How to get data in json form - php

Question

Currently, I am making an API and I want the data in JSON format. But I am not able to get it. It is coming in normal form. But how to convert it into JSON. If I am writing json_encode($response) outside the loop then I am getting the data in json format but only one data.
If i wriing the json encode inside the loop then i am getting all the data but not in JSON form. How to solve this. I am not able to get the perfect solution for ths question.
$tsym = strtolower($_REQUEST['tsym']);
$time = strtolower($_REQUEST['time']);
$mil = $time;
$seconds = $mil / 1000;
$normal_date = date("Y-m-d H:i:s", $seconds);
$sql = "SELECT * FROM `forex` where pair='".$tsym."' and date >= '".$normal_date."' order by date limit 0,10";
$result = mysqli_query($conn, $sql);
$response = array();
while($rows = mysqli_fetch_assoc($result)){
$from_sym = $rows['pair'];
//if(!isset($response[$from_sym]))
{
$response[$from_sym] = $rows;
//echo json_encode($response, true);
//}
print_r( json_encode($response)); //this prints all the data but not
in json form
}
print_r( json_encode($response)); //this prints single data but in
json form
I want all the data but in json form. how to get it? Thank you for the help.
I want data like this:
{
"CHFJPY": {
"id": "33",
"pair": "CHFJPY",
"date": "2018-04-22 20:42:21",
"price": "110.413",
"change_rate": "0",
"fetched": "1"
}
},
{
"CHFJPY": {
"id": "75",
"pair": "CHFJPY",
"date": "2018-04-22 20:42:29",
"price": "110.413",
"change_rate": "0",
"fetched": "1"
}
},
{
"CHFJPY": {
"id": "117",
"pair": "CHFJPY",
"date": "2018-04-23 11:25:47",
"price": "110.585",
"change_rate": "0",
"fetched": "1"
}
},
{
"CHFJPY": {
"id": "159",
"pair": "CHFJPY",
"date": "2018-04-23 12:34:54",
"price": "110.816",
"change_rate": "0",
"fetched": "1"
}
},
{
"CHFJPY": {
"id": "201",
"pair": "CHFJPY",
"date": "2018-04-23 12:35:04",
"price": "110.825",
"change_rate": "0",
"fetched": "1"
}
}
But i am getting only one data.

Answer 1

You have to append your row to the final array $response in your while loop
$response = array();
while($rows = mysqli_fetch_assoc($result)){
$from_sym = $rows['pair'];
$res[$from_sym] = $rows;
$response[] = $res;
}
print_r(json_encode($response));

Answer 2

You need to move the json_encode into the outside of while loop
<?php
$tsym_escaped = mysqli_real_escape_string($conn, $_REQUEST['tsym']);
$date = date("Y-m-d H:i:s", $_REQUEST['time']/1000);
$sql = sprintf(
"SELECT * FROM `forex` WHERE `pain`='%s' AND `date`>='%s' ORDER BY `date` LIMIT 0,10",
$tsym_escaped,
$date
);
$result = mysqli_query($conn, $sql);
$response = array();
while($row = mysqli_fetch_assoc($result)){
$response[$row['pair']] = $row;
}
echo json_encode($response);
Also, the way you're passing the data into the SQL query is insecure and can lead into a SQL injection.

Answer 3

is not clear why you have not json so .. in more clear way try
while($rows = mysqli_fetch_assoc($result)){
$response[$rows['pair']] = $rows;
}
$myJSON = json_encode($response);
var_dump($myJSON);
this should buil a $reponse array with the related vleus for each 'pair' index
but could be you need all the rows result so try
while($rows = mysqli_fetch_assoc($result)){
$response[] = $rows;
}
$mySecondJSON = json_encode($response);
var_dump($mySecondJSON);
or
while($rows = mysqli_fetch_assoc($result)){
$response[] = $rows['pair'];
}
$myOtherJSON = json_encode($response);
var_dump($myOtherJSON);