小编Rog*_*uri的帖子

首先,我正在运行MacOSX 10.7.1.我安装了所有正确的Xcode 4和所有库,以使用C语言.

我在shell中运行gprof命令时遇到了麻烦.我将逐步解释我正在做的事情以及我收到的输出.

步骤1:

~ roger$ cd Path/to/my/workspace
~ roger$ ls

输出(步骤1):

queue.c         queue.h         testqueue.c

第2步:

~ roger$ gcc -c -g -pg queue.c
~ roger$ ls

输出(第2步):

queue.c         queue.h         queue.o         testqueue.c

第3步:

~ roger$ gcc -o testqueue -g -pg queue.o testqueue.c
~ roger$ ls

输出(步骤3):

queue.c         queue.h         queue.o         testqueue       testqueue.c

第4步:

~ roger$ ./testqueue
~ roger$ ls

输出(步骤4):

enqueue element 16807
head=0,tail=1
enqueue element 282475249
head=0,tail=2
enqueue element 1622650073
head=0,tail=3
enqueue element 984943658
head=0,tail=4
enqueue element 1144108930
head=0,tail=5
enqueue …

PhpMyAdmin不接受过程中的参数"id_jugador_IN".

这是我们正在执行的程序:

SET GLOBAL event_scheduler = 1;

use traviandatabase;

DELIMITER $$

DROP PROCEDURE IF EXISTS sp_monitoritzacio_jugador $$

CREATE PROCEDURE sp_monitoritzacio_jugador(IN id_jugador_IN INT(10))

BEGIN

    CREATE OR REPLACE 
    VIEW dades_ususari AS 
    SELECT j.nom AS jugador, j.sexe AS sexe, j.edat AS edat, j.correu AS correu, a.nom AS alianca, p.nom AS pais, c.nom_ciutat AS ciutat
    FROM jugador AS j
    JOIN alianca AS a
    JOIN pais AS p
    JOIN ciutat AS c
    ON j.id_jugador = id_jugador_IN
    GROUP BY id_jugador_IN;

END $$

DELIMITER ;

CALL sp_monitoritzacio_jugador(1); …

我正在努力让左按钮正常工作并模仿后退按钮.

我创建按钮的代码:

UIBarButtonItem *backButton = [self customBarButton:@"back_button" imageHiglighted:@"settings_button_highlighted" x:20 y:0 widthDivider:2.6 heightDivider:2.6];

backButton.target = self;
backButton.action = @selector(buttonPressed:);

self.navigationItem.leftBarButtonItem = backButton;

这里调用创建自定义按钮的方法:

- (UIBarButtonItem *)customBarButton:(NSString *)imageDefault imageHiglighted:(NSString *)imageHighlighted x:(float)x y:(float)y widthDivider:(float)widthDivider heightDivider:(float)heightDivider {

UIImage *customImageDefault = [UIImage imageNamed:imageDefault];
UIImage *customImageHighlighted = [UIImage imageNamed:imageHighlighted];

CGRect frameCustomButton = CGRectMake(x, y, customImageDefault.size.width/widthDivider, customImageDefault.size.height/heightDivider);

UIButton *customButton = [[UIButton alloc] initWithFrame:frameCustomButton];

[customButton setBackgroundImage:customImageDefault forState:UIControlStateNormal];
[customButton setBackgroundImage:customImageHighlighted forState:UIControlStateHighlighted];

UIBarButtonItem *barCustomButton =[[UIBarButtonItem alloc] initWithCustomView:customButton];

return barCustomButton;

}

行动:

-(void)buttonPressed:(id) sender{

    NSLog(@"Entered");
    SearchViewController *ViewController = [[SearchViewController alloc] init]; …

我想限制一个INSERT与另一个表的行数如下:

INSERT INTO pacte (alianca_en_pacte, tipus)
SELECT id_alianca_en_pacte1 ,tipus_pacte1
FROM c_alianca
LIMIT 0, SELECT COUNT(tipus_pacte1) FROM c_alianca WHERE tipus_pacte1 IS NOT NULL

以下查询:

SELECT COUNT(tipus_pacte1) FROM c_alianca WHERE tipus_pacte1 IS NOT NULL

返回4

运行脚本时,出现以下错误:

您的SQL语法有错误; 检查与MySQL服务器版本对应的手册,以便在第4行的"SELECT*FROM c_alianca"附近使用正确的语法