MySQL:第行出现错误1072（42000）。。键列UserID不'；表中不存在

我正在尝试用两个表创建一个简单的数据库，第一个表用于用户信息，第二个表用于上传，因为这是教员的项目，我有一些作业。。。一种是使用外键

但是我不知道怎么做，当我用phpMyAdmin做的时候，一切都很好，但是当我导出它并把它放到我的服务器上时，我有一些错误；请帮帮我

CREATE TABLE IF NOT EXISTS `korisnici` (
  `UserID` int(11) NOT NULL AUTO_INCREMENT,
  `username` varchar(12) COLLATE utf8_bin NOT NULL,
  `password` varchar(32) COLLATE utf8_bin NOT NULL,
  `email` varchar(32) COLLATE utf8_bin NOT NULL,
  `telefon` varchar(16) COLLATE utf8_bin NOT NULL,
  PRIMARY KEY (`UserID`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_bin ROW_FORMAT=COMPACT AUTO_INCREMENT=4;

CREATE TABLE IF NOT EXISTS `slike` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(200) COLLATE utf8_bin NOT NULL,
  `size` int(11) NOT NULL,
  `type` varchar(200) COLLATE utf8_bin NOT NULL,
  `file_path` varchar(200) COLLATE utf8_bin NOT NULL,
  `username` varchar(12) COLLATE utf8_bin NOT NULL,
  `naslov` varchar(32) COLLATE utf8_bin NOT NULL,
  `adresa` varchar(80) COLLATE utf8_bin NOT NULL,
  `opis` varchar(1200) COLLATE utf8_bin NOT NULL,
  `datum` date NOT NULL,
  `UserID` int(11) NOT NULL,
  PRIMARY KEY (`id`),
  KEY `UserID` (`UserID`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_bin AUTO_INCREMENT=4 ;


ALTER TABLE `slike`
  ADD CONSTRAINT `slike_ibfk_1` FOREIGN KEY (`UserID`) REFERENCES `korisnici` (`UserID`);

错误：

第77行出现错误1072（42000）：表中不存在键列“UserID”

如果有人能修改我的代码并解释一下，我将不胜感激

编辑新错误：

第42行出现错误1054（42S22）：“字段列表”中的未知列“UserID”

同样，当我插入某些内容时：

SET FOREIGN_KEY_CHECKS=0;
SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";

CREATE TABLE IF NOT EXISTS `korisnici` (
  `username` varchar(12) COLLATE utf8_bin NOT NULL,
  `password` varchar(32) COLLATE utf8_bin NOT NULL,
  `email` varchar(32) COLLATE utf8_bin NOT NULL,
  `telefon` varchar(16) COLLATE utf8_bin NOT NULL,
  PRIMARY KEY (`username`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

INSERT INTO `korisnici` (`username`, `password`, `email`, `telefon`) VALUES('Aleksa2', 'c4b4d0dea25b6f2f38fef63330ea15c8', 'sss@gmail.com', '0649999999');

INSERT INTO `korisnici` (`username`, `password`, `email`, `telefon`) VALUES('Aleksa123', 'a4b4d0dea25b6f2f38fef63330ea15c8', 'aaaa@gmail.com', '0649999999');

CREATE TABLE IF NOT EXISTS `slike` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(200) COLLATE utf8_bin NOT NULL,
  `size` int(11) NOT NULL,
  `type` varchar(200) COLLATE utf8_bin NOT NULL,
  `file_path` varchar(200) COLLATE utf8_bin NOT NULL,
  `username` varchar(12) COLLATE utf8_bin NOT NULL,
  `naslov` varchar(32) COLLATE utf8_bin NOT NULL,
  `adresa` varchar(80) COLLATE utf8_bin NOT NULL,
  `opis` varchar(1200) COLLATE utf8_bin NOT NULL,
  `datum` date NOT NULL,
  PRIMARY KEY (`id`),
  KEY `USERNAME_ID` (`username`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_bin AUTO_INCREMENT=3 ;

INSERT INTO `slike` (`id`, `name`, `size`, `type`, `file_path`, `username`, `naslov`, `adresa`, `opis`, `datum`) VALUES(15, '4eb40ffebd3fbec4b1acf34a4fe8cb2d.png', 43395, 'image/png', 'C:/Program Files (x86)/EasyPHP-DevServer-14.1VC11/data/localweb/images/4eb40ffebd3fbec4b1acf34a4fe8cb2d.png', 'Aleksa123', 'asdasdasd', 'dadasdas', 'Opisite problem u najvise 1200 karaktera.', '2014-09-05');
INSERT INTO `slike` (`id`, `name`, `size`, `type`, `file_path`, `username`, `naslov`, `adresa`, `opis`, `datum`) VALUES(16, '92bc0821619c053c503694666f2717ee.png', 32461, 'image/png', 'C:/Program Files (x86)/EasyPHP-DevServer-14.1VC11/data/localweb/images/92bc0821619c053c503694666f2717ee.png', 'Aleksa123', 'dadasdasdas', 'dadasdasdasdasdas', 'Opisite problem u najvise 12dasdasdasdas00 karaktera.', '2014-09-05');
INSERT INTO `slike` (`id`, `name`, `size`, `type`, `file_path`, `username`, `naslov`, `adresa`, `opis`, `datum`) VALUES(17, '89d0507bf3ed3f492647b7fd2f39047a.png', 162203, 'image/png', 'C:/Program Files (x86)/EasyPHP-DevServer-14.1VC11/data/localweb/images/89d0507bf3ed3f492647b7fd2f39047a.png', 'Aleksa123', '1111111111111a', 'dasdasdasda', 'Opisite problsdasdasdasdasaem u najvise 1200 karaktera.', '2014-09-05');

ALTER TABLE `slike`
  ADD CONSTRAINT `slike_ibfk_1` FOREIGN KEY (`username`) REFERENCES `korisnici` (`username`);
SET FOREIGN_KEY_CHECKS=1;

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是否愿意分享您遇到的错误？第77行的错误1072（42000）：表中不存在关键列“UserID”。问题中的代码运行良好：。我要自杀，因为我需要上传的服务器*。sql说我有错误：（无论如何，谢谢：）也许现在有人知道我为什么有错误了？我现在没有表格输出。。。