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

[ 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。。。
, ~4 N- ~$ z8 Z# f, U* q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。
- a% E5 A2 P# X9 F) ^tag_t part1;
UF_PART_load_status_t error_status;
int type=UF_feature_type;
tag_t feature=NULL_TAG;
tag_t * fea_tags=0;
int I_count=0,feature_count;
3 ?9 _9 Z, d/ `$ hchar * feature_name;
, p$ }! Y9 `% \- j4 z& Dchar * feature_type;
char * parent_name;
char main_body[20];
tag_t main_body_tag;
1 }' d+ Z3 m$ W  z# T) Jchar msg[100];
char buffer[20];
" d/ c7 t8 m& Q' w; M7 T3 e, _ part1=UF_PART_ask_nth_part(0);
* Y) Y& m6 r. U UF_OBJ_cycle_objs_in_part(part1,type,&feature);
- l! N% K7 r3 Q: V while(feature!=NULL_TAG)
{  I_count++;
$ G5 S8 }2 h4 K, i, q+ t8 b; t0 [  UF_OBJ_cycle_objs_in_part(part1,type,&feature);
) Y; B. X. a/ E* z" x8 O# i4 G }
5 B3 X' G. h) P8 ^9 L* z) ?6 jfeature_count=I_count;
$ K* F  ^5 ?7 t* u* F' Vfea_tags =new tag_t[feature_count];
char* fea_names[500];
7 F: b& x3 J8 j) x8 k) W: tchar* fea_types[500];
for(int i=0;i< feature_count;i++)
; S8 `+ K" [4 S5 J{
fea_names=new char[133];
6 Z1 R# k& S, I    fea_types=new char[133];
}
+ X( y: U& @9 t( ]7 o8 ?% E feature=NULL_TAG;
1 _2 ~# \( G# \& f4 v  for (i=0;i< feature_count;i++)
  {
7 Y/ `1 F+ [0 {2 V( G, a    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);
9 B1 z/ N  C' m  o9 U    uc1601(_itoa( fea_tags, buffer, 10 ),1);
1 @/ Y4 x7 P+ k& N# F    uc1601(fea_names,1);
: |# N& p; Y0 q    uc1601(fea_types,1);
' u+ E$ ^' _$ ]0 O3 N0 @/ {  }
//above routine gets all the features in a part, including tags,names and type.
for (i=0;i<feature_count;i++)
% x6 @" }( z9 G) P{
5 H; K* E9 t! {- e, D2 @" ?- O+ k if(strcmp(fea_types,"SKETCH")==0)
8 t) }) Z+ k+ A1 ]) G1 l# f- f% k {
  strcpy(main_body,fea_names);
       main_body_tag=fea_tags;
3 |* @" Z" Y- A  break;
}
}
* z  A3 W' p' S1 q$ @strcpy(msg,"main body is ");
) _3 N8 n7 t% s4 d. Fuc1601(strcat(msg,main_body),1);
strcpy(msg,"main body tag is ");
6 |3 j, T5 T( K4 ]- u  Luc1601(strcat(msg,_itoa( main_body_tag, buffer, 10 )),1);
7 t, `7 t6 N3 e5 H//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;  
tag_t sketch_line= NULL_TAG;  
( f" r! X+ y6 `tag_t sketch_body_line;  
uf_list_t *sketch_line_list,*extrude_sket;  
$ p  m& C) X7 i" R3 C) d( Z; iUF_MODL_create_list(&sketch_line_list);
  
  D7 @4 v- Y1 M7 O5 f5 SUF_OBJ_cycle_objs_in_part( main_body_tag, type, &sketch_line );  
while ( sketch_line!= NULL_TAG )  
{  
     count_1++;  
     UF_OBJ_cycle_objs_in_part( main_body_tag, type, &sketch_line );  
: f& j% @; E7 }) i6 E     if(count_1==33)  
8 h1 `. t8 `9 ?; V* P    {  
  
. O# b+ n8 B2 w7 x: j( v3 V  ^          UF_MODL_ask_feat_body(sketch_line,&sketch_body_line);
  Z# y; t0 w9 i- [1 R0 P          UF_DISP_set_highlight(sketch_line,1);  
9 A6 k% U" O  c# t* S6 r% q, h2 y         UF_MODL_put_list_item(sketch_line_list,sketch_body_line);
   UF_MODL_ask_list_item(sketch_line_list,0,&list_item);
5 ^/ l% {% r  b# {9 [   }   
}  
0 V. D, w! T! v% V//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,
5 d- v3 c$ {; i6 J7 G
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


& U: C( T) t6 G. y1 x我现在只想用下草图里的 特征标识做下拉伸,
" |; `( j, e# E3 f$ O) T2 I
* x1 f5 V1 x8 C  F- r& `5 i" k求教...

dyks000

UF_OBJ_cycle_objs_in_part( main_body_tag, type, &sketch_line );  

- l# p" l% y! k9 D) QUF_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?