Question

我有一个UITextView我输入了一些文字。我有UIButton。我想点击按钮并打开我应在UITextView中输入的文字的网址。我正在为IOS 7编写一本词典应用程序，当我想让人们翻译我的文本时需要这个功能。我知道如何打开URL，但我的文本应该出现在URL中的框中。有人可以帮我编码吗？

SozdikViewController.h
@interface SozdikViewController : UIViewController
@property (strong, nonatomic) IBOutlet UITextView *textView1;
@property (weak, nonatomic) IBOutlet UIButton *Button_Ask;
-(IBAction)ask;
@end

这是我的SozdikViewController.m文件

#import "SozdikViewController.h"
@interface SozdikViewController ()
@end
@implementation SozdikViewController
-(IBAction)ask
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.audaru.kz/?product=SoylemMT&word="]];
}
...

@end

Answer 1

-(IBAction)ask
{
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.audaru.kz/?product=SoylemMT&word=%@",textView1.text]];
}

Answer 2

如果我理解正确，您只想翻译在UITextField中输入的文本。所以你必须简单使用

[[UIApplication sharedApplication] openURL:[NSURL URLWithString: [NSString stringWithFormat:@"http://www.audaru.kz/?product=SoylemMT&word=%@", textView1.text]]];

将%@移动到您网址中的正确位置。希望它有所帮助

Answer 3

试试这个

//Append your word in the text field to the url

 NSString *urlString = [NSString stringWithFormat:@"http://www.audaru.kz/?product=SoylemMT&word=%@",textField.text];            
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:urlString]];

Answer 4

您可以从文本视图中获取URL：

NSString *urlString = self.textView.text;

然后你可以打开网址：

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:urlString]];

Answer 5

我在代码中添加了注释以解释每个步骤，但您应该能够使用NSString *构建stringWithFormat:并将其传递给您（http://www.audaru.kz/?product=SoylemMT&word=%@）的URL和值[[self textView1] text]然后您需要做的就是将其传递给NSURL *的实例并使用[[UIApplication sharedApplication] canOpenURL:]检查您是否可以打开该网址。检查完毕并且如果是，则可以使用[[UIApplication sharedApplication] openURL:]打开它。同时确保所有IBOutlets都正确关联，否则无法执行任何操作。

<强> SozdikViewController.h

// Start of interface file
@interface SozdikViewController : UIViewController

// Property declarations
// Note the lower cases at the beginning of the property variable names
@property (strong, nonatomic) IBOutlet UITextField *textView1;
@property (weak, nonatomic) IBOutlet UIButton *buttonAsk; 

// Method declarations
- (IBAction)ask;

@end
// end of interface file

<强> SozdikViewController.m

// Start of implementation file
#import "SozdikViewController.h"

// You are clearly not using the class extention so why is it there
// Removed it as it is pointless being there if you aren't using it

@implementation SozdikViewController

// Synthesize is done for you automatically

- (IBAction)ask
{
    // We are going to do it like everyone says but add a few checks in
    // First lets check to make sure that the textView1s text isn't nil or an empty string
    if([[self textView1] text] != nil && ![[[self textView1] text] isEqualToString:@""]) {
        // Construct the URL with the word that has been passed in
        NSURL *urlString = [NSURL URLWithString:[NSString stringWithFormat:@"http://www.audaru.kz/?product=SoylemMT&word=%@", [[self textView1] text]]];
        // Next check to make sure that the URL can actually be opened
        // otherwise forget about it as it would be pointless.
        if([[UIApplication sharedApplication] canOpenURL:urlString]) {
            // We can open it so lets open it
        [[UIApplication sharedApplication] openURL:urlString];
        } else {
            // else do something to tell the use it can't be done.
        }
    }
}

@end
// end of implementation file

如何通过按钮打开URL？

5 个答案: