Json API with Php

Question

I'm new to JSON API and i was just testing it out to understand how this works so this is an javascript example of json

$(document).ready(function() {
    var url = 'https://api.themoviedb.org/3/movie/',
        mode = 'search/',
        input,
        movieName,
        key = '?api_key=e9dfeccf734a61b9a52d9d7660f0d0a1';

    $('button').click(function() {
        var input = $('#movie').val(),
            movieName = encodeURI(input);
        $.ajax({
            url: url + mode + movieName + key, 
            dataType: 'jsonp',
            success: function(data) {
             console.log(data);
            }
        });
    });
});

And HTML

<input id="movie" type="text" /><button>Search</button>

I'm trying to use that API using PHP but failed..

<?php
$json = file_get_contents("url");
$obj = json_decode($json);
$response = array();
$json_response = json_encode($response);
echo $json_response;
header("Content-type: application/json");
?>

Anyone can help me with using search query as a name of movie to fetch out the details..

Example: https://api.themoviedb.org/3/movie/297762?api_key=e9dfeccf734a61b9a52d9d7660f0d0a1&language=en-US

I made this work but i need to style it..

<?php
  //$url = file_get_contents('https://api.themoviedb.org/3/movie/297762?api_key=e9dfeccf734a61b9a52d9d7660f0d0a1&language=en-US');
  //echo $url;
  $json_url = "https://api.themoviedb.org/3/movie/297762?api_key=e9dfeccf734a61b9a52d9d7660f0d0a1&language=en-US";
  $json = file_get_contents($json_url);
  $data = json_decode($json, TRUE);
  echo "<pre>";
  print_r($data);
  echo "</pre>";
?>

Benjaco · Accepted Answer · 2017-06-30 01:21:38Z

1

The api is a little different when you search for a movie compared to when you grab information for a movie.

Here is the url for search for a movie http://api.themoviedb.org/3/search/movie?api_key=e9dfeccf734a61b9a52d9d7660f0d0a1&query=Jumanji:%20Welcome%20to%20the%20Jungle

With that in mind, you can change the javascript so it matches the url, no need for php for this api.

$(document).ready(function() {
    var url = 'https://api.themoviedb.org/3/',
        mode = 'search/movie/',
        key = 'e9dfeccf734a61b9a52d9d7660f0d0a1';

    $('button').click(function() {
        var input = $('#movie').val();

        $.ajax({
            url: url + mode, 
            type: "get",
            data: {
               "api_key": key,
               "query": input
            },
            dataType: 'jsonp',
            success: function(data) {
                console.log(data);
            }
        });
    });
});

answered Jun 30, 2017 at 1:21

Benjaco

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3 Comments

Luke Joshua Park Over a year ago

@NoorQureshi If you aren't capable of changing the console.log call to something that modifies the DOM then you are in over your head and need to go back to basics.

user5896280 Over a year ago

@LukePark I'm sorry bro but im not a web developer im just completing this code to update my small application.. I can't go to basics right now.. :/

Benjaco Over a year ago

@NoorQureshi I answered exactly what you asked about, so how about you "fix" your question instead. Or even better, post a new one, because it has nothing to do with the API. Link the new question to me, and I will answer it. Show the json structure, and how you want the html structure to be, I can't read your brain or guess it. Do you use a template engine, or do you want a plain jQuery solution? Provide all the necessary information.

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