2012-10-30 43 views
0

有人告訴我如何使用java代碼爲iOS創建鈦移動模塊? https://github.com/cashlo/JPEG-compression-for-titaninum-如何使用java代碼爲iOS創建鈦移動模塊?

如何編譯爲iOS移動應用下面的代碼(鈦)??

package com.cashlo.jpglib; 

import java.io.ByteArrayOutputStream; 

import org.appcelerator.kroll.KrollModule; 
import org.appcelerator.kroll.annotations.Kroll; 

import org.appcelerator.titanium.TiBlob; 
import org.appcelerator.titanium.TiContext; 
import org.appcelerator.titanium.util.Log; 
import org.appcelerator.titanium.util.TiConfig; 

import android.graphics.Bitmap; 
import android.graphics.Bitmap.CompressFormat; 
import android.graphics.BitmapFactory; 

@Kroll.module(name = "Jpglib", id = "com.cashlo.jpglib") 
public class JpglibModule extends KrollModule { 

    // Standard Debugging variables 
    private static final String LCAT = "JpglibModule"; 
    private static final boolean DBG = TiConfig.LOGD; 

    // You can define constants with @Kroll.constant, for example: 
    // @Kroll.constant public static final String EXTERNAL_NAME = value; 

    public JpglibModule(TiContext tiContext) { 
     super(tiContext); 
    } 

    @Kroll.method 
    public TiBlob compress(TiBlob image) { 

     BitmapFactory.Options opts = new BitmapFactory.Options(); 
     opts.inJustDecodeBounds = true; 
     byte[] imageBytes = image.getBytes(); 
     BitmapFactory.decodeByteArray(imageBytes, 0, imageBytes.length, opts); 
     Integer imageSize = Math.max(opts.outWidth, opts.outHeight); 
     if (imageSize > 600) 
      opts.inSampleSize = imageSize/600; 
     opts.inJustDecodeBounds = false; 
     Bitmap resized = BitmapFactory.decodeByteArray(image.getBytes(), 0, 
       imageBytes.length, opts); 
     ByteArrayOutputStream stream = new ByteArrayOutputStream(); 
     resized.compress(CompressFormat.JPEG, 70, stream); 
     TiBlob blob = TiBlob.blobFromData(getTiContext(), stream.toByteArray(), 
       "image/jpeg"); 

     return blob; 
    } 

} 
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你能請張貼相關,而不僅僅是鏈接到github上,也部分代碼 - 接受大家的一些問題,如果你想有一個更好的反應。 – david99world

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本機iOS模塊是用Obj-C而不是Java編寫的。你的意思是將這個Java代碼移植到Obj-C? –

回答