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

$sth = mysqli_query($conn, "SELECT ...");
$rows = array();
while($r = mysqli_fetch_assoc($sth)) {
    $rows[] = $r;
}
print json_encode($rows);

The function json_encode needs PHP >= 5.2 and the php-json package - as mentioned here

NOTE: mysql is deprecated as of PHP 5.5.0, use mysqli extension instead http://php.net/manual/en/migration55.deprecated.php.

Answer 2

Try this, this will create your object properly

 $result = mysql_query("SELECT ...");
 $rows = array();
   while($r = mysql_fetch_assoc($result)) {
     $rows['object_name'][] = $r;
   }

 print json_encode($rows);

Answer 3

http://www.php.net/mysql_query says "mysql_query() returns a resource".

http://www.php.net/json_encode says it can encode any value "except a resource".

You need to iterate through and collect the database results in an array, then json_encode the array.

Answer 4

Thanks.. my answer goes:

if ($result->num_rows > 0) {
            # code...
            $arr = [];
            $inc = 0;
            while ($row = $result->fetch_assoc()) {
                # code...
                $jsonArrayObject = (array('lat' => $row["lat"], 'lon' => $row["lon"], 'addr' => $row["address"]));
                $arr[$inc] = $jsonArrayObject;
                $inc++;
            }
            $json_array = json_encode($arr);
            echo $json_array;
        }
        else{
            echo "0 results";
        }

Answer 5

The code below works fine here!

<?php

  $con=mysqli_connect("localhost",$username,$password,databaseName);

  // Check connection
  if (mysqli_connect_errno())
  {
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

  $query = "the query here";

  $result = mysqli_query($con,$query);

  $rows = array();
  while($r = mysqli_fetch_array($result)) {
    $rows[] = $r;
  }
  echo json_encode($rows);

  mysqli_close($con);
?>

Answer 6

The above will not work, in my experience, before you name the root-element in the array to something, I have not been able to access anything in the final json before that.

$sth = mysql_query("SELECT ...");
$rows = array();
while($r = mysql_fetch_assoc($sth)) {
    $rows['root_name'] = $r;
}
print json_encode($rows);

That should do the trick!

Answer 7

When using PDO

Use fetchAll() to fetch all rows as an associative array.

$stmt = $pdo->query('SELECT * FROM article');
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($rows);

When your SQL has parameters:

$stmt = $pdo->prepare('SELECT * FROM article WHERE id=?');
$stmt->execute([1]);
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($rows);

When you need to rekey the table you can use foreach loop and build the array manually.

$stmt = $pdo->prepare('SELECT * FROM article WHERE id=?');
$stmt->execute([1]);

$rows = [];
foreach ($stmt as $row) {
    $rows[] = [
        'newID' => $row['id'],
        'Description' => $row['text'],
    ];
}

echo json_encode($rows);

When using mysqli

Use fetch_all() to fetch all rows as an associative array.

$res = $mysqli->query('SELECT * FROM article');
$rows = $res->fetch_all(MYSQLI_ASSOC);
echo json_encode($rows);

When your SQL has parameters you need to perform prepare/bind/execute/get_result.

$id = 1;
$stmt = $mysqli->prepare('SELECT * FROM article WHERE id=?');
$stmt->bind_param('s', $id); // binding by reference. Only use variables, not literals
$stmt->execute();
$res = $stmt->get_result(); // returns mysqli_result same as mysqli::query()
$rows = $res->fetch_all(MYSQLI_ASSOC);
echo json_encode($rows);

When you need to rekey the table you can use foreach loop and build the array manually.

