在我的应用程序中,我有一个十六进制类型的图像;我想转换成blob类型并保存到数据库中;
这是如何可能的。
关于K L白酒
发布于 2012-04-18 19:36:37
据我所知,它是base64编码的,假设你从web服务获得图像数据,你需要将接收到的字符串数据编码成base64编码。
NSData* data=[[NSData alloc] initWithBase64EncodedString:base64ImageString];然后,您可以轻松地将数据检索到UIImageView中,
UIImageView* imageView=[[UIImageView alloc] initWithImage:[UIImage imageWithData:data]];并将图像数据作为Blob数据类型保存到sqlite中,如下代码段所示
if(insertStmt == nil) {
const char *sql = "insert into TableName values(?,?)";
if(sqlite3_prepare_v2(database, sql, -1, &insertStmt, NULL) != SQLITE_OK)
NSAssert1(0, @"Error while creating update statement. '%s'", sqlite3_errmsg(database));
}
sqlite3_bind_text(insertStmt, 1, [imageName UTF8String], -1, SQLITE_TRANSIENT);
NSData *imgData = UIImagePNGRepresentation(imageView.image);
int returnValue = -1;
returnValue = sqlite3_bind_blob(insertStmt, 2, [imgData bytes], [imgData length], NULL);
if(returnValue != SQLITE_OK)
NSLog(@"Not OK!!!");
if(SQLITE_DONE != sqlite3_step(insertStmt))
NSAssert1(0, @"Error while inserting. '%s'", sqlite3_errmsg(database));
sqlite3_reset(insertStmt);
[data release];
[imageView release];从数据库检索图像数据
if(detailStmt == nil) {
const char *sql = "Select * from TableName";
if(sqlite3_prepare_v2(database, sql, -1, &detailStmt, NULL) != SQLITE_OK)
NSAssert1(0, @"Error while creating detail view statement. '%s'", sqlite3_errmsg(database));
}
if(SQLITE_DONE != sqlite3_step(detailStmt)) {
NSData *data = [[NSData alloc] initWithBytes:sqlite3_column_blob(detailStmt, 2) length:sqlite3_column_bytes(detailStmt, 2)];
if(data == nil)
NSLog(@"No image found.");
else
imageview.image = [UIImage imageWithData:data];
} 致以敬意,
马约尔
https://stackoverflow.com/questions/10206864
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