Object of class stdClass could not be converted to string - laravel - php

I am following this documentation
to implement export to Excel in my laravel 4 project.
So am trying to generate excel file from array like this:
//$results is taken with db query
$data = array();
foreach ($results as $result) {
$result->filed1 = 'some modification';
$result->filed2 = 'some modification2';
$data[] = $result;
}
Excel::create('Filename', function($excel) use($data) {
$excel->sheet('Sheetname', function($sheet) use($data) {
$sheet->fromArray($data);
});
})->export('xls');
But this raises exception:
Object of class stdClass could not be converted to string
What am I doing wrong ?
UPDATE:
Tried this:
$data = get_object_vars($data);
which results in:
get_object_vars() expects parameter 1 to be object, array given
This:
$data = (array)$data;
Results in the initial error.

Try this simple in one line of code:-
$data= json_decode( json_encode($data), true);
Hope it helps :)

$data is indeed an array, but it's made up of objects.
Convert its content to array before creating it:
$data = array();
foreach ($results as $result) {
$result->filed1 = 'some modification';
$result->filed2 = 'some modification2';
$data[] = (array)$result;
#or first convert it and then change its properties using
#an array syntax, it's up to you
}
Excel::create(....

You might need to change your object to an array first. I dont know what export does, but I assume its expecting an array.
You can either use
get_object_vars()
Or if its a simple object, you can just typecast it.
$arr = (array) $Object;

If you have a Collection of stdClass objects, you could try with this:
$data = $data->map(function ($item){
return get_object_vars($item);
});

I was recieving the same error when I was tring to call an object element by using another objects return value like;
$this->array1 = a json table which returns country codes of the ip
$this->array2 = a json table which returns languages of the country codes
$this->array2->$this->array1->country;// Error line
The above code was throwing the error and I tried many ways to fix it like; calling this part $this->array1->country in another function as return value, (string), taking it into quotations etc. I couldn't even find the solution on the web then i realised that the solution was very simple. All you have to do it wrap it with curly brackets and that allows you to target an object with another object's element value. like;
$this->array1 = a json table which returns country codes of the ip
$this->array2 = a json table which returns languages of the country codes
$this->array2->{$this->array1->country};
If anyone facing the same and couldn't find the answer, I hope this can help because i spend a night for this simple solution =)

This is easy all you need to do is something like this Grab your contents like this
$result->get(filed1) = 'some modification';
$result->get(filed2) = 'some modification2';

