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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?

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When you decoded your json using json_decode(), did you return it as an array or object? If object, try using $p2->name. – Adrian Crepaz Mar 30 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 Mar 30 at 18:21
LOL @AdrianCrepaz beat me to it – thaJeztah Mar 30 at 18:22
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2 Answers

up vote 2 down vote accepted

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($item['venue']['name']);
            }
        }
    }
}
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1  
Well spotted, looks like the right answer to me – thaJeztah Mar 30 at 18:25
So what should my last foreach look like? I tried about every combination I can think of. Why am I accessing venue.id.name. I thought id and name were separate elements. I just want to get id and name as separate items. – Dave Mar 30 at 18:42
@Dave - Check out my example. There is only a single venue for each item, so you cannot iterate them. You should also try to use more descriptive names for your variables to improve readability. This makes it a lot easier to catch errors like this. – elusive Mar 30 at 18:45
If I do 'echo varDumpToString($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? – Dave Mar 30 at 18:47
@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. – elusive Mar 30 at 18:50
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up vote 0 down vote

Try changing this line,

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

to

 foreach($p1['venue'] as $p2Key => $p2)
share|improve this answer
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. – elusive Mar 30 at 18:27
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