Efficient way to pass JSON from PHP

Question

Here is a JSONArray I have:

 [{"items_id":"13","total":"1"}, {"items_id":"216","total":"0"},{"items_id":"16","total":"1"}]

Sometimes, each object has more than two properties (attributes?). But I am just showing the principle here. In Java, I only need to grab "total". I don't need "items_id".

I assume it shows up because here is my MySQL query in PHP:

$count_query_result=mysql_query("
SELECT items.items_id, 
COUNT(ratings.item_id) AS total
FROM `items`
LEFT JOIN ratings ON (ratings.item_id = items.items_id)
WHERE items.cat_id = '{$cat_id}'  AND items.spam < 5
GROUP BY items_id ORDER BY TRIM(LEADING 'The ' FROM items.item) ASC;");

Here is my JSON output (I have only displayed one of three queries above):

print(json_encode(array($output,$output2,$output3)));

I only want three properties encoded in JSON (one in each of the three output variables). I want the properties "total", "rate" ,and "item".

So my question is, can I get rid of the unneeded items_id property? Or do I even NEED to? (I know I need it in the SQL to make the query work -- but how can I remove it in the JSONArray?)

I am thinking if I have a list with hundreds or thousands of items, I can save half the space (and time?) by only outputing the JSON property I need -- is this thinking correct?

Edit: More code as requested:

while($row=mysql_fetch_assoc($count_query_result))
    $output[]=$row;
while($row=mysql_fetch_assoc($average_query_result))
    $output2[]=$row;
while($row=mysql_fetch_assoc($items_query_result))
    $output3[]=$row;

print(json_encode(array($output,$output2,$output3)));
mysql_close();

RaphMclee · Accepted Answer · 2012-07-31 16:09:32Z

0

Use the JSONObject to parse the String. Just put the string as an parameter in the constructor of the JSONObject.

jObject = new JSONObject(myString);

After you can retrieve all variables out of the JSONObject.

For the php side:

Use json_decode end json_encode to switch betweeen json and mixed variable. Once you have you mixed variable, delete the stuff you don't need and convert it back to json.

edited Jul 31, 2012 at 16:09

answered Jul 31, 2012 at 15:57

RaphMclee

1,62313 silver badges16 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

TheLettuceMaster Over a year ago

I know that is how I get the single object I need in Java/Android - but I am asking about the "handoff" method in PHP. IS there a way to addres this before it even gets to the Java?

TheLettuceMaster Over a year ago

Ok, fair question. At the time, I thought it made sense. But after I asked it, it became clear that this was much more php oriented. I suppose I can remove it.

Joe Simpson · Accepted Answer · 2012-07-31 16:12:52Z

0

http://php.net/unset

Unset the unneeded values, then output json

answered Jul 31, 2012 at 16:12

Joe Simpson

2,5944 gold badges30 silver badges47 bronze badges

6 Comments

TheLettuceMaster Over a year ago

Well this almost works unless I am doing wrong. I am doing this unset($row['items_id') when I loop through results. I did it only for first $output and it worked fine. Second $output2 said '[false]'. So it somehow adversely affects the other outputs?

Joe Simpson Over a year ago

Hm, can you post more of your code how you get from SQL to $output, $output2 etc?

TheLettuceMaster Over a year ago

Ok, I have done that. I also am figuring out more of what I am trying to do and added that explanation as well...

TheLettuceMaster Over a year ago

Not sure that will work since items_id is required for the query to work.

Joe Simpson Over a year ago

It will, as you're telling mysql: Grab for me the ID and Count(ratings.itemid). Just try it and report back if it didn't work :)

|

Robin Kanters · Accepted Answer · 2012-11-29 15:01:05Z

0

I don't think you need it in your SQL query. If you need data from a database just in the context of the query, you don't need to SELECT it. Try doing just the:

SELECT COUNT(*) FROM ...

answered Nov 29, 2012 at 15:01

Robin Kanters

5,1372 gold badges23 silver badges36 bronze badges

Collectives™ on Stack Overflow

Efficient way to pass JSON from PHP

3 Answers 3

2 Comments

6 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

6 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related