
R112820221010678
R107400302010052
I107400210080345
I107400132000567
R101620120090042
R101620111262995
I101620401000003
I101200504070002
I101200414080002
R200100404010009
R100060403040099
I100060530070088
I100060307080004
D10022
I101200344010009
R1012003110X0067
R101200134040004
R101620205080002
R101620145080001
I101620213000005
R101620114010042
C13218TRUE COLOURS        THE CORBETT CENTRE;OLDBURY;WEST MIDS; ;                     0000500590001500
I101890112070032
I101890104050010
D11525
I100060214070009
R100060403040402

The file is a text file and each line is terminated by a newline character!

Avoid terminating your if() statements.

Code:

if(fgets(string,104,input_ptr));
{
   /* ... */
}

Check return values.

Code:

database = malloc(num_of_recs * (sizeof(union records)));
/* no check of (database != NULL) before database is dereferenced */

Be sure not to go beyond the bounds of arrays.

Code:

/* ... */
cust_code[6],
/* ... */
cust_limit[8];
/* ... */
database[x].c.cust_code[6]='\0';
/* ... */
database[x].c.cust_limit[8]='\0';

Avoid magic numbers and reserved words (string).

Code:

char string[104];
/* ... */
while(fgets(string,104,input_ptr))
{
   /* ... */
}

/* instead consider*/

char buffer[104];
while(fgets(buffer,sizeof(buffer),input_ptr))
{
   /* ... */
}

Look up the function strncpy() and consider it as a replacement for the loops.

Code:

cust_bal[10],
/* ... */
for(y=0;y<9;y++)
   database[x].c.cust_bal[y]=string[y+86];
database[x].c.cust_bal[9]='\0';

And again consider sizeof() instead of hard-coding the size in.

Some functions have misleading names. For example, atof returns a double.

Code:

float bal_total=0;
/* ... */
bal_total+=((atof(database[x].c.cust_bal))/100);

Might you want bal_total to be a bal_double?

[EDIT]
Remember to free malloc'd memory.

For the following,

Code:

int sort_function(const void *a, const void *b)
{
	const union records *pa = a;
	const union records *pb = b;
	if (strncmp( pa->c.cust_code, pb->c.cust_code, 5) < 0) return -1;
	if (strncmp( pa->c.cust_code, pb->c.cust_code, 5) > 0) return +1;
		/*else they are equal*/
	return (pb->c.type - pa->c.type);
}

why not

Code:

int sort_function(const void *a, const void *b)
{
   const union records *pa = a;
   const union records *pb = b;
   return strncmp(pa->c.cust_code, pb->c.cust_code, sizeof(pa->c.cust_code) - 1);
}

Why have database and database2? You are using x to index into the database array, so you could reuse database after the sort.
[/EDIT]

[EDIT]
Reminded by a recent thread on exit(), you may also want to use exit(EXIT_FAILURE) instead of exit(1) when fopen() fails. And adding a perror() to the failure condition might provide a better reason for the failure. Whether or not you choose to use perror(), you might consider attaching an identifier to the file name strings.

Code:

const char InputFilename[]  = "503442VF.dat";
const char OutputFilename[] = "503442SF.dat";
const char PrintFilename[]  = "PRINT.DAT";

Then you could have something like

Code:

   input_ptr= fopen(InputFilename,"rt");
   if ( input_ptr == NULL )
   {
      printf("\nError in opening input file!!");
      perror(InputFilename);
      exit(EXIT_FAILURE);
   }

Using offsetof() instead a hard-coded offset (in the case of string[y+86], for example) was another idea. But since strctures may contain padding that the text file might not match, this could be bad. But if your compiler and options could circumvent padding, then it might be possible to encapsulate these values using the structure definition itself.

Other thoughts might be creating structure-specific parsing routines -- as separate functions -- to be called once the type is evaluated. Or maybe fscanf() would be useful for parsing the input file.

Other thoughts might be to use a switch on string[0] instead of successive if comparisons, to minimize repetition in the C code as a maintenance concern.

You might have print_headers return the number of lines to increment. That way

Code:

   print_headers(page_num);
   line_count+=4;

could be done as

Code:

   line_count+=print_headers(page_num);

[/EDIT]