Related

Filter API response with PHP

Not sure what issue was with first pastebin code, here is another attempt.
I am connecting to Vimeo Live API, in doing so the response is huge > 500kb in total - I have an example with only one object here -> there are over 20 it returns. I have a better idea of what Im doing in JS than PHP, but returning the huge array or json to the browser doesn't seem like a good idea and its use of repeated ajax calls isn't good either. So this question is two fold, both in theory and practice. Is this a good design and how do I filter the result in PHP and only send what I need back to browser.
Here is the design, or at least what I think is best:
page loads, sends ajax request to PHP script
PHP script connects to API and gets response in an array (example of one object)
Search through the array for 'metadata->connections->live_video' for one that has an associated array containing [status] => streaming'
If one (there will only be one at a time) is found, return that whole object and that object only, not the entire array.
At this time I do not have a complete understanding of how this data should be returned or formatted for ease of sifting through. Ive tried using json_encode on the array, which gets nicely formatted JSON but I can't iterate through it and can only get single objects like data[0]->metadata->connections->live_video. Ive tried json_encode, then json_decode and Im back to a similar array structure of what is originally sent.
However, in the browser I am able to return the whole array and in the success function of the ajax call sift through it via JS like so:
let live_stream = json.data.filter(function(value, key) {
let connection = value.metadata.connections['live_video'];
return connection && connection.status === 'streaming';
});
I know this isn't the right way, I know I need to sift through the array, find the object / key Im looking for and only return that. Any advice is appreciated, once I get this figured out, I can apply it in a range of ways for this project.
The closest I can get in PHP is:
function live_event() {
global xxx;
$lib = xxx;
$response = xxx;
$body = $response['body'];
header('Content-Type: application/json');
$jsonstr = json_encode($body);
$json = json_decode($jsonstr);
foreach ($json->data as $item) {
if ($item->uri == "/live_events/2354796") {
echo"one";
}
}
}
and this:
function live_event() {
$global xxx;
$lib = xxx;
$response = xxx;
$body = $response['body'];
header('Content-Type: application/json');
$jsonstr = json_encode($body);
$json = json_decode($jsonstr,true);
$results = array_filter($json['data'], function($item) {
return $item['metadata']['connections']['live_video']['status'] == "streaming";
});
var_dump($results);
}
last one gets me this error "Warning: Trying to access array offset on value of type null in /var/www/vhosts/mysite.com/httpdocs/SSI/Vimeo.php on line 31" is there something similar to optional chaining in PHP? if its null I don't want it to log an error.
This at least output "one" as there is only one object with [uri]=>"/live_events/2354796". I can't get it to return that entire object or search one more nested array deeper.
There are a few things that come to mind.
OBJECTS:
If you're using PHP 8 you could use the safe access operator
function live_event() {
//...
$results = []
foreach ($json->data as $item) {
if($item?->metadata?->connections?->live_video?->status == 'streaming'){
$results[] = $item;
}
}
return $results;
}
ARRAYS:
For lower versions of PHP it's better to work with arrays when dealing with keys that may or may not exists.
Generally speaking whatever fetch library you're working with will allow you set the output to either array or object. If Object is your only option then yes doing the old json_decode(json_encode($data), true) is the way to go.
checking first isset(). Assuming that's it's desired to filter out results without that key.
function live_event() {
//...
$results = array_filter($json['data'], function($item) {
if(!isset($item['metadata']['connections']['live_video']['status']))
return false;
return $item['metadata']['connections']['live_video']['status'] == "streaming";
});
header('Content-Type: application/json');
echo json_encode([
"status" => true,
"data" => $results
]);
}
or you can always just use the age old trick of suppressing the error with the # symbol. TBH i'm not sure where the # symbol would go to suppress the error, perhaps in front of the filter function definition.
function live_event() {
//...
$results = array_filter($json['data'], function($item) {
return #$item['metadata']['connections']['live_video']['status'] == "streaming";
});
return $results;
}

How to convert an object properties into an array in php

Those data in $data variable. When I'm using dd($data); I got this:
CurlHandle {#1918 ▼
+url: "https://example.com"
+content_type: "application/json; charset=UTF-8"
+http_code: 200
+header_size: 887
+namelookup_time_us: 139522
+pretransfer_time_us: 326662
+redirect_time_us: 0
+starttransfer_time_us: 668686
+total_time_us: 668752
}
I want to convert this data to an array.
I'm using this: $arr = json_decode($data,true);
But, this is not working. Now, how can I convert this?
You can use below solution
$yourArray = json_decode(json_encode($yourObject), true);
and this convert object to an array for more info
Objects are, in PHP, maybe iterable. Notice, you may iterate through an object's public fields only via the foreach loop. So the following works:
$array = [];
foreach ($object as $property) {
$array[] = $property; // Stores public field only
}
var_dump($array);
Simply to get an array of object properties, you may use the get_object_vars() function.
var_dump(get_object_vars($object));
You may cast an object to be an array as #MoussabKbeisy said. And this would be the easiest way:
$array = (array) $object;
Here is another way is using ArrayIterator:
$iterator = new ArrayIterator($object);
var_dump(iterator_to_array($iterator));
while this is a PHP Object so you can deal with it by 2 ways:
1- if you want to get on of it's parameters, you can simply use
$yourObject->parameter;
2- if you need to use convert and use it as array, then there is different ways to convert an object to an array in PHP
//1
$array = (array) $yourObject;
//2
$array = json_decode(json_encode($yourObject), true);
Also see this in-depth blog post:
Fast PHP Object to Array conversion