$stmt = $mysqli->prepare('SELECT * FROM article WHERE id=?');
$stmt->bind_param('s', $id);
$stmt->execute();
$res = $stmt->get_result();

$rows = [];
foreach ($res as $row) {
    $rows[] = [
        'newID' => $row['id'],
        'Description' => $row['text'],
    ];
}

echo json_encode($rows);

When using mysql_* API

Please, upgrade as soon as possible to a supported PHP version! Please take it seriously. If you need a solution using the old API, this is how it could be done:

$res = mysql_query("SELECT * FROM article");

$rows = [];
while ($row = mysql_fetch_assoc($res)) {
    $rows[] = $row;
}

echo json_encode($rows);

Answer 8

My simple fix to stop it putting speech marks around numeric values...

while($r = mysql_fetch_assoc($rs)){
    while($elm=each($r))
    {
        if(is_numeric($r[$elm["key"]])){
                    $r[$elm["key"]]=intval($r[$elm["key"]]);
        }
    }
    $rows[] = $r;
}   

Answer 9

Sorry, this is extremely long after the question, but:

$sql = 'SELECT CONCAT("[", GROUP_CONCAT(CONCAT("{username:'",username,"'"), CONCAT(",email:'",email),"'}")), "]") 
AS json 
FROM users;'
$msl = mysql_query($sql)
print($msl["json"]);

Just basically:

"SELECT" Select the rows    
"CONCAT" Returns the string that results from concatenating (joining) all the arguments
"GROUP_CONCAT" Returns a string with concatenated non-NULL value from a group

Answer 10

<?php
define('HOST','localhost');
define('USER','root');
define('PASS','');
define('DB','dishant');

$con = mysqli_connect(HOST,USER,PASS,DB);


  if (mysqli_connect_errno())
  {
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

 $sql = "select * from demo ";

 $sth = mysqli_query($con,$sql);

$rows = array();

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

 $row_array['id'] = $r;

    **array_push($rows,$row_array);**
}
echo json_encode($rows);

mysqli_close($con);
?>

aarray_push($rows,$row_array); help to build array otherwise it give last value in the while loop

this work like append method of StringBuilder in java

Answer 11

One more option using FOR loop:

 $sth = mysql_query("SELECT ...");
 for($rows = array(); $row = mysql_fetch_assoc($sth); $rows[] = $row);
 print json_encode($rows);

The only disadvantage is that loop for is slower then e.g. while or especially foreach

Answer 12

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 13

I have the same requirement. I just want to print a result object into JSON format so I use the code below. I hope you find something in it.

// Code of Conversion
$query = "SELECT * FROM products;";
$result = mysqli_query($conn , $query);

if ($result) {
echo "</br>"."Results Found";

// Conversion of result object into JSON format
$rows = array();
while($temp = mysqli_fetch_assoc($result)) {
    $rows[] = $temp;
}
echo "</br>" . json_encode($rows);

} else {
    echo "No Results Found";
}

Answer 14

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 15

we could simplify Paolo Bergantino answer like this

$sth = mysql_query("SELECT ...");
print json_encode(mysql_fetch_assoc($sth));

Answer 16

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);

Answer 17

$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 18

B"H

Considering there's not really any NESTED json objects in mysql in general etc., it's fairly easy to make your own encoding function

First, the function to retrieve the mysqli results in an array:

function noom($rz) {
    $ar = array();
    if(mysqli_num_rows($rz) > 0) {
        while($k = mysqli_fetch_assoc($rz)) {
            foreach($k as $ki=>$v) {
                $ar[$ki] = $v;
            }
        }
    }
    return $ar;
}

Now, function to encode array as json:

function json($ar) {
    $str = "";
    $str .= "{";
    $id = 0;
    foreach($ar as $a=>$b) {
        $id++;
        $str .= "\"".$a."\":";
        if(!is_numeric($b)) {
            $str .= "\"".$b."\"";
        } else {
            $str .= $b;
        }
        
        if($id < count($ar)) {
            $str .= ",";
        }
    }
    $str .= "}";
    return $str;
}

Then to use it:

<?php
$o = new mysqli(
    "localhost",
    "root",""
);
if($o->connect_error) {
    echo "DUDE what are you/!";
} else {
    $rz = mysqli_query($o,
        "SELECT * FROM mydatabase.mytable"
    );
    $ar = noom($rz);
    echo json($ar);
}
?>

Answer 19

$rows = json_decode($mysql_result,true);

as simple as that :-)