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问题 We have an images folder on our web servers that we may publish via a CDN. Sometimes we append query-string like syntax to URL's to help us freshen content that has changed, even though it rarely does. Example: /images/file.png?20090821 will URL's like this work with your average content-delivery-network? 回答1: Yes, We use Akamai, which keeps a cached copy of each distict url requested including the querystring. So the first request for /images/file.png?20090821 will go to the origin server.

问题 We have an images folder on our web servers that we may publish via a CDN. Sometimes we append query-string like syntax to URL's to help us freshen content that has changed, even though it rarely does. Example: /images/file.png?20090821 will URL's like this work with your average content-delivery-network? 回答1: Yes, We use Akamai, which keeps a cached copy of each distict url requested including the querystring. So the first request for /images/file.png?20090821 will go to the origin server.

问题 I am trying to open a website in my app, but for some reason one line keeps returning nil, heres my code: let url = URL(string: "http://en.wikipedia.org/wiki/\(element.Name)")! if #available(iOS 10.0, *) { UIApplication.shared.open(url, options: [:], completionHandler: nil) } else { UIApplication.shared.openURL(url) } } It's the first line ( let url = URL... ) that keeps on returning this error: fatal error: unexpectedly found nil while unwrapping an Optional value. What should I do to fix

问题 http://www.example.com/some_folder/some file [that] needs "to" be (encoded).zip urlencode($myurl); The problem is that urlencode will also encode the slashes which makes the URL unusable. How can i encode just the last filename ? 回答1: Try this: $str = 'http://www.example.com/some_folder/some file [that] needs "to" be (encoded).zip'; $pos = strrpos($str, '/') + 1; $result = substr($str, 0, $pos) . urlencode(substr($str, $pos)); You're looking for the last occurrence of the slash sign. The part

问题 http://www.example.com/some_folder/some file [that] needs "to" be (encoded).zip urlencode($myurl); The problem is that urlencode will also encode the slashes which makes the URL unusable. How can i encode just the last filename ? 回答1: Try this: $str = 'http://www.example.com/some_folder/some file [that] needs "to" be (encoded).zip'; $pos = strrpos($str, '/') + 1; $result = substr($str, 0, $pos) . urlencode(substr($str, $pos)); You're looking for the last occurrence of the slash sign. The part

问题 http://www.example.com/some_folder/some file [that] needs "to" be (encoded).zip urlencode($myurl); The problem is that urlencode will also encode the slashes which makes the URL unusable. How can i encode just the last filename ? 回答1: Try this: $str = 'http://www.example.com/some_folder/some file [that] needs "to" be (encoded).zip'; $pos = strrpos($str, '/') + 1; $result = substr($str, 0, $pos) . urlencode(substr($str, $pos)); You're looking for the last occurrence of the slash sign. The part

问题 I need to get my current route without params in Angular 2, I found a way to get the current route with params as follows: this.router.url and then split it: this.router.url.split(';')[0] But this looks as workaround, I think there should be better way? 回答1: parseTree from Router helps fetching the segments without any knowledge about url structure. import { Router } from '@angular/router'; ... constructor(private router: Router) {} ... const urlTree = this.router.parseUrl(url); const

问题 I need to get my current route without params in Angular 2, I found a way to get the current route with params as follows: this.router.url and then split it: this.router.url.split(';')[0] But this looks as workaround, I think there should be better way? 回答1: parseTree from Router helps fetching the segments without any knowledge about url structure. import { Router } from '@angular/router'; ... constructor(private router: Router) {} ... const urlTree = this.router.parseUrl(url); const

问题 I'm working on a VueJS project and am trying to load an image on a carousel. I am using the standard setup and have the image in the assets folder. I reference the image URL with <v-carousel-item src="@/assets/promo1.jpg"> But this throws an Image Load Failed error when I run the server using npm run serve . console.js?66f6:36 [Vuetify] Image load failed src: @/assets/promo1.jpg found in ---> <VImg> <VCarouselItem> <VCarousel> <Home> at src/views/Home.vue <VApp> <App> at src/App.vue <Root> If

问题 I'm working on a VueJS project and am trying to load an image on a carousel. I am using the standard setup and have the image in the assets folder. I reference the image URL with <v-carousel-item src="@/assets/promo1.jpg"> But this throws an Image Load Failed error when I run the server using npm run serve . console.js?66f6:36 [Vuetify] Image load failed src: @/assets/promo1.jpg found in ---> <VImg> <VCarouselItem> <VCarousel> <Home> at src/views/Home.vue <VApp> <App> at src/App.vue <Root> If