Coverting php array to json array

Question

I am fetching records from php & saving them in associative arrays. i want to create a json format like this

{
"contacts": [
    {
            "qid": "c201",
            "question": "what is your name?,

    },
    {
            "qid": "c202",
            "question": "what is your age?

    },
   ]}

I am using following code to get the data in php

 <?
$response = array();
require("dbconfig.php");
$sql_get = 'SELECT qquesid,question FROM qhistory_data';
$retval = mysql_query( $sql_get, $conn );
if(!$retval )
{

            $response["success"] = 0;
            $response["message"]= mysql_error();
            echo json_encode($response);
}
else
{

$num_rows=mysql_num_rows($retval);
if($num_rows>=0)
    {
        $i=0;
    while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
        {
            $result_array[]= $row['question'];
            $result_id[]=$row['qquesid'];

        }

You're json_encoding the response if there are no records, so why not json_encode the response when there are? — Mark Baker
– Mark Baker, Commented Jan 31, 2015 at 11:53
I am having two arrays $result_array,$result_id i want to create a json response like i have mentioned — TechChain
– TechChain, Commented Jan 31, 2015 at 11:54
I am not able to get the mentioned json format.I want the response like i have mentioned?@mark baker — TechChain
– TechChain, Commented Jan 31, 2015 at 11:55

acontell · Accepted Answer · 2015-01-31 11:53:50Z

1

If you want to get the array in the form you mention, you have to parse it accordingly before doing the json_encode:

$result = array("contacts" => array());
while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) {
    $result["contacts"][]= array("qid" => $row['qquesid'], "question" =>$row['question']);
}

echo json_encode($result);

answered Jan 31, 2015 at 11:53

acontell

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1 Comment

TechChain Over a year ago

question id strats from 2 but it should start from 1.first record is missing

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