Display multiple records in json using php mysql query

Question

I have written a script to query a mysql database and encode the data in json format.

I have added $encoded = json_encode($encodable[0]); which removes the other [ ] brackets but then only displays 1 record. Is there a way to still remove those brackets but display for example every record that I am querying?

Sorry not sure how to describe the problem in a better way!

Might be missing something here but if you remove the [ ] brackets without parsing the data surely the JSON will become invalid? — Clive
– Clive, Commented Oct 8, 2011 at 14:01

Benno · Accepted Answer · 2011-10-08 14:18:03Z

0

Encode the result set of the query as JSON? Or encode data that is being sent to the database??

$encodeable[0] is the first element of the array, so its clear that that is what will be encoded. Depending on the format of your data, encodeable[0] would be [{"key":"value"},{"key":"value"},{"key":"value"}] by default as thats an indexed array with keys of 0,1,2...

If your goal is to just output the json as a JS object {}, instead of an array [] you can use json_encode($array, JSON_FORCE_OBJECT); which will encode indexed arrays as {"0":"value"} or {"0":{"0":"value"}} instead of just ["value"] or [["value"]]. If that's not what you're after, you may just need to loop through each encodeable element of your array and encode that way (use a for loop)

http://php.net/manual/en/function.json-encode.php

answered Oct 8, 2011 at 14:18

Benno

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6 Comments

Steve Over a year ago

Ok this is the code I am using could you tell me where I put JSON FORCE OBJECT? $encodable = array(); mysql_connect(localhost, $username, $password); mysql_select_db($database) or die("Unable to select database"); $result = mysql_query($sql); while($obj = mysql_fetch_object($result)) { $encodable[] = $obj; } $encoded = json_encode($encodable); echo $encoded; mysql_close();

Steve Over a year ago

Thanks I have added that but now it just echos "ArrayJSON_FORCE_OBJECT"

Benno Over a year ago

o_O Sorry about that, somehow I put a dot instead of a comma lol; $encoded = json_encode($encodable, JSON_FORCE_OBJECT) (thats the reason why it says array, it was concatenating it)

Steve Over a year ago

Warning: json_encode() expects exactly 1 parameter, 2 given in

Benno Over a year ago

Ah... you need PHP 5.3.0 or greater to use the JSON_FORCE_OBJECT option... Try this (stackoverflow.com/questions/2634009/…) if you can't upgrade your PHP version. json_encode((object)$encodable);

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