Finding titles in JSON

I currently have this large JSON file: hastebin
But just want the titles of the posts.
I've tried this...
$json = $page;
$o = json_decode($json, true);
echo($json);
$titles = $o["*"]["*"]["*"]["*"]["title"];
var_dump($titles);
But it isn't working - it's returning NULL! Sometimes it just doesn't return anything.
If anyone is wondering, yes this is from Reddit.
This should do it:
$titles = array_map(function($post) {
return $post['data']['title'];
}, $o['data']['children']);
I'm not sure what you expected using "x" indices, but you should probably read about arrays.
PHP can't use wildcards like * in array keys. Whatever string you use to reference the key, it's going to try to find a key with that exact string. So what you tried can't work because there aren't any * keys.
You can get it by iterating all the levels, or iterating the outer level and referring to the proper nested key. But if you're just looking for all instances of 'title' a recursive method may be an easier way to get them.
array_walk_recursive($o, function($value, $key) use (&$titles) {
if ($key == 'title') $result[] = $value;
});
var_dump($titles);
This will get any value of 'title' regardless of its depth in the array, so if that's not what you want, then you'll need to iterate it and specifically reference the proper ones.
It's very hard to deal directly with such a long JSON document. The returned result from the page is not a valid JSON. It contains some HTML tags, but if you take the posts data and insert it in a file you can do the following according to the structure of your JSON (You can find your JSON in an external link here):
<?php
header("Content-Type:application/json");
$posts=file_get_contents('json.php');
//decode your JSON STRING
$posts=json_decode($posts,true);
//create a title variable to store your titles
$titles=array();
foreach($posts['data']['children'] as $child)
{
array_push($titles,$child['data']['title']);
}
echo json_encode($titles);
?>
You can even use this approach using a URL but ensure that it will return a valid JSON with no html

Access JSON values in PHP

{"coord":{"lon":73.69,"lat":17.8},"sys":{"message":0.109,"country":"IN","sunrise":1393032482,"sunset":1393074559},"weather":[{"id":800,"main":"Clear","description":"Sky is Clear","icon":"01n"}],"base":"cmc stations","main":{"temp":293.999,"temp_min":293.999,"temp_max":293.999,"pressure":962.38,"sea_level":1025.86,"grnd_level":962.38,"humidity":78},"wind":{"speed":1.15,"deg":275.503},"clouds":{"all":0},"dt":1393077388,"id":1264491,"name":"Mahabaleshwar","cod":200}
I am trying to fetch description from the weather from the json above but getting errors in php. I have tried the below php code:
$jsonDecode = json_decode($contents, true);
$result=array();
foreach($jsonDecode as $data)
{
foreach($data{'weather'} as $data2)
{
echo $data2{'description'};
}
}
Any help is appreciated. I am new in using json.
You have to use square brackets ([]) for accessing array elements, not curly ones ({}).
Thus, your code should be changed to reflect these changes:
foreach($data['weather'] as $data2)
{
echo $data2['description'];
}
Also, your outer foreach loop will cause your code to do something completely different than you intend, you should just do this:
foreach($jsonDecode['weather'] as $data2)
{
echo $data2['description'];
}
Your $jsonDecode seems to be an array, so this should work-
foreach($jsonDecode['weather'] as $data)
{
echo $data['description'];
}
You can access data directly with scopes
$json = '{"coord":{"lon":73.69,"lat":17.8},"sys":{"message":0.109,"country":"IN","sunrise":1393032482,"sunset":1393074559},"weather":[{"id":800,"main":"Clear","description":"Sky is Clear","icon":"01n"}],"base":"cmc stations","main":{"temp":293.999,"temp_min":293.999,"temp_max":293.999,"pressure":962.38,"sea_level":1025.86,"grnd_level":962.38,"humidity":78},"wind":{"speed":1.15,"deg":275.503},"clouds":{"all":0},"dt":1393077388,"id":1264491,"name":"Mahabaleshwar","cod":200}';
$jsonDecode = json_decode($json, true);
echo $jsonDecode['weather'][0]['description'];
//output Sky is Clear
As you can see wheater` is surrounded with scopes so that means it is another array. You can loop throw that array if you have more than one result
foreach($jsonDecode['weather'] as $weather)
{
echo $weather['description'];
}
Live demo
If the result of decode is an array, use:
$data['weather']
If the result is an object, use:
$data->weather
you have to access "weather" with "[]" operator
like this,
$data["weather"]
There is several things worth answering in your question:
Q: What's the difference between json_decode($data) and json_decode($data, true)?
A: The former converts JSON object to a PHP object, the latter creates an associative array: http://uk1.php.net/json_decode
In either case, there is no point on iterating over the result. You probably want to access just the 'weather' field:
$o = json_decode($data) => use $weather = $o->weather
$a = json_decode($data, true) => use $weather = $a['weather']
Once you have the 'weather' field, look carefully what it is:
"weather":[{"id":800,"main":"Clear","description":"Sky is Clear","icon":"01n"}]
It's an array, containing a single object. That means you will either need to iterate over it, or use $clearSky = $weather[0]. In this case, it does not matter which approach of json_decode did you choose => JSON array is always decoded to a PHP (numeric indexed) array.
But, once you get $clearSky, you are accessing the object and it again matters, which approach you chose - use arrow or brackets, similarly to the first step.
So, the correct way to get for exaple the weather description would be either of these:
json_decode($data)->weather[0]->description
json_decode($data, true)['weather'][0]['description']
Note: In the latter case, dereferencing result of the function call is supported only in PHP 5.4 or newer. In PHP 5.3 or older, you have to create a variable.
Note: I also encourage you to always check if the expected fields are actually set in the result, using isset. Otherwise you will try to access undefined field, which raises an error.

