XMLWrangler

Easily deal with XMLs in Swift.

Add the following dependency to your Package.swift:

.package(url: "https://github.com/sersoft-gmbh/xmlwrangler", from: "6.0.0"),

Every element in an XML is represented by the XMLElement struct. It has three properties, name which reflects the element's tag name, attributes which contains all attributes of the element and content which describes the content of the element. The content is an collection whose Element is an enum. The enum has two cases: .string and .element. The order in the collection is the order in which the content has been found. So if an element first contains some text, then contains a child element and finally again some text, content will contain a .string whose associated StringPart is the first text. Next there would be a .element whose associated XMLElement would be the child element. Finally, there would be another .string with the last text.

While you can create an XMLElement with a content of [.string("abc"), .string("def"), .element(XMLElement(name: "test"))], and it would also lead to valid XML, it could be cleaned up to [.string("abcdef"), .element(XMLElement(name: "test"))]. To achieve that, it's recommended to use the various append functions on XMLElement.content or even XMLElement directly when you can't assure that the content is cleaned upon creation. If your element was created with an empty content ([]), and you append each of the content elements above, the append functions make sure that they append the "def" string to the first "abc" string instead of adding another .string to the content. If for some reason you still end up with a situation where your content has consecutive .string elements, there's a convenience function compress() (or it's non-mutating sibling compressed()), which merges these .string elements into one.

An XMLElement can be compared to another element and is considered equal if all three properties (name, attributes and content) are equal. This means that for a big tree, all children of the root element will be compared. So be careful when comparing big trees and fall back to manually comparing name and/or attributes if necessary. XMLElement also conforms to Identifiable and uses the name as id.

Both, serializing and parsing XMLs with XMLWrangler relies on XMLElement.

Parsing existing XMLs can be done using the static functions on XMLElement. You can parse either a given Data object or a String containing the XML. If parsing succeeds, the parsed root object is returned. Otherwise whatever error happend along the way is thrown. Errors thrown are the ones created by Foundation.XMLParser.

do {
    let xml = """
              <?xml version='1.0' encoding='UTF-8'?>
              <root myattr='myvalue'>
                  <child1/>
                  <child2>some text</child2>
              </root>
              """
    let root = try XMLElement.parse(xml)
} catch {
    print("Something went wrong while parsing: \(error)")
}

In this example, root.name.rawValue would of course be "root". root.content would contain two .elements. The first would have a associated XMLElement with a name of "child1" and an empty content. The name of XMLElement of the second .element would be "child2" and its content would contain one .string having "some text" associated. root.attributes would contain the value "myvalue" for the key "myattr".

Since you can parse XMLs, you can also convert an XMLElement to a String. For this, there are two functions on XMLElement. The first one just converts an XMLElement into a String. This happens by creating an opening and ending tag (where the beginning tag contains the attributes if available) and putting the content of the element in between. Also, content is compressed (using the aforementioned compress function) before being serialized.

var root = XMLElement(name: "root", attributes: ["myattr": "myvalue"])
root.content.append(element: "child1")
root.content.append(element: XMLElement(name: "child2", content: "some text"))

let xmlString = xml.serialize() // -> "<root myattr=\"myvalue\"><child1/><child2>some text</child2></root>"

If the traditional XML header should also be added, there's a second function which takes a version and a document encoding as additional parameters, but otherwise follows the same rules:

var root = XMLElement(name: "root", attributes: ["myattr": "myvalue"])
root.content.append(element: "child1")
root.content.append(element: XMLElement(name: "child2", content: "some text"))

let xmlDocumentString = root.serializeAsDocument(at: DocumentVersion(major: 1), using: .utf8)
// -> "<?xml version=\"1.0\" encoding=\"UTF-8\"?><root myattr=\"myvalue\"><child1/><child2>some text</child2></root>"

Please note that XMLWrangler does not escape the string based on the given encoding. It simply uses it the generate the document header.

Both functions can take an additional parameter options which contains a set of options to control the serialization behaviour. Currently the following options are possible:

• .pretty: Use pretty formatting. This adds newlines around the tags to make the resulting XML more readable. This is usually not needed for processing XML.
• .singleQuoteAttributes: When this option is present, then attributes of elements will be enclosed in single quotes (') instead of double quotes (").
• .explicitClosingTag: This option forces empty elements to be serialized with an explicit closing tag instead of using the shorthand /> syntax.

XMLWrangler will always extract all content and attributes as String internally. This is because XML itself does not differentiate between types like e.g. JSON does. However, there are many helper functions to safely look up and convert content and attributes of an XMLElement:

• First, there are helpers to extract all child elements with a given name: XMLElement.elements(named:)
• Next, there are helpers to extract an element at a given path: XMLElement.element(at:)
• Another helper allows to extract attributes of an element: XMLElement.attribute(for:).
• It is then also possible to convert those attributes (for some types like e.g. RawRepresentable you don't need to pass a converter): XMLElement.convertedAttribute(for:converter:)
• Last but not least you can extract the string content of an Element: XMLElement.stringContent()
• And of course as you can with attributes, you can also convert string content: XMLElement.convertedStringContent(converter:)

All these methods throw an error (XMLElement.LookupError) when something went wrong instead of returning optionals. If you prefern an optional, you can always use try?. For more information also check the header docs which describe these methods a little closer.

While not yet integrated, the following features might provide added value and could make it into XMLWrangler in the future:

• Indention support for serializing and parsing.
• Extracting "KeyPaths": It could be useful to directly extract a path. It would not be necessary to extract every single element then.

The API is documented using header doc. If you prefer to view the documentation as a webpage, there is an online version available for you.

If you find a bug / like to see a new feature in XMLWrangler there are a few ways of helping out:

• If you can fix the bug / implement the feature yourself please do and open a PR.
• If you know how to code (which you probably do), please add a (failing) test and open a PR. We'll try to get your test green ASAP.
• If you can do neither, then open an issue. While this might be the easiest way, it will likely take the longest for the bug to be fixed / feature to be implemented.

See LICENSE file.

Copyright © 2016-2023 ser.soft GmbH.