我刚开始开发一个类似于Contact的ios应用程序。 它包括姓名,生日,小组(学校,工作),血型,图像等。
其中,可以通过键入来插入其他数据。 但是,图像可以来自拍照,从Facebook下载等等。
将图像输入xcode是不可能的。 Bcoz,另一个图像可以一次又一次地出现。
我该怎么办?请给我建议。
答案 0 :(得分:1)
你只需将libSqlite3.dylib添加到Linked FrameWork和Lilbraries并在.h文件中声明数据库变量
//Database Variables
@property (strong, nonatomic) NSString *databasePath;
@property (nonatomic)sqlite3 *contactDB;
拖放UIImageView并将其命名为...我声明为imgView。
转到.m文件,您只需复制并粘贴该代码
- (void)viewDidLoad
{
[super viewDidLoad];
// Do any additional setup after loading the view, typically from a nib.
NSString *docsDir;
NSArray *dirPaths;
// Get the documents directory
dirPaths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
docsDir = dirPaths[0];
// Build the path to the database file
_databasePath = [[NSString alloc] initWithString: [docsDir stringByAppendingPathComponent: @"images.db"]];
//docsDir NSPathStore2 * @"/Users/gayathiridevi/Library/Application Support/iPhone Simulator/7.0.3/Applications/B5D4D2AF-C613-45F1-B414-829F38344C2A/Documents" 0x0895e160
NSFileManager *filemgr = [NSFileManager defaultManager];
if ([filemgr fileExistsAtPath: _databasePath ] == NO)
{
const char *dbpath = [_databasePath UTF8String];
if (sqlite3_open(dbpath, &_contactDB) == SQLITE_OK)
{
char *errMsg;
const char *sql_stmt = "CREATE TABLE IF NOT EXISTS IMAGETB (ID INTEGER PRIMARY KEY AUTOINCREMENT,URL TEXT UNIQUE, IMAGE BLOB)";
if (sqlite3_exec(_contactDB, sql_stmt, NULL, NULL, &errMsg) != SQLITE_OK)
{
NSLog( @"User table Not Created Error: %s", errMsg);
}
else
{
NSLog( @"User table Created: ");
}
sqlite3_close(_contactDB);
}
else {
NSLog( @"DB Not Created");
}
}
[self saveImage];
[self showImage];
}
- (void)saveImage
{
sqlite3_stmt *statement;
const char *dbpath = [_databasePath UTF8String];
if (sqlite3_open(dbpath, &_contactDB) == SQLITE_OK)
{
NSString *insertSQL=@"INSERT INTO IMAGETB(image) VALUES(?)";
if(sqlite3_prepare_v2(_contactDB, [insertSQL cStringUsingEncoding:NSUTF8StringEncoding], -1, &statement, NULL)== SQLITE_OK)
{
UIImage *image = [[UIImage alloc] initWithData:[NSData dataWithContentsOfURL:[NSURL URLWithString:@"https://lh6.googleusercontent.com/-vJBBGUtpXxk/AAAAAAAAAAI/AAAAAAAAADQ/nfgVPX1n-Q8/photo.jpg"]]];
NSData *imageData=UIImagePNGRepresentation(image);
//sqlite3_bind_text(statement, 1, [@"" UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_blob(statement, 1, [imageData bytes], [imageData length], SQLITE_TRANSIENT);
NSLog(@"Length : %lu", (unsigned long)[imageData length]);
}
if (sqlite3_step(statement) == SQLITE_DONE)
{
NSLog( @"Insert into row id %lld",(sqlite3_last_insert_rowid(_contactDB)));
}
else {
NSLog( @"Error IN INSERT" );
}
sqlite3_finalize(statement);
sqlite3_close(_contactDB);
}
}
- (void)showImage
{
sqlite3_stmt *statement;
const char *dbpath = [_databasePath UTF8String];
int i = 1;
if(sqlite3_open(dbpath,&_contactDB)==SQLITE_OK)
{
NSString *insertSQL = [NSString stringWithFormat:@"Select IMAGE FROM IMAGETB WHERE ID = %d",i];
if(sqlite3_prepare_v2(_contactDB,[insertSQL cStringUsingEncoding:NSUTF8StringEncoding], -1, &statement, NULL) == SQLITE_OK) {
while(sqlite3_step(statement) == SQLITE_ROW) {
int length = sqlite3_column_bytes(statement, 0);
NSData *imageData = [NSData dataWithBytes:sqlite3_column_blob(statement, 0) length:length];
NSLog(@"Length : %lu", (unsigned long)[imageData length]);
if(imageData == nil)
NSLog(@"No image found.");
else
_imgView.image = [UIImage imageWithData:imageData];
NSLog(@"image found.");
}
}
sqlite3_finalize(statement);
}
sqlite3_close(_contactDB);
}
我的代码。您可以通过更改i值来修复后退和前进按钮以获取详细信息 - (无效)showImage 功能
答案 1 :(得分:0)
关于拍摄照片或从图书馆抓取照片,请参阅UIImagePickerController Class Reference和Camera Programming Topics for iOS。要从Facebook上抓取照片,你应该参考它的API,因为这是一个完全不同的鱼。