如何得到一个草图内的图形?(已解决)

pppflak

我试了很多种方法都没法得到一个草图内的图形。不知道那位高手会的。教教小弟。谢谢了

$ V# |( V" B' ^[ 本帖最后由 pppflak 于 2007-7-27 01:53 编辑 ]

bluelighting

请再描述的详细些。

pppflak

我在零件中建立了一个草图。里面只画了一个矩形框。我现在希望写个程序得到里面4条线的TAG。我在之前的程序中已经能得到草图的TAG。但是用UF_OBJ_cycle_in part和UF_SKET_ask_sketch_features扫描草图得到的都是NULL_TAG。请问这个程序应该怎么写呢？谢谢。谢谢。在线等。急啊。。。
另外。UF_FEATURE_TYPE具体包括哪些类型呢？请问在哪儿能查到？
谢谢各位大侠了

pppflak

以下是我写的代码。从扫描零件开始。直到读取草图内容。刚开始学习UG。程序难免繁琐混乱。望大家见谅。尽力帮帮小弟。
/ k( s4 o, V" htag_t part1;
UF_PART_load_status_t error_status;
int type=UF_feature_type;
0 b$ Y) _$ r$ {: p  m4 R7 _tag_t feature=NULL_TAG;
/ q3 b2 ?+ C* atag_t * fea_tags=0;
2 q5 O/ |, ~" Q' h- E- v7 {, s2 i) pint I_count=0,feature_count;
, u! N& T% u5 X) |: |char * feature_name;
char * feature_type;
7 r' l/ X4 G$ Z8 d, Kchar * parent_name;
char main_body[20];
( p; n9 B5 ^) U% |, C& Qtag_t main_body_tag;
; e& \  y2 C4 l2 x6 Cchar msg[100];
/ L' d7 q, H; ~( m' k8 v- S4 G- vchar buffer[20];
3 b. Q- o* b1 ` part1=UF_PART_ask_nth_part(0);
  S7 l( c9 I, j4 W% U  K UF_OBJ_cycle_objs_in_part(part1,type,&feature);
4 L3 H) L5 @& H0 X- x while(feature!=NULL_TAG)
{  I_count++;
  UF_OBJ_cycle_objs_in_part(part1,type,&feature);
( @- J/ Q. q5 F' ^ }
feature_count=I_count;
fea_tags =new tag_t[feature_count];
char* fea_names[500];
" z( Z9 G, c/ m$ Q$ I; vchar* fea_types[500];
2 P8 B& @+ k$ ~. u" Jfor(int i=0;i< feature_count;i++)
{
fea_names=new char[133];
    fea_types=new char[133];
}
- _% A1 B  ~' ]7 M$ m feature=NULL_TAG;
  for (i=0;i< feature_count;i++)
  {
    UF_OBJ_cycle_objs_in_part(part1,type,&feature);
    fea_tags=feature;
    UF_MODL_ask_feat_name(feature,&feature_name);
    strcpy(fea_names,feature_name);
    UF_MODL_ask_feat_type(feature,&feature_type);
    strcpy(fea_types,feature_type);
8 O: `3 ]" K" ]& L5 p0 n    uc1601(_itoa( fea_tags, buffer, 10 ),1);
    uc1601(fea_names,1);
4 g/ J) J$ i2 t    uc1601(fea_types,1);
  x) a, }! A" P  }
8 k6 \% H8 ^/ V" ^! ^$ L7 P//above routine gets all the features in a part, including tags,names and type.
6 q) p" ]3 u8 @for (i=0;i<feature_count;i++)
{
if(strcmp(fea_types,"SKETCH")==0)
/ }0 i4 K9 U9 B {
  strcpy(main_body,fea_names);
       main_body_tag=fea_tags;
  break;
}
}
strcpy(msg,"main body is ");
0 n4 U% y/ v! z! |uc1601(strcat(msg,main_body),1);
' V+ j7 l& q6 a: H! @2 astrcpy(msg,"main body tag is ");
, V4 R$ P  `0 s# ouc1601(strcat(msg,_itoa( main_body_tag, buffer, 10 )),1);
! g! @' j/ ^( n/ P* W9 p% v5 b) I//suppose the first sketch found in the part is named main_body. which in a part only has a sketch, this points to
the sketch. i.e. main_body_tag is the tag of the sketch.
int count_1 = 0;  
3 r% B0 A5 V7 j/ f' [6 J1 N0 S0 ytag_t sketch_line= NULL_TAG;  
tag_t sketch_body_line;  
. c. m' s, H8 r4 `; F+ F' j' U6 uuf_list_t *sketch_line_list,*extrude_sket;  
UF_MODL_create_list(&sketch_line_list);
  
  z$ }. q) R+ iUF_OBJ_cycle_objs_in_part( main_body_tag, type, &sketch_line );  
# e) |4 ^( A+ U- Swhile ( sketch_line!= NULL_TAG )  
{  
     count_1++;  
& _4 y# T5 |: b8 ^/ w/ R8 o  I/ h     UF_OBJ_cycle_objs_in_part( main_body_tag, type, &sketch_line );  
' V8 J3 @( _6 m# n. ?     if(count_1==33)  
    {  
  
7 `, d  {; T8 D% W          UF_MODL_ask_feat_body(sketch_line,&sketch_body_line);
! P2 Q7 c* |6 ?. H. @! E$ k( M' w          UF_DISP_set_highlight(sketch_line,1);  
         UF_MODL_put_list_item(sketch_line_list,sketch_body_line);
) M  T7 `" P9 V& L  M6 z   UF_MODL_ask_list_item(sketch_line_list,0,&list_item);
   }   
8 k2 ~7 l3 S! o3 J4 I}  
7 B6 ?+ [1 a& S//here this part of the progrom was supposed to get the features in a sketch and put into list,sketch_line_list.

cebx

11111111111

sinyonxieliang

看看

lzq641039

我是新手,看不懂

l3278

thanks

dyks000

不知道LZ是怎么解决的啊,

1 t; j/ n  @' Z( R3 g% K* b$ j/ q遍历完了得到的是整个 part 文件里的 tag ,如何区分出草图里的 tag呢
' C7 X- f( W' K- ^) s
7 @- w0 N& A3 _1 T
我现在只想用下草图里的 特征标识做下拉伸,

求教...

dyks000

UF_OBJ_cycle_objs_in_part( main_body_tag, type, &sketch_line );  
5 D9 w, m5 p1 Z6 i- Q
UF_OBJ_cycle_objs_in_part()第一个参数不用非得是 .prt文件的 tag吗,可以用草图的tag吗?