How to get name if I put BTC ?
$url = file_get_contents(https://api.coinmarketcap.com/v1/ticker/);
$json = json_decode($url,true);
and result
[
{
"id": "bitcoin",
"name": "Bitcoin",
"symbol": "BTC",
},
{
},
{
"id": "ethereum",
"name": "Ethereum",
"symbol": "ETH",
},
{
"id": "bitcoin-cash",
"name": "Bitcoin Cash",
"symbol": "BCH",
},
]
eg.
when I put "BTC"
and need output = "Bitcoin"
or put "ETH"
output = "Ethereum"
You can find proper record using:
$key = array_search('BTC', array_column($json, 'symbol'));
And then (if $key found) get the record like this:
$record = $json[$key];
var_dump($record);
Probalby u need to use full coins name (lowercase) as described in official docs https://coinmarketcap.com/api/ look into paragraph Ticker (Specific Currency)
Related
i wanna convert a json file has text like this design json_text https://phponlines.com/result/e2j9RgRnm7
$json_text = '
[
{"id": "1", "name": "john", "bd": []},
{"id": "2", "name": "gary", "bd": [1, 2]}
]';
$json_decoded = json_decode($json_text, true);
var_dump(json_encode($json_decoded));
then after i edit it to get same design again but i get different deisng instead in one line
[{"id":"1","name":"john","bd":[]},{"id":"2","name":"gary","bd":[1,2]}]
but what i want is
[
{"id": "1", "name": "john", "bd": []},
{"id": "2", "name": "gary", "bd": [1, 2]}
]
As stated in my comments above, this is not particularly advisable, and writing any portion of a JSON file by hand is even more inadvisable.
But if you do not wish to see the light, then at least the use the minimal amount of darkness.
// polyfill for PHP<8.1
if (!function_exists('array_is_list')) {
function array_is_list(array $a) {
return $a === [] || (array_keys($a) === range(0, count($a) - 1));
}
}
function do_weird_json(array $input) {
if( ! array_is_list($input) ) {
throw new Exception('Input data must be a list');
}
return sprintf("[\n%s\n]\n", implode(",\n", array_map('json_encode', $input)));
}
$json_text = '
[
{"id": "1", "name": "john", "bd": []},
{"id": "2", "name": "gary", "bd": [1, 2]}
]';
$json_decoded = json_decode($json_text, true);
$json_decoded[1]['name'] = 'bill';
var_dump(do_weird_json($json_decoded));
Output:
string(74) "[
{"id":"1","name":"john","bd":[]},
{"id":"2","name":"bill","bd":[1,2]}
]
"
Again, the best thing to do would be to leave the JSON as the default, not-human-friendly format and use a client-side tool like jq, etc to reformat the JSON for you for viewing purposes.
I'm trying to get JSON by name and save it as a variable with everything related to that nest,
what would the best way to accomplish this?
(in my scenario the names are unique)
[{
"name": "Joe",
"age": "30",
"gender": "male"
},
{
"name": "Logan",
"age": "27",
"gender": "male"
}]
to something like this
{
"Joe": [{
"name": "Joe",
"age": "30",
"gender": "male"
}],
"Logan": [{
"name": "Logan",
"age": "27",
"gender": "male"
}]
}
i need to be able to search on the name to get the correct one, the API switches order so can't get it from just id
Hi This might helps you
$list = '[
{
"name":"Joe",
"age":"30",
"gender":"male"
},
{
"name":"Logan",
"age":"27",
"gender":"male"
}]';
$newjson = json_decode($list, true);
$final = [];
foreach ($newjson as $key => $value) {
$final[$value['name']][]=$value;
}
$finaloutput = json_encode($final, true);
echo "<pre>";
print_r($finaloutput);
echo "</pre>";
exit;
The output is
{"Joe":
[{"name":"Joe","age":"30","gender":"male"}],
"Logan":
[{"name":"Logan","age":"27","gender":"male"}]}
The Laravel Collections Library is a very useful lib to use cases like yours.
For this problem you could use the keyBy method!
$json = '[
{
"name":"Joe",
"age":"30",
"gender":"male"
},
{
"name":"Logan",
"age":"27",
"gender":"male"
}
]';
$array = collect(json_decode($json, true))->keyBy('name')->all();
print_r($array); // will be the array with the keys defined by the name!
See https://laravel.com/docs/5.5/collections#method-keyby
I have an URL that returns the JSON object below:
{
"addressList": {
"addresses": [
{
"id": 0000000,
"receiverName": "Name Example",
"country": {
"id": "BRA",
"name": "Brasil"
},
"state": {
"id": "SP"
},
"city": "São Paulo",
"zipcode": "00000000",
"type": "Residential",
"street": "000, St Example",
"number": 00,
"neighborhood": "Example",
"hash": "1bf09357",
"defaultAddress": false,
"notReceiver": false
}
]
}
}
I want to get the state value, how can I retrieve that with PHP?
I tried, something like this, but I couldn't get the state value, that should be SP in this case.
$string = '{ "addressList": { "addresses": [ { "id": xxxxxx, "receiverName": "XXXXX XXXXX", "country": { "id": "BRA", "name": "Brasil" }, "state": { "id": "SP" }, "city": "São Paulo", "zipcode": "03164xxx", "type": "Residential", "street": "Rua xxx", "number": xx, "neighborhood": "xxxxx", "hash": "xxxxx", "defaultAddress": false, "notReceiver": false } ] } }';
$json_o = json_decode($string);
$estado = $json_o->state;
How can I achieve the result I want?
Your JSON is not valid - you can validate it on jsonlint.com (it's invalid due to incorrectly formatted numeric values - "id" : 000000).
