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@omarrr
Last active October 10, 2015 19:48
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Split any well-formed URI into its parts
// parseUri 1.2.2
// (c) Steven Levithan <stevenlevithan.com>
// MIT License
// via: http://blog.stevenlevithan.com/archives/parseuri
function parseUri (str) {
var o = parseUri.options,
m = o.parser[o.strictMode ? "strict" : "loose"].exec(str),
uri = {},
i = 14;
while (i--) uri[o.key[i]] = m[i] || "";
uri[o.q.name] = {};
uri[o.key[12]].replace(o.q.parser, function ($0, $1, $2) {
if ($1) uri[o.q.name][$1] = $2;
});
return uri;
};
parseUri.options = {
strictMode: false,
key: ["source","protocol","authority","userInfo","user","password","host","port","relative","path","directory","file","query","anchor"],
q: {
name: "queryKey",
parser: /(?:^|&)([^&=]*)=?([^&]*)/g
},
parser: {
strict: /^(?:([^:\/?#]+):)?(?:\/\/((?:(([^:@]*)(?::([^:@]*))?)?@)?([^:\/?#]*)(?::(\d*))?))?((((?:[^?#\/]*\/)*)([^?#]*))(?:\?([^#]*))?(?:#(.*))?)/,
loose: /^(?:(?![^:@]+:[^:@\/]*@)([^:\/?#.]+):)?(?:\/\/)?((?:(([^:@]*)(?::([^:@]*))?)?@)?([^:\/?#]*)(?::(\d*))?)(((\/(?:[^?#](?![^?#\/]*\.[^?#\/.]+(?:[?#]|$)))*\/?)?([^?#\/]*))(?:\?([^#]*))?(?:#(.*))?)/
}
};
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