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[Former Member]

Hi all,

My requirement is i want to convert 2900(means 29 min and 00 sec) into

00:29:00 ( 00 -

---

hrs,29 -

---

min, 00 -

---

seconds.).Pls explain with a sample code..

Regards & Thanks,

Alex

[Former Member]

data p1(6) type c value '2900'.

data p2(8) type c.

data len type i.

datat off_set type i.

data dummy(6) type c value '0000'.

<b>* operation</b>

len = strlen( p1 ).

<i>* len = 4.</i>

off_set = 6 - len.

• offset = 2.

if off_set <> 0.

CONCATENATE dummy(off_set) p1 into p1.

endif.

<i>* p1 = 002900.</i>

CONCATENATE p1(2) ':' into p2.

<i>* p2 = 00:2900.</i>

CONCATENATE p1+3(2) ':' into p2.

<i>* p2 = 00:29:00.</i>

Message was edited by:

vickram s

[Former Member]

hi Alex,

I am giving a psuedo code. u might need to do adjustment for syntax etc.

1) l_time has 2900.

2) take l_time in string format l_time_str

3) length = length( l_time_str ) - 2.

4) l_seconds = l_time_str+length(2).

5) length = length - 2.

6) l_minutes = l_time_str+length(2).

So on. need to put If condition for length >= 0.

Regards,

Vivek.

[Former Member]

Hi Alex,

try this

data x type t value '002900'.

write (8) x using edit mask '__:__:__'.

regards,

pankaj singhreward points if problem is solved

[Former Member]

<b>* declaration part</b>

data p1(6) type c value '2900'.

data p2(8) type c.

data len type i.

datat off_set type i.

data dummy(6) type c value '0000'.

<b>

• converting 2900 to 00:29:00.</b>

len = strlen( p1 ).

<i>* len = 4.</i>

off_set = 6 - len.

<i>* offset = 2.</i>

if off_set <> 0.

CONCATENATE dummy(off_set) p1 into p1.

endif.

<i>* p1 = 002900.</i>

CONCATENATE p1(2) ':' into p2.

<i>* p2 = 00:2900.</i>

CONCATENATE p1+3(2) ':' into p2.

<i>* p2 = 00:29:00.</i>

write: p2.

[Former Member]

try this.

data : pdat(6) type n value '2900'.

data : phr(2),

pmin(2),

psec(2),

ptime(8).

phr = pdat(2).

pmin = pdat+2(2).

psec = pdat+4(2).

concatenate phr pmin psec into ptime separated by ':'.

regards

shiba dutta