how do I assign a field of a json_encoded result to a variable. I have the following:
$jsonres = json_encode($result); //where result is an array holding fields including name (string), properties (object type or array)
I tried the following:
echo $jsonres['properties']; // failed with "Illegal string offset 'properties'"
var_dump ($jsonres->properties); //"Notice: Trying to get property of non-object in..."
I need to be able to use the value of 'properties' in my form.
Thanks
Simply encode the properties property, not the entire object:
$jsonres = json_encode($result['properties']);
echo $jsonres;
Related
I am storing data in json format in mysql database table column like
table A
column-user_details
"{\"name\":\"sadasfsf\",\"phone\":\"7896521747\",\"address_1\":\"dvgsdsd\",\"state_name\":\"g\",\"city_name\":\"sdgds\",\"zip_code\":\"ghdfh\"}"
I am fetching data like
json_decode($variable, true);
<?php echo $variable['name'];?>
However I am getting error like
Illegal string offset 'name'
the first parameter of json_decode is not a parameter passing by reference, so if you want to get the json code as php array, you mast adding the result of json_decode into a variable.
<?php
$variable = "{\"name\":\"sadasfsf\",\"phone\":\"7896521747\",\"address_1\":\"dvgsdsd\",\"state_name\":\"g\",\"city_name\":\"sdgds\",\"zip_code\":\"ghdfh\"}";
$array = json_decode($variable, true);
echo $array['name'];
NB: json_decode() is a php code, so you must use it after <?php tag, not outside of it
Hi am having a little from with json array. when i try to get the values out of json array.
Controller
$data['payment'] = $this->admin->get_payment_settings();
$value = $data['payment'][0]->json;
echo $value['username'];
Hi am having a json array in database. using my controller i am getting the filed json. When i do a var_dump($value) array look like this
{"username":"foodi1.lch.co","password":"FBUEWQH6X4D","signature":"AFcWxV21C7fd0v3bZnWKgr9on9AmTuhyd4MVq","currency":"CAD"}
i want to get each value out of this array.
echo $value['username'];
echo $value['password'];
When i try to do this i get the error
A PHP Error was encountered
Severity: Warning
Message: Illegal string offset 'username'
Filename: controllers/administrator.php
Line Number: 620
Can some one help me to get the values out of json array. tnx..
You need to decode the string to a valid array, use json_decode as follows:
$json = json_decode($value, true);
echo $json['username'];
It seems your $data['payment'][0] is json string.So you need to decode it.
Try this way
$value=json_decode($data['payment'][0]->json,true);//second parameter true will return as array.
echo $value['username'];
Please how do I echo or obtain the values of a field of type object?
I first encoded, replaced string and decoded the code as below:
$mongorow = json_encode($mongorow);
$mongorow= preg_replace("/_DOT_/", ".", $mongorow);
$mongorow = json_decode($mongorow);
And then in my form I'm trying to use the fields as below:
value="<?php echo $mongorow->name;?>" //this works producing value of name
value="<?php echo $mongorow->properties;?>" //this produces an error where properties is of type object
Error message I get is (symfony)
Catchable Fatal Error: Object of class stdClass could not be converted to string in ...
First try to debug the type of the $mongorow->properties you can debug the object by using the var_dumb function.
Try this:
var_dump($mongorow);
And you will get the exact data.
For more read here about the var_dump.
You can use var_dump() or print_r() function to display the object or hashtype field
echo "<pre>"; var_dump($mongorow);
print_r($mongorow);
If you want to assign object to the variable as a string use:
serialize($obj);
If you want to just dump value on screen use:
var_dump($obj);`
I know there already are questions like this, but It didn't help me.
I get the follow error on my site:
Warning: Illegal string offset 'networkConnections' in
/var/www/bitmsg/templates/header.php on line 25 {
The line is
<?= $bmstatus["networkConnections"] ?> p2p nodes
if I print_r $bmstatus, then I get:
{
"numberOfBroadcastsProcessed": 2308,
"networkStatus": "connectedAndReceivingIncomingConnections",
"softwareName": "PyBitmessage",
"softwareVersion": "0.4.1",
"networkConnections": 52,
"numberOfMessagesProcessed": 22888,
"numberOfPubkeysProcessed": 8115
}
How to I fetch the information from this array?
I've tried both $bmstatus['networkConnections'] and $bmstatus->networkConnections
but both is returning that error?
$bmstatus contains a JSON string. You have to decode it first to be able to extract the required information out of it. For this purpose, you can use the built-in function json_decode() (with the second parameter set as TRUE to get an associative array, instead of an object):
$json = json_decode($bmstatus, true);
echo $json['networkConnections'];
It's a json string. You need to decode your json response using json_decode with second parameter true to get as an associative array.
$bmstatusArray = json_decode($bmstatus,true);
echo $bmstatusArray["networkConnections"];
I'm trying to output the value of the email value of an array, but have problems doing so.
The array is based on json_decode()
This is the error I receive
Fatal error: Cannot use object of type stdClass as array in /home/.... line 57
JSON (value of: $this->bck_content)
{"email":"test#email.com","membership_id":"0","fname":"Kenneth","lname":"Poulsen","userlevel":"1","created":"2012-04-23 10:57:45","lastlogin":"2012-04-23 10:58:52","active":"y"}
My code
# Display requested user details
$details_array = json_decode($this->bck_content);
$value = $details_array['email'];
print $value;
You need to use the second argument to json_decode to force array structures on JS objects.
json_decode($this->bck_content, true);
This will make sure all JS objects in the json are decoded as associative arrays instead of PHP StdObjects.
Of course that is assuming you want to use array notation to access them. If you're fine with using object notation then you can just use:
$value = $details_array->email;
try this one
$value = $details_array->email;
or
json_decode($json, true);
or
$details_array = (array)json_decode($json);
what have you done wrong is writen in error description