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This seems more like a PHP programming question, but yes as you guessed in PHP if you want the value of a variable you just need $var. echo $var outputs it and will cause a syntax error how you've used it. Otherwise everything else looks ok, although obviously I can't say if that data structure will give you valid JSON-LD ;-) EDIT: Also it's good to give ...


Try following way .................... URL: /wp-json/wp/v2/posts?_embed image: json["_embedded"]["wp:featuredmedia"][0]["source_url"], It's working fine.try

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