php traverse object using variables for property names

I have a case where I may recive one of two json objects, in this case from either the google geocode api, or the places api.
Acccessing values from the geocoding api it will look like this:
$coordinates = $data->results[0]->geometry->location;
$cache_value['lat'] = (string) $coordinates->lat;
$cache_value['lng'] = (string) $coordinates->lng;
and an almost identical structure for places.
$coordinates = $data->result->geometry->location;
$cache_value['lat'] = (string) $coordinates->lat;
$cache_value['lng'] = (string) $coordinates->lng;
In my code I've two functions to handle each case, but they are almost idential with the exception of the result vs results[0] and I'd like to combine them. I've tried to passing a varriable but it throws errors:
$result = ($place) ? 'result' : 'results[0]';
$coordinates = $data->$result->geometry->location;
Gives the following:
Notice: Undefined property: stdClass::$result[0]
I'd like to know the correct syntax to achive what im after, and any pointers on nominclature, as Im afraid this question title is a bit inpercise.
Just do:
$result = $place ? $data->result : $data->results[0];
$coordinates = $result->geometry->location;
What your code is doing, is this: It tries to resolve a property of $data object with a name of results[0], and there is none; Once again - it does not resolve 0 index of the results property, but it tries to find a property with a literal name results[0]; It would work if your object looked like this:
$obj = (object)array( 'results[0]' => 'hey there' );
If for any reason you would like to play with that, you could create a silly property like this: $data->{'results[0]'} = 5; - but it's stupid, don't do that :)
i believe php is looking for a key named results[0], its not smart enough to know that the properties name is results and you want the first member of the collection [0]
The issue is the reference of the variable name, rather than its value.
$result = ($place) ? 'result' : 'results[0]';
$coordinates = $data->$result->geometry->location;
$result is just a string and should be the actual value of either $data->result or $data->result[0].
To correct it, simply use $result to hold the value of result.
$result = ($place) ? $data->result : $data->results[0];
$coordinates = $result->geometry->location;

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