Created
November 26, 2016 14:54
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Use SEO friendly urls using angular 1.x and ui-router
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// assuming you have constant that looks like that: | |
const serverSEOUrls = { | |
'super-seo-friendly-category-url': {type: 'category', id: 1}, | |
'super-seo-friendly-tag-url': {type: 'tag', id: 24} | |
}; | |
angular.module('app', []) | |
.config(($stateProvider, $urlMatcherFactoryProvider) => { | |
$urlMatcherFactoryProvider.type('dbType', { | |
encode: function (urlDef) { | |
return urlDef && _.findKey(serverSEOUrls, urlDef); | |
}, | |
decode: function (url) { | |
return serverSEOUrls[url]; | |
}, | |
is: function (urlDef) { | |
return urlDef && _.findKey(serverSEOUrls, urlDef); | |
} | |
}); | |
.state('front', { | |
url: '/{def:dbType}', | |
templateProvider: ($stateParams) => `<${urlTypeToComponent[$stateParams.def.type]}></${urlTypeToComponent[$stateParams.def.type]}>` | |
}); | |
}); | |
// Then you can use ui-sref with that | |
<a ui-sref="front({def: {type:'category', id:1}})">categorty</a> | |
// that will navigate to `super-seo-friendly-category-url` | |
// the controller of the component will get access to the object def by injecting $stateParams and accessing $stateParams.def.type for type or $stateParams.def.id for id. |
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