我在桌子上有两张图片。现在我必须将它们作为blob存储在SQLite中。但我很难找到一种方法来获得它在两个URL的情况下。如果它是一个我可以解决它。任何人都可以建议我如何解决这个问题吗?
我正在尝试的代码是:
private void callInsertion(String bid, String bbookId,String bname, String image1, String image2, String bdesc) throws IOException {
// TODO Auto-generated method stub
DefaultHttpClient mHttpClient = new DefaultHttpClient();
HttpGet mHttpGet1 = new HttpGet(image1);
HttpGet mHttpGet2 = new HttpGet(image2);
HttpResponse mHttpResponse = mHttpClient.execute(mHttpGet1,mHttpGet2);
if (mHttpResponse.getStatusLine().getStatusCode() == HttpStatus.SC_OK)
{
HttpEntity entity = mHttpResponse.getEntity();
if ( entity != null)
{
insertData(bid,bbookId,bname,EntityUtils.toByteArray(entity),EntityUtils.toByteArray(entity),bdesc);
}
}
}
但我在客户端执行时遇到错误。
答案 0 :(得分:1)
尝试这样做!!
private void callInsertion(String bid, String bbookId,String bname, String image1, String image2, String bdesc) throws IOException {
// TODO Auto-generated method stub
byte[] FirstImage = null;
byte[] SecondImage = null;
DefaultHttpClient mHttpClient1 = new DefaultHttpClient();
DefaultHttpClient mHttpClient2 = new DefaultHttpClient();
HttpGet mHttpGet1 = new HttpGet(image1);
HttpGet mHttpGet2 = new HttpGet(image2);
HttpResponse mHttpResponse1 = mHttpClient1.execute(mHttpGet1);
HttpResponse mHttpResponse2 = mHttpClient2.execute(mHttpGet2);
Log.i("calling112221","______________");
if (mHttpResponse1.getStatusLine().getStatusCode() == HttpStatus.SC_OK)
{
HttpEntity entity1 = mHttpResponse1.getEntity();
FirstImage = EntityUtils.toByteArray(entity1);
Log.i("calling11111111","______________");
}
if (mHttpResponse2.getStatusLine().getStatusCode() == HttpStatus.SC_OK)
{
HttpEntity entity2 = mHttpResponse2.getEntity();
SecondImage = EntityUtils.toByteArray(entity2);
Log.i("calling","______________");
}
insertData(bid,bbookId,bname,FirstImage,SecondImage,bdesc);
}