From then on, you can decode the value and access your data:
$json_o = json_decode($string);
$estado = $json_o->addressList->addresses[0]->state->id;
If you don't have access to the code that generates the JSON, you can attempt to run a regex to match, replace & wrap the numerical values with ":
$valid_json = preg_replace("/\b(\d+)\b/", '"${1}"', $str);
Note: The above is just an example - you'll have to figure out a case where a numerical value is already wrapped by ".
Your JSON has a couple of syntax errors:
"id": 0000000
"number": 00
JSON doesn't support leading zeros. If precise formatting is important, use strings:
"number": "00"
"id": "0000000"
Alternatively, use well-formed integers in the JSON (saves space) and convert them to formatted strings in PHP.
Once you've fixed your JSON, you can access the state->id value of the first address as I do below. When you decode JSON from an untrusted source, be prepared to do some error handling:
$json_string ="..."; //your source, to be decoded
$json_o= json_decode($json_string);
if(is_null($json_o)): //there was an error decoding
$errno = json_last_error();
$err_msg = json_last_error_msg();
die("Json decode failed: Error #$errno: $err_msg");
endif;
//we get here if json_decode succeeded. To get "SP", do...
$stateID = $json_o->addressList->addresses[0]->state->id;
Try:
$json_o = json_decode($string, true);
$estado = $json_o['addressList']['addresses'][0]['state']['id'];
I have accessed this json file on my website:
[{
"01:03:2016": "410",
"02:03:2016": "200",
"03:03:2016": "380"
}]
I want to use PHP to change the formatting to something like this, where the dates and counts are all values:
[{
"date": "01:03:2016",
"count": "410"
},
{
"date": "02:03:2016",
"count": "200"
},
{
"date": "03:03:2016",
"count": "380"
}]
PHP
$string = file_get_contents("data.json"); // your current json data
$json_a = json_decode($string, true);
$new_json = array();
foreach ($json_a as $key => $value) {
$new_json[] = array("date"=>$key,'count'=>$value);
}
echo json_encode($new_json); //output
data.json file (your current data)
[{
"01:03:2016": "410",
"02:03:2016": "200",
"03:03:2016": "380"
}]
How can I read several values with the same name from a JSON from Facebook, when I use this code:
$urlamandar = 'https://graph.facebook.com/'.$user."?access_token=".$access_token;
$content = file_get_contents($urlamandar);
$obj = json_decode($content, true);
$obj = json_decode($content);
It displays this:
{
"id": "XXXXXXXX",
"name": "Ricardo Capistran",
"first_name": "Ricardo",
"last_name": "Capistran",
"gender": "male",
"locale": "es_MX",
"username": "richycapy"
}
And since because there is only one ID, one NAME, one USERNAME, etc, Its quite simple place them in vars like so:
$fbuserid = $obj->{'id'};
$fbname = $obj->{'name'};
$fbusername = $obj->{'username'};
$fbemail = $obj->{'email'};
$fbbirthday = $obj->{'birthday'};
But when it comes to larger files like this code:
$urlamandar = 'https://graph.facebook.com/'.$user.'/albums?access_token='.$access_token;
Its displays a way bigger array like so:
{
"data": [
{
"id": "10150732237818223",
"from": {
"name": "Ricardo Capistran",
"id": "XXXXXXX"
},
"name": "EnterateNorte.com Photos",
"link": "https://www.facebook.com/album.php?fbid=10150732237818223&id=743158222&aid=457026",
"privacy": "custom",
"count": 31,
"type": "app",
"created_time": "2012-03-11T02:44:42+0000",
"updated_time": "2014-01-07T03:13:24+0000",
"can_upload": false
},
{
"id": "440168313222",
"from": {
"name": "Ricardo Capistran",
"id": "743158222"
},
"name": "Timeline Photos",
"link": "https://www.facebook.com/album.php?fbid=440168313222&id=743158222&aid=220377",
"cover_photo": "10151730849598223",
"privacy": "everyone",
"count": 175,
"type": "wall",
"created_time": "2010-06-30T22:38:45+0000",
"updated_time": "2014-01-01T02:09:11+0000",
"can_upload": false
},
{
"id": "10150797320378223",
"from": {
"name": "Ricardo Capistran",
"id": "743158222"
},
"name": "Instagram Photos",
"link": "https://www.facebook.com/album.php?fbid=10150797320378223&id=743158222&aid=466555",
"cover_photo": "10151695050098223",
"privacy": "friends",
"count": 37,
"type": "app",
"created_time": "2012-04-09T23:50:08+0000",
"updated_time": "2013-12-29T08:29:15+0000",
"can_upload": false
}
],
"paging": {
"cursors": {
"after": "NDM1NjY5NjI4MjIy",
"before": "MTAxNTA3MzIyMzc4MTgyMjM="
},
"next": "https://graph.facebook.com/*my_id*/albums?access_token=*access_token*&limit=25&after=NDM1NjY5NjI4MjIy"
}
}
And I would need the “name” and “id” off all the albums, so then I can repeat this same procedure with the containing pictures
Obviously it has way more albums, I just cut it after 3 just to explain my self…
Is there a way to place them in vars? With a php “for each” some how
Thanks!
You can do something like
$obj = json_decode($content, true);
$array_album = array();
foreach($obj['data'] as $key=>$val){
$array_album[] = array("id"=>$val["id"],"name"=>$val["name"]);
echo "ID : ".$val["id"]. " NAME : ".$val["name"] ;
echo "<br />";
}
print_r($array_album); // This will have all the id and names
If you dont want to store in array then the names and id will appear in the above loop and you can do whatever you want with those values. Or use the array $array_album and loop through if you want to use it later.