I printed this json from a get_file_contents output.
{"status":"INVALID_CREDENTIALS"}
Now I want to echo the content of "status" and I used
echo $status
but not working. Please help.
First, use json_decode to convert to php object:
$json = '{"status":"INVALID_CREDENTIALS"}';
$obj = json_decode($json);
Then:
echo $obj->status;
Related
I am trying to access data from json in PHP but it seems not working.
code:
$raw =file_get_contents("http://api.mydomain.com/data.json");
$data = json_decode($raw->list);
echo $data;
I'm getting error that list is not an object.
Here is my json
{ "list" : [ { "data1":" my data"}, {"data2": "my data 2"}]};
What did u do wrong? Also how can i access data1 and others?
You don't need the $raw->list bit and you are getting an object back from json_decode so use print_r and not echo
$raw = file_get_contents("http://api.mydomain.com/data.json");
$data = json_decode($raw);
print_r($data);
I have a .txt file called 'test.txt' that is a JSON array like this:
[{"email":"chrono#gmail.com","createdate":"2016-03-23","source":"email"}]
I'm trying to use PHP to decode this JSON array so I can send my information over to my e-mail database for capture. I've created a PHP file with this code:
<?php
$url = 'http://www.test.com/sweeps/test.txt';
$content = file_get_contents($url);
$json = json_decode($content,true);
echo $json;
?>
For some reason, it's not echoing the decoded JSON when I visit my php page. Is there a reason for this and can anyone shed some light? Thanks!
You use echo to print scalar variables like
$x = 'Fred';
echo $x;
To print an array or object you use print_r() or var_dump()
$array = [1,2,3,4];
print_r($array);
As json_decode() takes a JSON string and converts it to a PHP array or object use print_r() for example.
Also if the json_decode() fails for any reason there is a function provided to print the error message.
<?php
$url = 'http://www.test.com/sweeps/test.txt';
$content = file_get_contents($url);
$json = json_decode($content,true);
if ( json_last_error() !== JSON_ERROR_NONE ) {
echo json_last_error_msg();
exit;
}
You'll need to split that json string into two separate json strings (judging by the pastebin you've provided). Look for "][", break there, and try with any of the parts you end up with:
$tmp = explode('][', $json_string);
if (!count($tmp)) {
$json = json_decode($json_string);
var_dump($json);
} else {
foreach ($tmp as $json_part) {
$json = json_decode('['.rtrim(ltrim($json_string, '['), ']').']');
var_dump($json);
}
}
I'm having difficulties grabbing any of the JSON information from this URL.
I've tried other JSON snippets and they seem to work so I'm not sure if it's the way that the URL is structured or something.
Basic example below.
<?php
$json = file_get_contents('http://nhs-sh.cfpreview.co.uk/api/version/fetchLatestData?dataType=Clinics&versionNumber=-1&uuID=website&dt=');
$obj = json_decode($json);
echo "Body: " . $obj->Body;
?>
The link provided starts with
{ data :
which is valid javascript but invalid json. You can test it on http://jsonlint.com. To fix this we can replace the data with "data" :
$json = file_get_contents('http://nhs-sh.cfpreview.co.uk/api/version/fetchLatestData?dataType=Clinics&versionNumber=-1&uuID=website&dt=');
$obj = json_decode($json);
if (json_last_error() !== JSON_ERROR_NONE) { //check if there was an error decoding json
$json = '{ "data" :'. substr(trim($json), 8); // replace the first 8-1 characters with { "data" :
$obj = json_decode($json);
}
print_r($obj->data); //show contents of data
Please note that this fix is dependent on the data source e.g. if they change data to dataset. The correct measure would be to ask the developers to fix their json implementation.
Hi Guys I have a JSON data I need to convert this data Treeview a json data url: http://torrent2dl.ml/json.php
recovered state = http://torrent2dl.ml/json.php?tree
I tried to do http://torrent2dl.ml/hedef.php
how to convert this data a php function or code ?
json_decode($jsonObject, true);
Use json_decode() :
<?php
$url = 'http://torrent2dl.ml/json.php';
$JSON = file_get_contents($url);
// echo the JSON (you can echo this to JavaScript to use it there)
echo $JSON;
// You can decode it to process it in PHP
$data = json_decode($JSON);
var_dump($data);
?>
Source : https://stackoverflow.com/a/8344667/4652564
<?php
$handle = fopen("https://graph.facebook.com/search?q=sinanoezcan#hotmail.com&type=user&access_token=2227472222|2.mLWDqcUsekDYK_FQQXYnHw__.3600.1279803900-100001310000000|YxS1eGhjx2rpNYzzzzzzzLrfb5hMc.", "rb");
$json = stream_get_contents($handle);
fclose($handle);
echo $json;
$obj = json_decode($json);
print $obj->{'id'};
?>
Here is the JSON: {"data":[{"name":"Sinan \u00d6zcan","id":"610914868"}]}
It echos the JSON but I was unable to print the id.
Also I tried:
<?php
$obj = json_decode($json);
$obj = $obj->{'data'};
print $obj->{'id'};
?>
Note that there is an array in the JSON.
{
"data": [ // <--
{
"name": "Sinan \u00d6zcan",
"id": "610914868"
}
] // <--
}
You could try $obj = $obj->{'data'}[0] to get the first element in that array.
data is an array, so it should be:
print $obj[0]->{'id'};
It looks like the key "data" is an array of objects, so this should work:
$obj = json_decode($json);
echo $obj->data[0]->name;
Have you tried $obj->data or $obj->id?
Update: Others have noted that it should be $obj->data[0]->id and so on.
PS You may not want to include your private Facebook access tokens on a public website like SO...
It's a bit more complicated than that, when you get an associative array out of it:
$json = json_decode('{"data":[{"name":"Sinan \u00d6zcan","id":"610914868"}]}', true);
Then you may echo the id with:
var_dump($json['data'][0]['id']);
Without assoc, it has to be:
var_dump($json->data[0]->id);