parsing json array in PHP

Question

I have been trying to parse a json array with no success. I can get a root element but not any array elements. Below is the beginning of my json array from Foursquare which has re-occurring venue elements.

     response: {
          keywords: {}
          suggestedRadius: 10000
          headerLocation: "here"
          headerFullLocation: "here"
          headerLocationGranularity: "unknown"
          headerMessage: "Suggestions for Friday evening"
          totalResults: 214
             groups: [
               {
                  type: "Recommended Places"
                  name: "recommended"
                  items: [
                      {
                       reasons: {
                       count: 0
                       items: [ ]
                        }
                       venue: {
                            id: "4b799a05f964a520b1042fe3"
                            name: "Green Gables"
                            contact: {
                            phone: "3097472496"
                            formattedPhone: "(309) 747-2496"
                                  }
                            location: {
                            address: "17485 East 2500 North Rd"

Below is my PHP code to try to get the name of the restaurants.

   $obj = json_decode($uri, true);

   foreach($obj['response']['groups'] as $p)
   {
   if(isset($p['items']))
   {
     foreach($p['items'] as $p1)
 {
  if(isset($p1['venue']))
  {
//   echo varDumpToString($p1['venue']);  // this dump works ok and shows the elements
 foreach($p1['venue'] as $p2)
 {

     echo varDumpToString($p2['name']);   //  This is where I get the error
   }
   }
   }
   }   
   }

The program drills down to the name element and then gives me an error saying "Unefined index: name" and also 'Illegal string offset name". This message appears 14 times which is one time for each item in the array. So why is "name" not recognized?

When you decoded your json using json_decode(), did you return it as an array or object? If object, try using $p2->name. — Adrian
– Adrian, Commented Mar 30, 2013 at 18:21
Are you sure that node is converted to an array, not a stdObject? In that case you can access the name via $p2->name — thaJeztah
– thaJeztah, Commented Mar 30, 2013 at 18:21

jwueller · Accepted Answer · 2013-03-30 18:44:42Z

2

response.groups[x].items[y].venue is not an array. You are currently trying to access response.groups[x].items[y].venue[z].name, but you are actually accessing response.groups[x].items[y].venue.id.name, which does not exist. That last foreach iterates over the venue properties, not the venues. There is only one venue for each item.

This is what it should look like:

$obj = json_decode($uri, true);

foreach ($obj['response']['groups'] as $group) {
    if (isset($group['items'])) {
        foreach ($group['items'] as $item) {
            if (isset($item['venue'])) {
                echo varDumpToString($item['venue']['name']);
            }
        }
    }
}

edited Mar 30, 2013 at 18:44

answered Mar 30, 2013 at 18:21

jwueller

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5 Comments

Dave Over a year ago

So what should my last foreach look like? I tried about every combination I can think of. Why am I accessing venue.id.name. I thought id and name were separate elements. I just want to get id and name as separate items.

jwueller Over a year ago

@Dave - Check out my example. There is only a single venue for each item, so you cannot iterate them. You should also try to use more descriptive names for your variables to improve readability. This makes it a lot easier to catch errors like this.

Dave Over a year ago

If I do 'echo varDumpToString($p2['venu']['name'] I will get the name but it repeats itself 14 times for each venue. How can I just get the name as a singleton?

jwueller Over a year ago

@Dave - I am not sure what exactly you are doing, but I suspect the following: If all items have the same venue, the name is obviously displayed multiple times.

Dave Over a year ago

@elusive - Thanks very much for your suggestion. It worked great! I am a newbie to PHP and it was impossible to find an example of what I was trying to do. Now I know and I really appreciate your help.

Adrian · Accepted Answer · 2013-03-30 18:25:23Z

0

Try changing this line,

 foreach($p1['venue'] as $p2)

to

 foreach($p1['venue'] as $p2Key => $p2)

answered Mar 30, 2013 at 18:25

Adrian

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1 Comment

jwueller Over a year ago

This is not going to help. $p2 will contain 4b799a05f964a520b1042fe3 in the first case, so accessing the name property of that string does not make a lot of sense. This is exactly why OPs code does not work.

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