0

我想做以下列表。

  1. 使用 UIWebView
  2. 点击 webview 上的按钮，然后执行本机代码（http 请求）并获取响应。

我可以得到回应。但是，在回调函数（冻结）中不能正确地“警告”。

为什么要那样做？

谢谢你。

文本视图控制器

#import "TestViewController.h"
#import "AFNetworking.h"

@interface TestViewController ()

@end

@implementation TestViewController

- (id)initWithNibName:(NSString *)nibNameOrNil bundle:(NSBundle *)nibBundleOrNil
{
    self = [super initWithNibName:nibNameOrNil bundle:nibBundleOrNil];
    if (self) {
        // Custom initialization]
        [self initWebView] ;
    }
    return self;
}

- (void)viewDidLoad
{
    [super viewDidLoad];
    // Do any additional setup after loading the view.
}

- (void)didReceiveMemoryWarning
{
    [super didReceiveMemoryWarning];
    // Dispose of any resources that can be recreated.
}
- (void)initWebView{
    self.webView = [[UIWebView alloc] initWithFrame:SCREEN_APPLICATIONFRAME] ;
    [self.view addSubview:self.webView] ;
    self.webView.delegate = self ;
    [self.webView loadRequest:[NSURLRequest requestWithURL:[NSURL URLWithString:@"SOME_HTML"]]] ;
}

-(BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)
request navigationType:(UIWebViewNavigationType)navigationType
{
    if ([ request.URL.scheme isEqualToString:@"native" ]) {
        if ([request.URL.host isEqualToString:@"foo"]) {            
            NSURL *url = [[NSURL alloc]initWithString:@"SOME_JSON_RESPONSE_URL"] ;
            NSURLRequest *request = [NSURLRequest requestWithURL:url];

            AFJSONRequestOperation *operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:request success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {
                NSLog(@"SUCCESS") ;
                [self.webView stringByEvaluatingJavaScriptFromString:@"cbFoo('OK')"] ;
            } failure:^(NSURLRequest *request, NSURLResponse *response, NSError *error, id JSON){
                NSLog(@"ERROR: %@",error) ;
                [self.webView stringByEvaluatingJavaScriptFromString:@"cbFoo('NG')"] ;
            }];
            [operation start] ;

        } 
        return NO;
    }
    // 通常のschemeの場合は、フックせずそのまま処理を続ける
    else {
        return YES;
    }
}


@end

SOME_HTML 是这样的。

<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset= UTF-8">
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
    <script type="text/javascript">
    // callback from native
        function cbFoo(result){
        alert(result) ; // FREEZE!!
        alert('1st') ;
        alert('2nd') ;
        }

    </script>
</head>
<body>
<a href="native://foo" >Native Func</a><br>
</body>
</html>
4

2 回答 2

0

使用 setTimeout，它工作正常。但是，我不想使用 setTimeout 来解决这个问题......

function cbFoo(result){
    setTimeout(function(){
        alert(result) ;
        alert('1st') ;
        alert('2nd') ;
    },0) ;
}
于 2013-07-18T00:52:30.400 回答
0

最后我写了以下代码。

NSString *str = [NSString stringWithFormat:@"setTimeout(function(){cbGetCommentData('%@');},0)", text] ;
[self.webView stringByEvaluatingJavaScriptFromString:str] ;
于 2013-07-22T04:51:18.113 回答