将图像编码为base64,获取无效的base64string(使用base64EncodedStringWithOptions的ios)

这是我的代码。

UIImage *img = [UIImage imageNamed:@"white.jpeg"]; NSData *imageData = UIImagePNGRepresentation(img); NSString *imageString = [imageData base64EncodedStringWithOptions:0]; NSLog(@"%@", imageString); 

而且我总是得到包含空间的无效base64string。

 iVBORw0KGgoAAAANSUhEUgAAARMAAAC3CAIAAAC MS2jAAAAHGlET1QAAAACAAAAAAAAAFwAAAAoAAAAXAAAAFsAAAMC oRdmgAAAs5JREFUeAHs1tFKI1EABFH//5NNVh02uA LQhwoCZ0zj5LbTlffInl58yCAwM8JvPz8iBMIIPDGHJcAgTMEmHOGmjMIMMcdQOAMAeacoeYMAsxxBxA4Q4A5Z6g5gwBz3IGHIPD zfMQL/fVSzDnKyr 9usEvhHn/ddf5N5/yJx7SflcSoA5KV7hz0Lgn0gPW9h3zsNO81wvdr1e/3w R3PmPNf82p4mcLlcDnlu/txCbvKcDqwP s6pCcu/i8Bhzn/yMOcucD705ARe/z43eY5fbQcN5jz5lVB/k4Bfa5u7alUTYE5NWP4mAeZs7qpVTYA5NWH5mwSYs7mrVjUB5tSE5W8SYM7mrlrVBJhTE5a/SYA5m7tqVRNgTk1Y/iYB5mzuqlVNgDk1YfmbBJizuatWNQHm1ITlbxJgzuauWtUEmFMTlr9JgDmbu2pVE2BOTVj JgHmbO6qVU2AOTVh ZsEmLO5q1Y1AebUhOVvEmDO5q5a1QSYUxOWv0mAOZu7alUTYE5NWP4mAeZs7qpVTYA5NWH5mwSYs7mrVjUB5tSE5W8SYM7mrlrVBJhTE5a/SYA5m7tqVRNgTk1Y/iYB5mzuqlVNgDk1YfmbBJizuatWNQHm1ITlbxJgzuauWtUEmFMTlr9JgDmbu2pVE2BOTVj JgHmbO6qVU2AOTVh ZsEmLO5q1Y1AebUhOVvEmDO5q5a1QSYUxOWv0mAOZu7alUTYE5NWP4mAeZs7qpVTYA5NWH5mwSYs7mrVjUB5tSE5W8SYM7mrlrVBJhTE5a/SYA5m7tqVRNgTk1Y/iYB5mzuqlVNgDk1YfmbBJizuatWNQHm1ITlbxJgzuauWtUEmFMTlr9JgDmbu2pVE2BOTVj JgHmbO6qVU2AOTVh ZsEmLO5q1Y1AebUhOVvEmDO5q5a1QSYUxOWv0mAOZu7alUTYE5NWP4mAeZs7qpVTeADAAD//66TqMcAAAIISURBVO3TsQ0AAAjDMP4/GnEDmc3exSKzjgCBv8D8JxYECKxyPAGBIqCcomZDQDl gEARUE5RsyGgHD9AoAgop6jZEFCOHyBQBJRT1GwIKMcPECgCyilqNgSU4wcIFAHlFDUbAsrxAwSKgHKKmg0B5fgBAkVAOUXNhoBy/ACBIqCcomZDQDl gEARUE5RsyGgHD9AoAgop6jZEFCOHyBQBJRT1GwIKMcPECgCyilqNgSU4wcIFAHlFDUbAsrxAwSKgHKKmg0B5fgBAkVAOUXNhoBy/ACBIqCcomZDQDl gEARUE5RsyGgHD9AoAgop6jZEFCOHyBQBJRT1GwIKMcPECgCyilqNgSU4wcIFAHlFDUbAsrxAwSKgHKKmg0B5fgBAkVAOUXNhoBy/ACBIqCcomZDQDl gEARUE5RsyGgHD9AoAgop6jZEFCOHyBQBJRT1GwIKMcPECgCyilqNgSU4wcIFAHlFDUbAsrxAwSKgHKKmg0B5fgBAkVAOUXNhoBy/ACBIqCcomZDQDl gEARUE5RsyGgHD9AoAgop6jZEFCOHyBQBJRT1GwIKMcPECgCyilqNgSU4wcIFAHlFDUbAsrxAwSKgHKKmg0B5fgBAkVAOUXNhoBy/ACBIqCcomZDQDl gEARUE5RsyGgHD9AoAgop6jZEFCOHyBQBJRT1GwIKMcPECgCB/GnRcSfso/fAAAAAElFTkSuQmCC 

如此迷茫。

当我尝试删除base64string中的空间,并解码它。

我无法获得原始图像。

我想你应该replace你的选项参数

更改:

 NSString *imageString = [imageData base64EncodedStringWithOptions:0]; 

至:

 NSString *imageString = [imageData base64EncodedStringWithOptions:NSDataBase64EncodingEndLineWithLineFeed]; 

以防万一你喜欢它:

 - (NSString *)imageToNSString:(UIImage *)image { NSData *data = UIImagePNGRepresentation(image); return [data base64EncodedStringWithOptions:NSDataBase64EncodingEndLineWithLineFeed]; } - (UIImage *)stringToUIImage:(NSString *)string { NSData *data = [[NSData alloc]initWithBase64EncodedString:string options:NSDataBase64DecodingIgnoreUnknownCharacters]; return [UIImage imageWithData:data]; } 

请记住,这是一个iOS 7 API。

邑,我发现问题是我不编码的url。

在发布数据中,我的base64“+”字符被翻译成“”字符。

所以我得到无效的base64string。

感谢Logan。

replace+%2B的 iosreplace所有的+“”空间,使图像无效

后端也必须处理它

 - (NSString *)base64String:(UIImage*)image {return [[UIImageJPEGRepresentation(image,1) base64EncodedStringWithOptions:NSDataBase64EncodingEndLineWithLineFeed] stringByReplacingOccurrencesOfString:@"+" withString:@"%2B"];} 

从Fadi Abuzant的启发回答这里是Swift 3版本

 stringBase64 = stringBase64.replacingOccurrences(of: "+", with: "%2B")