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I am trying to build a restful web service for my website. I have a php mysql query using the following code:

function mysql_fetch_rowsarr($result, $taskId, $num, $count){
  $got = array();
  if(mysql_num_rows($result) == 0)
    return $got;
  mysql_data_seek($result, 0);
  while ($row = mysql_fetch_assoc($result)) {
    $got[]=$row;
}
  print_r($row)
  print_r(json_encode($result));
  return $got;

which returns the following using the print_r($data) in the code above

Array ( [0] => Array ( [show] => Blip TV Photoshop Users TV [region] => UK [url] => http://blip.tv/photoshop-user-tv/rss [resourceType] => RSS / Atom feed [plugin] => Blip TV ) [1] => Array ( [show] => TV Highlights [region] => UK [url] => http://feeds.bbc.co.uk/iplayer/highlights/tv [resourceType] => RSS / Atom feed [plugin] => iPlayer (UK) ) )

Here is the json it returns:

 [{"show":"Blip TV Photoshop Users TV","region":"UK","url":"http:\/\/blip.tv\/photoshop-user-tv\/rss","resourceType":"RSS \/ Atom feed","plugin":"Blip TV"},{"show":"TV Highlights","region":"UK","url":"http:\/\/feeds.bbc.co.uk\/iplayer\/highlights\/tv","resourceType":"RSS \/ Atom feed","plugin":"iPlayer (UK)"}]

I am using the following code to add some items to the array then convert it to json and return the json.

 $got=array(array("resource"=>$taskId,"requestedSize"=>$num,"totalSize"=>$count,"items"),$got);

using the following code to convert it to json and return it.

 $response->body = json_encode($result);
 return $response;

this gives me the following json.

[{"resource":"video","requestedSize":2,"totalSize":61,"0":"items"},[{"show":"Blip TV Photoshop Users TV","region":"UK","url":"http:\/\/blip.tv\/photoshop-user-tv\/rss","resourceType":"RSS \/ Atom feed","plugin":"Blip TV"},{"show":"TV Highlights","region":"UK","url":"http:\/\/feeds.bbc.co.uk\/iplayer\/highlights\/tv","resourceType":"RSS \/ Atom feed","plugin":"iPlayer (UK)"}]]

The consumers of the API want the json in the following format and I cannot figure out how to get it to come out this way. I have searched and tried everything I can find and still not get it. And I have not even started trying to get the xml formatting 

 {"resource":"video", "returnedSize":2, "totalSize":60,"items":[{"show":"Blip TV Photoshop Users TV","region":"UK","url":"http://blip.tv/photoshop-user-tv/rss","resourceType":"RSS / Atom feed","plugin":"Blip TV"},{"show":"TV Highlights","region":"UK", "url":"http://feeds.bbc.co.uk/iplayer/highlights/tv","resourceType":"RSS / Atom feed","plugin":"iPlayer (UK)"}]}

I appreciate any and all help with this. I have setup a copy of the database with readonly access and can give all the source code it that will help, I will warn you that I am just now learning php, I learned to program in basic, fortran 77 so the php is pretty messy and I would guess pretty bloated.

OK The above about json encoding was answered. The API consumers also want the special character "/", not to be escaped since it is a URL. I tried the "JSON_UNESCAPED_SLASHES " in the json_encode and got the following error. 

 json_encode() expects parameter 2 to be long
share|improve this question
Apply json_decode to the desired json output and see what array you need to assemble to get the same – zerkms Apr 11 '12 at 23:47
how can I get the desired json into php? can I make it a string then json)decode the string? – Moltra Apr 11 '12 at 23:48
"The consumers of the API want the json in the following format " --- this is the desired format, the following one. The one your consumers want is obviously a desired – zerkms Apr 11 '12 at 23:49

1 Answer

up vote 0 down vote accepted

Your $result line should look like

$result=array(
    "resource"=>$taskId,
    "requestedSize"=>$num,
    "totalSize"=>$count,
    "items" => $got
);
share|improve this answer
funny how easy it is once someone puts the correct answer under your nose. – Moltra Apr 11 '12 at 23:51
You would get the same if you used json_decode with the line you want to achieve (the last json in your question) – zerkms Apr 11 '12 at 23:53
I made a small change to the question, I tried to add the JSON_UNESCAPED_SLASHES to the json_encode and get the following. Warning: json_encode() expects parameter 2 to be long, string given in – Moltra Apr 11 '12 at 23:59
@Moltra: as it said in documentation (nz.php.net/json_encode) that constant was introduced in php 5.4. Obviously you use some older version – zerkms Apr 12 '12 at 0:05
Yes I am using PHP Version: PHP 5.3 Is there anyway around this using php 5.3 that is the current php on godaddy 4GH. – Moltra Apr 12 '12 at 0:07
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