我正在尝试制作我的第一个 Android 游戏,为此我制作了一个 OBJ 加载程序,它采用资源 id 并使用 Context.getResources().openRawResource() 返回的 InputStream。
我成功读取了顶点信息和面信息,并且可以成功加载没有任何纹理的网格。
看到问题是纹理没有正确映射。我通过将“vt”后面的所有数字以及第一个“/”后面的 f 中的索引(即“f 1/1 ...”)放入数组中,设法从文件中读取信息
我不太确定我应该如何处理这些数字...如果我只是将所有内容放入字节缓冲区并继续,就像我从这两个教程中学到的方式一样( jayway ,和nehe端口)它仍然会出现错误。
我的想法是文件中的“vt”信息乱序(因此需要“f”数据中的“/x”),并且我应该在创建之前以某种方式将其按顺序排列来自数组的字节缓冲区....我不知道该怎么做:(
所以我想问一下,有人可以帮我吗?我知道互联网上还有其他 obj 加载器,但我已经制作了这个,并且我知道它是如何工作的,不仅如此,它对我来说也是一种更好的学习体验。
这是我正在使用的代码:
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Vector;
import android.content.Context;
import android.util.Log;
import com.snakeinalake.catchthemoney.Mesh.CMeshInfo;
public class CMeshParserOBJ extends CMeshParser{
public CMeshParserOBJ(){
type = "OBJ";
}
@Override
public CMeshInfo loadMesh(Context context, int FileID){
CMeshInfo ret = new CMeshInfo(type);
String buildWord = "";
int data;
int found = -1;
Vector<Float> vert = new Vector<Float>();
Vector<Float> uv = new Vector<Float>();
Vector<Short> ind = new Vector<Short>();
Vector<Short> map = new Vector<Short>();
InputStream is = context.getResources().openRawResource(FileID);
InputStreamReader rd = new InputStreamReader(is);
try {
data = rd.read();
while(data!=-1){
char c = (char)data;
if(found == -1){
if(c == 'v'){ //found vertex data
data = rd.read();
if((char)data == ' '){
found = 1;
}else if((char)data == 't'){ //found uv map data
data = rd.read();
if((char)data == ' '){
found = 3;
}
}else{
found = -1;
}
}else if(c == 'f'){ //found index data (faces)
data = rd.read();
if((char)data == ' '){
found = 2;
}else{
found = -1;
}
}else{
found = -1;
}
}
if(found == 1){
float x = 0,y = 0,z = 0;
for(int repeat = 0; repeat<3; repeat++){
buildWord="";
do{
data = rd.read();
c = (char)data;
buildWord+=c;
}while((char)data!=' ' && (char)data!='\n');
if(repeat==0)
x = Float.parseFloat(buildWord.trim());
else if(repeat==1)
y = Float.parseFloat(buildWord.trim());
else if(repeat==2)
z = Float.parseFloat(buildWord.trim());
}
vert.add(x);
vert.add(y);
vert.add(z);
found = -1;
}
if(found == 2){
short v1 = 0, v2 = 0, v3 = 0;
short map1 = 0, map2 = 0, map3 = 0;
boolean uvdata = false;
for(int repeat = 0; repeat<3; repeat++){
buildWord="";
do{
data = rd.read();
if(!uvdata && (char)data == '/'){
uvdata = true;
}else{
c = (char)data;
buildWord+=c;
}
}while((char)data!=' ' && (char)data!='\n' && (char)data!='/');
if(repeat==0){
v1 = Short.parseShort(buildWord.trim());
}else if(repeat==1){
v2 = Short.parseShort(buildWord.trim());
}else if(repeat==2){
v3 = Short.parseShort(buildWord.trim());
}
if(uvdata){
uvdata = false;
buildWord = "";
do{
data = rd.read();
c = (char)data;
buildWord+=c;
if(!uvdata && (char)data == '/'){
uvdata = true;
}
}while((char)data!=' ' && (char)data!='\n');
if(repeat == 0)
map1 = Short.parseShort(buildWord.trim());
else if(repeat == 1)
map2 = Short.parseShort(buildWord.trim());
else if(repeat == 2)
map3 = Short.parseShort(buildWord.trim());
else{
map1 = 0; map2 = 0; map3 = 0;
}
}
}
ind.add(v1);
ind.add(v2);
ind.add(v3);
map.add(map1);
map.add(map2);
map.add(map3);
found = -1;
}
if(found == 3){
float uvx = 0, uvy = 0;
for(int repeat = 0; repeat<2; repeat++){
buildWord="";
do{
data = rd.read();
c = (char)data;
buildWord+=c;
}while((char)data!=' ' && (char)data!='\n');
if(repeat==0)
uvx = Float.parseFloat(buildWord.trim());
else if(repeat==1)
uvy = Float.parseFloat(buildWord.trim());
}
uv.add(uvx);
uv.add(uvy);
found = -1;
}
data = rd.read();
}
rd.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
ret.vertices = new float[vert.size()];
for(int i=0; i<ret.vertices.length; i++){
ret.vertices[i] = vert.get(i);
Log.d("CATCH", Float.toString(ret.vertices[i]));
}
ret.uvMappings = new short[map.size()];
for(int i=0; i<ret.uvMappings.length; i++){
ret.uvMappings[i] = (short) (map.get(i)-1);
Log.d("CATCH", Float.toString(ret.uvMappings[i]));
}
ret.uvtex = new float[uv.size()];
for(int i=0; i<ret.uvtex.length; i++){
ret.uvtex[i] = uv.get(i);
Log.d("CATCH", Float.toString(ret.uvtex[i]));
}
ret.indices = new short[ind.size()];
for(int i=0; i<ret.indices.length; i++){
ret.indices[i] = (short) (ind.get(i)-1);
}
ret.MapUVCoordinates();
ret.fillBuffers();
return ret;
}}
“ret.MapUVCoordiantes”方法不起作用,这只是我自己尝试订购它们......我不会发布它,因为它只是不做任何事情。 方法“ret.fullBuffers”如下。它只是从数组中创建字节缓冲区:
public void fillBuffers(){
ByteBuffer vbb = ByteBuffer.allocateDirect(vertices.length * 4);
ByteBuffer ibb = ByteBuffer.allocateDirect(indices.length * 2);
ByteBuffer ubb = ByteBuffer.allocateDirect(uvtex.length * 4);
Log.d("CMeshInfo::fillBuffers", Float.toString(vertices.length * 4));
Log.d("CMeshInfo::fillBuffers", Float.toString((indices.length * 2)));
Log.d("CMeshInfo::fillBuffers", Float.toString(uvtex.length * 4));
vbb.order(ByteOrder.nativeOrder()); //Vertices
ubb.order(ByteOrder.nativeOrder()); //UV coordinates
ibb.order(ByteOrder.nativeOrder()); //Indices
vertBuff = vbb.asFloatBuffer();
uvBuff = ubb.asFloatBuffer();
IndBuff = ibb.asShortBuffer();
vertBuff.put(vertices);
uvBuff.put(uvtex);
IndBuff.put(indices);
vertBuff.position(0);
uvBuff.position(0);
IndBuff.position(0);
}
最佳答案
如果我没记错的话,这些数字是顶点数组的索引。
关于java - 帮助我的 Android obj 加载器上的 UV,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4558220/