I have below json data and decode it to display on the screen. When I check the type of the value, it shows array instead of object. How to get actual type of value in PHP.
JSON is
{ "allData" : { "image" : [], "contents": {.., "box": {}, "text":[]} } }
When I decode and parse the above JSON data the "allData", "contents", "box" type are shows as array instead of object. How can I get those type as object and "image" type as array. Please help.
Thanks, Guru
This normally occurs when you are using the true option in the json_decode function.
For eg.,
$str = '{"allData":{"image":["img1.png"],"contents":{"title":"title name","box":{"name":["sample text 1","sample text2"]},"text":[]}}}';
var_dump(json_decode($str, true));
Just try to remove the true in the json_decode function and you should get an object.
Hope this helps.
foreach on arrays not on objects. if you want to use foreach loop then you'll have to convert your JSON to array.
foreach like that).If you use json_decode with the true option, it will return the result as an array.
Maybe you can check this link.
If you get "Cannot use object of type stdClass as array" the error, you can look at this answer.
<?php
$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
var_dump(json_decode($json));
var_dump(json_decode($json, true));
?>Extract from the RFC 7159 (JSON) : These are the six structural characters:
begin-array = ws %x5B ws ; [ left square bracket
begin-object = ws %x7B ws ; { left curly bracket
end-array = ws %x5D ws ; ] right square bracket
end-object = ws %x7D ws ; } right curly bracket
..
However: php allows you to treat the result as an array (of arrays)
so:
json_decode($json, true); // return as array
returns the result as an array.
and
json_decode($json)
gives you the result as Objects AND arrays . So given your example:
"allData" : { "image" : [], ..
returns a stdClass-Object with the field "image" of type array. The array is empty for your example.
So to retrieve all images, use something like:
$result=json_decode($json);
foreach($result->allData->image as $img) {
echo "found image: $img.";
}
