JSON encode MySQL results

Question

How do I use the json_encode() function with MySQL query results? Do I need to iterate through the rows or can I just apply it to the entire results object?

Answer 1

$rows = json_decode($mysql_result,true);

as simple as that :-)

Answer 2

I solved like this

$stmt->bind_result($cde,$v_off,$em_nm,$q_id,$v_m);
	$list=array();
	$i=0;
	while ($cresult=$stmt->fetch()){	
				
	
		$list[$i][0]=$cde;
		$list[$i][1]=$v_off;
		$list[$i][2]=$em_nm;
		$list[$i][3]=$q_id;
		$list[$i][4]=$v_m;
		$i=$i+1;
	}
	echo json_encode($list);		

This will be returned to ajax as result set and by using json parse in javascript part like this :

obj = JSON.parse(dataX);

Answer 3

$array = array();
$subArray=array();
$sql_results = mysql_query('SELECT * FROM `location`');

while($row = mysql_fetch_array($sql_results))
{
    $subArray[location_id]=$row['location'];  //location_id is key and $row['location'] is value which come fron database.
    $subArray[x]=$row['x'];
    $subArray[y]=$row['y'];


 $array[] =  $subArray ;
}
echo'{"ProductsData":'.json_encode($array).'}';

Answer 4

How to create JSON using data from MySQL database

JSON (JavaScript Object Notation) is more preferred nowadays over XML as it’s lightweight, readable and easily manageable for exchanging data across various platforms. we’ll see how JSON Data can be created from Employee table stored in MySQL database.

 echo json_encode($data);

Live : [Example ]

Answer 5

For example $result = mysql_query("SELECT * FROM userprofiles where NAME='TESTUSER' ");

1.) if $result is only one row.

$response = mysql_fetch_array($result);
echo json_encode($response);

2.) if $result is more than one row. You need to iterate the rows and save it to an array and return a json with array in it.

$rows = array();
if (mysql_num_rows($result) > 0) {
    while($r = mysql_fetch_assoc($result)) {
       $id = $r["USERID"];   //a column name (ex.ID) used to get a value of the single row at at time
       $rows[$id] = $r; //save the fetched row and add it to the array.
    }
}    
echo json_encode($rows);

Answer 6

Code:

$rows = array();

while($r = mysqli_fetch_array($result,MYSQL_ASSOC)) {

 $row_array['result'] = $r;

  array_push($rows,$row_array); // here we push every iteration to an array otherwise you will get only last iteration value
}

echo json_encode($rows);