小编Ser*_*ton的帖子

我正在使用storyboard作为iOS应用程序,我的故事板看起来像这样:http://d.pr/7yAY(droplr url)

问题是当我单击登录按钮时,我将捕获的用户名发送到事件表视图控制器.为此我使用prepareForSegue函数,但显然当我尝试设置用户名时抛出异常.

我的代码如下:

ViewController.h

#import <UIKit/UIKit.h>

@interface ViewController : UIViewController

- (IBAction) logintButton:(id)sender;
@property (weak, nonatomic) IBOutlet UITextField *username_tf; // textfield

@end

ViewController.m

#import "ViewController.h"
#import "EventsTableViewController.h"

@interface ViewController ()

@end

@implementation ViewController
@synthesize username_tf;


- (void) prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if([[segue identifier] isEqualToString:@"ToMainApp"])
    {
        EventsTableViewController * dest = (EventsTableViewController *)[segue destinationViewController];
        NSString * username = [[NSString alloc] initWithFormat:@"%@", username_tf.text];
        [dest setUsername:username];
    }
}


- (IBAction) logintButton:(id)sender
{
    //NSLog(@"Logint button pressed");
    [self performSegueWithIdentifier:@"ToMainApp" sender:sender];
} …

我正在努力学习Subquerys.
我有这个麻烦:

两个表:

CREATE TABLE DEPT
(DEPTNO NUMBER(2) CONSTRAINT DEPT_PRIMARY_KEY PRIMARY KEY,
LOC varchar2(3));

CREATE TABLE EMP
(ENAME varchar2(10),
JOB varchar2(9),
DEPTNO NUMBER(2) NOT NULL
CONSTRAINT EMP_FOREIGN_KEY REFERENCES DEPT (DEPTNO));

我想获得名称(emp.ename)和工作(emp.job),但只有"CHICAGO"中的工作也存在.

这就是我所做的:

SELECT emp1.ename, emp1.job
FROM emp emp1
WHERE emp1.job EXISTS (SELECT emp2.job
                      FROM emp emp2
                      FULL JOIN dept ON (emp2.deptno = dept.deptno)
                      WHERE dept.loc = 'CHICAGO');

我总是在第3行得到"无效的关系运算符"错误.

结果示例:

ENAME | JOB  | LOC 
JONES | SALE | CHICAGO 
FORD  | SALE | NEW YORK  //He doesn't sit in CHICAGO …

我的mysql表看起来像这样:

word1  word2  count
a      c      1
a      d      2
a      e      3
a      f      4
b      c      5
b      d      6
b      g      7
b      h      8

"a"和"b"是用户输入 - 从表中选择*,其中word1 ='a'或word1 ='b' - 得到~10000行

我需要一个查询来获取:word1是intput"a"的列

word1_是输入"b"的列

word2和word2_是同一列,因此可以忽略其中一个

我需要结合上表中的表格.例如这个查询:

select 
  t1.word1, t1.word2, t1.count, 
  t2.word1 as word1_, t2.word2 as word2_, t2.count as count_
from table t1
join table t2 on t1.word2 = t2.word2
where t1.word1 = 'a' and t2.word1 = 'b'

产生

word1   word2   count   word1_  word2_  count_  
a       c       1 …

当我插入度数符号时使用 PHP 输入表单，而不是将其保存为显示 MySQL 中的值"°"，例如，如果我将"37°C"其保存为"37°C"

这是我的代码示例

<input class="form-control" type="text" name="inc" placeholder="48 Hrs Aerobic, 37°C" value="48 Hrs Aerobic, 37°C">

"°C"当从数据库中获取数据时，也会显示该字符。

我有这个问题

SELECT Month(date) AS Monthly, Year(date) AS Annual, COUNT(idcad) AS NumCad, SUM(CONVERT(FLOAT, valorpag)) AS Valor FROM PI_AS
WHERE (status = 'Paid' OR status = 'Available') AND platform = 'Sales'
GROUP BY Year(date), Month(date)
ORDER BY Year(date), Month(date)

样本结果:

Monthly | Annual | NumCad | Valor
      3 | 2014   |     62 | 72534
      4 | 2014   |      7 |  8253.6
      5 | 2014   |     42 | 45356.39
      6 | 2014   |     36 | 33343.19
      7 | 2014   |      5 |  4414.6 …