Parsing error: Object of class stdClass could not be converted to string

Question

I'm using twitter api to retrieve fav tweets, then parse them using a template:

$favs_list = $this->connection->get('favorites/list');
$data_to_parse['fav_list'] = $fav_list;
$html = $this->parser->parse('templates/tweet_list', $data_to_parse, TRUE);
$output['html'] = $html;
$this->load->view('read', $output);

the template tweet_list is simply:

<h3>Favs:</h3>
{fav_list}
<h5>{text}</h5>
{/fav_list}

But I keep getting these two error, and repeated multiple times:

A PHP Error was encountered    
Severity: 4096     
Message: Object of class stdClass could not be converted to string     
Filename: libraries/Parser.php
Line Number: 143

A PHP Error was encountered   
Severity: Notice     
Message: Object of class stdClass to string conversion     
Filename: libraries/Parser.php    
Line Number: 143

The weird thing is that, under these errors, my html is shown in correct format. Any idea on what caused this?

I think $favs_list is standard object array and you much get it converted to simple array and you are done. — TheMohanAhuja, Feb 26 '14 at 05:48
@RakeshSharma $html id the basically my template `tweet_list` after parsing it. See: http://ellislab.com/codeigniter/user-guide/libraries/parser.html — iTurki, Feb 26 '14 at 05:48
I don't know optimal method but I do usually with json_encode(json_decode ($favs_list, true)); — TheMohanAhuja, Feb 26 '14 at 05:52
@TheMohanAhuja I got an error saying: `json_decode() expects parameter 1 to be string, array given`. I'm guessing `$favs_list` is an array after all. — iTurki, Feb 26 '14 at 05:55
sorry my mistake try this please json_decode(json_encode ($favs_list), true); — TheMohanAhuja, Feb 26 '14 at 05:58
i dont know your structure of this stdClass Object, but you also can do thinks like this (array)$stdClassValue ... this will also do this encode / decode job from your json functions ;) — dschibait, Feb 26 '14 at 06:33
@TheMohanAhuja This works perfectly. Although I chose to use [this way](http://stackoverflow.com/a/11577501/543711). Please post an answer so that I can accept it. — iTurki, Feb 26 '14 at 07:51

score 2 · Accepted Answer · answered Feb 26 '14 at 10:12

2

I don't know optimal method but I do usually with

json_decode(json_encode ($favs_list), true);

answered Feb 26 '14 at 10:12

TheMohanAhuja

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just curious, just how do u know this? – Conan Mar 09 '18 at 09:35
@Conan When I get data from mysql, I usually get data in object format, that I skip using this me way only. That's how I know this. – TheMohanAhuja Mar 09 '18 at 12:45

score 0 · Answer 2 · answered Nov 25 '16 at 13:30

You are trying to print out favs_list but the content is an object, and its having a hard time trying to convert that object.

Just convert the object to an array instead and use accordingly

$array = (array) $object;

in your case

$favs_list = $this->connection->get('favorites/list');
$data_to_parse['fav_list'] = (array) $fav_list;

Parsing error: Object of class stdClass could not be converted to string

2 Answers2