php - combining json with json_decode

Question

I'm trying to combine multiple JSON objects into a single one in PHP. I'm iterating through the JSON objets, decoding them, parsing out the parts I want to keep, and storing them in a property in my php class.

Supposing my json objects look like the following format:

{
    "lists" : {
        "list" : [
            {
                "termA" : 2 ,
                "termB" : "FOO" 
            } 
        ] 
    } 
}

I want to eventually combine everything into a JSON object like so.

{
    "lists" : {
        "list" : [
            {
                "termA" : 2 ,
                "termB" : "FOO" 
            },
            {
                "termA" : 2 ,
                "termB" : "FOO" 
            } 
        ] 
    } ,
    "lists" : {
        "list" : [
            {
                "termA" : 4 ,
                "termB" : "BAR" 
            },
            {
                "termA" : 4 ,
                "termB" : "BAR" 
            } 
        ] 
    } 
}

I'm trying to store Arrays in a property within my class in a function that gets called iteratrivley:

   private function parseData($json){
        $decodeData = json_decode($json);
            $list = $decodeData->lists;
        $this->output .= $list
    }

However I get the following error during the "$this->output .= $list" line.

Object of class stdClass could not be converted to string

Right now $this->output has no initial value. What might be the best way to store the "list" arrays temporarily, and then reformat them after going through all of the json objects?

Thanks.

cletus · Accepted Answer · 2009-11-10 06:57:58Z

7

You were close:

private function parseData($json){
  $decodeData = json_decode($json);
  $list = $decodeData['lists'];
  $this->output .= $list
}

answered Nov 10, 2009 at 6:57

cletus

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Comments

troelskn · Accepted Answer · 2009-11-10 08:33:30Z

2

{
    "lists" : {
      ...
    } ,
    "lists" : {
      ...
    } 
}

That's not valid/meaningful JSON. You have a hash with the same key (lists) in it twice. How would you address that?

answered Nov 10, 2009 at 8:33

troelskn

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1 Comment

troelskn Over a year ago

Because jsonlint is insufficient? The restriction is not syntactic, so that's probably why. See: rfc-4627, section 2.2 for the definition (Or think about it twice - It's quite intuitive).

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