Php json decode with slashes6/12/2023 ![]() Instead of using the brackets, use the object operator. You should not need to convert an object to an array to access its properties. Hopefully this will help you understand objects and why we want to keep an object an object. To access the first part of the object: print $response->. We will assume when you print_r the object you might see this: print_r($response) To access the object that has nested objects you could do the following: ![]() Sometimes when working with API you simply want to keep an object an object. P.S : Also, If you set the second parameter of the json_decode() to true, it will automatically convert the object to an array() Their are other options to print_r() as well, like var_dump() and var_export() One, can also use multiple keys to extract the sub elements incase if the object has nested arrays. ![]() you can access the elements of the object with the help of -> Operator. $a = (array)$object Īs mentioned earlier, when you use json_decode() function, it returns an Object of stdClass. Hence, we cannot pass an object inside of print_r(). The arguments, which are to be passed inside of print_r() should either be an array or a string. When we use json_decode(), we get an object of type stdClass as return type. Then in the inner loop, I list out the car’s properties, in this cae the make and colour or the car.Īgain I use the key/value pair, this time to name the prperty on-screen.Print_r - Prints human-readable information about a variable So each item in the array get’s a link to its own page. This was done just as an example how and why you may access the key in foreach. In a hypothetical scenario, I may have pulled data from a database in such a way that the array key matches the database ID of the item. In the outer loop I access the key to add a query string to a URL. But I choose those varable names, they are not built in, they are whatever you want them to be. I call the key $prop (my shorthand for property) as I don’t want to use $key again and cause confusion. In the inner loop I access the nested arrays. So in the outer loop I’m calling the key $key and the value (which is a nested array) $car, because it describes a car. Here you write foreach($car as $prop => $value) So the $car variable is used to grab the car property but then stores the value inside the $key variable? Seems odd. Logically, it would seem that the $car variable would hold the value which in this case would be mazda and lincoln. Last question, in this snippet, you write ($dec as $key => $car) and then inside the foreach you use the key variable instead of the $car variable. Like the reference JSON encoder, jsonencode () will generate JSON that is a simple value (that is, neither an object nor an array) if given a string, int, float or bool as an input value. So running stripslashes json decode whether in the first or second function returns the string as empty when saving to the db.Ĭould you elaborate as to how you can see printing out messages like the one in your post? Why it returns empty in the db after decoding it, and lastly, how would I access the data after decoding it? I tried decoding in the original function before sending the data as an argument to the second function and it returns empty in the database there as well. After your replace try to parse your string like the following example. (It’s the same data, same format, same everything?) This article describes how to remove backslashes from json strings in PHP and the. I find it interesting that I can save the data as a string to the db but when I simply pass the data as is in a string with the backslashes and all as an argument to another function I can’t save it to the db in the second function. I tried running this and again when I try to save $dec to the database it returns empty, also tried encoding after decoding and returns empty in the databae. How are you getting this printed information? With javascript I can use console.log($variable)
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