13 thoughts on “PHP and JSON”

1. I don´t have alternatives or addition but I wanted to thank you for this.
   I read it and implemented it yesterday and I should say that I was impressed! Thank you for the tip!
   I also cannot wait for the phpbnl11 to start! It will be great, I am sure!
   See you there hopefully!

   Reply ↓

2. Don’t forget you can also pass True to the second parameter of json_decode to get back an associative array instead of an object..

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3. I see json_decode() to be pretty useless without setting it’s second parameter to true to obtain an associative array.

   What can you do with an stdClass object? Iterate over it? get_class_vars() returns an associative array of what could be built directly by json_decode($str, true).

   A stdClass object simply is not really useful, and json_decode() has no way to define using another class like mysqli_fetch_object().

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   • I almost never use json_decode() without true as the second argument either–I find stdClass objects generally pretty useless.

     json_encode()’s JSON_FORCE_OBJECT in 5.3 is useful in dealing with PHP’s array/hash duality:

     json_encode(array())
     // -> “[]”
     json_encode(array(), JSON_FORCE_OBJECT)
     // -> “{}”

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     • thanks everyone who mentioned about the object/array flag. Since I was only passing it to print_r it didn’t occur to me to switch it!

       Michael: I didn’t know 5.3 was getting that new functionality, thanks for adding your comment :)

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4. Very nice information man, I am just starting my project. This will really help me, thanks a lot

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5. Thank you very much, I am very new to PHP Web services. Your article about PHP and JSON simply superb. Now I understand to write my own code.

   Reply ↓

6. What if I had something like

   {
   “results”: [
   “time_of_day”:{
   “lunch”: xxxxx,
   “dinner”: yyyyyy,
   }
   }

   How would I use php to extract the value for lunch and dinner?

   Reply ↓

   • Your sample isn’t valid JSON so PHP won’t decode it, but essentially you do json_decode() and pass in the value like you posted there, and I recommend you pass “TRUE” as the second parameter so that you get an array.

     Call print_r() on the array to see what’s in it, then you can do something like $array[‘results’][‘time_of_day’][‘lunch’] – depending on the structure of your (valid) JSON.

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7. oops, sorry, forgot to do the [code][/code] thing

   Hi, I am stumped on producing a specifically formatted JSON for a flot chart. Any help is much appreciated!

   I use this php code

   [code]
   $query = mysql_query(‘SELECT * FROM all_programs_extended’);

   while($row = mysql_fetch_assoc($query)) {
   $newData[] = array(
   ‘label’ => $row[‘program_type’],
   ‘data’ => array(
   “employee1” => $row[’employee1′],
   “employee2” => $row[’employee2′],
   “employee3” => $row[’employee3′]
   )
   );

   }

   print json_encode($newData);

   [/code]
   which produces this valid JSON

   [{“label”:”Program A”,”data”:{“employee1″:”5″,”employee2″:”3″,”employee3″:”1”,}},
   {“label”:”Program B”,”data”:{“employee1″:”0″,”employee2″:”4″,”employee3″:”2”,}},
   {“label”:”Program A”,”data”:{“employee1″:”4″,”employee2″:”2″,”employee3″:”4”,}}]

   but I need it look like this:

   {
   “label”: “Program A”,
   “data”: [[employee1, 5], [employee2, 3], [employee3, 1]],
   “label”: “Program B”,
   “data”: [[employee1, 0], [employee2, 4], [employee3, 2]],
   “label”: “Program C”,
   “data”: [[employee1, 4], [employee2, 2], [employee3, 4]]

   }

   I seem to have mixed up by brackets and braces, and I also need a comma separating my value pairs between brackets rather than the colon in the first JSON. Be nice to get ride of the quotation marks around the numbers too!

   Thanks very much!

   Reply ↓

   • For the data elements, try wrapping them in another level of array, so where you have “employee2” => $row[’employee2′] wrap an array() around the whole thing.

     I think that should get you closer to what you need.

     Reply ↓

8. Hey, great post. I have a quick question… So far my php code produces an array of elements in json format:
   PHP code (snippet):
   [code]
   $sql = “select * from customercarpark”;
   $result = mysqli_query($con, $sql) or die(“Error in Selecting ” . mysqli_error($con));
   $emparray = array();
   while($row = mysqli_fetch_assoc($result))
   {
   $emparray[] = $row;
   }
   echo json_encode($emparray);
   [/code]

   This gives me the output:
   [{“StudentID”:”13000000″,”TagNumber”:”111111111″,”FirstName”:”Megan”,”LastName”:”Smith”,”ContactNumber”:”07135274827″},{“StudentID”:”13011111″,”TagNumber”:”212355852″,”FirstName”:”John”,”LastName”:”Smith”,”ContactNumber”:”07523244423″}]

   How would I alter the php code so that this array will have a name/label so the output will be like this:
   { “Customers” : [ {“StudentID”:”13000000″,”TagNumber”:”111111111″,”FirstName”:”Megan”,”LastName”:”Smith”,”ContactNumber”:”07135274827″},{“StudentID”:”13011111″,”TagNumber”:”212355852″,”FirstName”:”John”,”LastName”:”Smith”,”ContactNumber”:”07523244423″}] }

   I’m sure it’s pretty simple but I’ve been searching for ages and cannot find anything, any help would be great :) thanks

   Reply ↓

   • The JSON output depends on how your array is structured – so you can change the line inside the while loop to say

     $emparray[“Customers”][] = $row;

     Then each individual record will go into a new element inside the Customers index in the array, and when you json_encode() it, you’ll get the structure you mentioned. Hope that makes sense …

     Reply ↓

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