I need to encode a table content to JSON in order to insert it into a file.
The output has to be as following :
{
"name1":[{"id":"11","name":"name1","k1":"foo","k2":"bar"}],
"name2":[{"id":"12","name":"name2","k1":"foo","k2":"bar"}],
}
Indeed, each JSON "line" corresponds to the content of the mysql row and the name of each JSON array is the name of the 'name' column.
The only thing I could manage for the moment is this :
$return_arr = array();
$sql = "SELECT * FROM bo_appart";
$result = mysql_query($sql) or die(mysql_error());
$index = 0;
while ($row = mysql_fetch_assoc($result)) {
$return_arr[$index] = $row;
$index++;
}
echo json_encode($return_arr);
And here is the output I get :
[
{"id":"11","name":"name1","k1":"foo","k2":"bar"},
{"id":"12","name":"name2","k1":"foo","k2":"bar"},
]
Thanks a lot !!!
UPDATED
Working code :
$return_arr = array();
$sql = "SELECT * FROM bo_appart";
$result = mysql_query($sql) or die(mysql_error());
while ($row = mysql_fetch_assoc($result)) {
$return_arr[ $row['nom_appart'] ][] = $row;
}
echo json_encode($return_arr);
}
You were close. I noticed you want final output to be an object not an array, because the outer brackets are {} not []. So you need a different object type, and you need to use each row's name as the key for storing that row.
$return_obj = new stdClass();
$sql = "SELECT * FROM bo_appart";
$result = mysql_query($sql) or die(mysql_error());
while ($row = mysql_fetch_assoc($result)) {
$name = $row['name'];
$return_obj->$name = [$row]; // for PHP < 5.4 use array($row)
}
echo json_encode($return_obj);
This loop is enough, to create the desired JSON:
$return_arr = array();
while ($row = mysql_fetch_assoc($result)) {
$return_arr[$row['name']][] = $row; #or $return_arr[$row['name']] = [$row];
}
echo json_encode($return_arr);
I was working on it a lot of time and Create mash/mysql-json-serializer package.
https://github.com/AndreyMashukov/mysql-json-serializer
You can select Json_array of json_objects and etc. It support ManyToMany, oneToMany, manyToOne relations
SELECT JSON_ARRAYAGG(JSON_OBJECT('id',est_res.est_id,'name',est_res.est_name,'advert_groups',(SELECT JSON_ARRAYAGG(JSON_OBJECT('id',adg.adg_id,'name',adg.adg_name)) FROM advert_group adg INNER JOIN estate est_2 ON est_2.est_id = adg.adg_estate WHERE est_2.est_id = est_res.est_id))) FROM (SELECT * FROM estate est LIMIT 1 OFFSET 2) est_res
<?php
$sql = "SELECT * FROM bo_appart";
$result = mysql_query($sql) or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
$return_arr[] = $row;
}
echo json_encode($return_arr);
?>
Related
I am designing a PHP service which fetches data for using it in android.
The issue is when I run the query with Where Clause (id < 30) it's working. but when I change (id <40) I am getting a blank screen:
$return_arr = array();
$fe = mysqli_query($conn,"SELECT id, name FROM cable_channels where id < 30;");
while ($row = mysqli_fetch_array($fe)) {
$row_array['id'] = $row['id'];
$row_array['name'] = $row['name'];
array_push($return_arr,$row_array);
}
echo json_encode($return_arr);
<?php
$return_arr = array();
$fe = mysqli_query($conn,"SELECT id, name FROM cable_channels where id < 30;");
while ($row = mysqli_fetch_assoc($fe)) {
$return_arr[] = $row;
}
echo json_encode($return_arr);
?>
Try above code, instead of using
mysql_fetch_array
use
mysql_fetch_assoc
and push all the values to array inside while loop.
i had this php code :
<?php
include "../mainmenu/koneksi.php";
// Start with the list of animals
$sql = "SELECT * FROM data_binatang";
$rows = array();
$res = mysql_query($sql);
for($i=0; $i<mysql_num_rows($res); ++$i){
$row1 = mysql_fetch_assoc($res);
$id_binatang = $row1['id_binatang'];
$sql = "SELECT * FROM data_waktu_vaksinasi WHERE id_binatang = $id_binatang AND (status_vaksin = 'belum' OR status_vaksin IS NULL) ORDER BY tanggal_vaksin ASC LIMIT 1";
$res2 = mysql_query($sql);
$row2 = mysql_fetch_assoc($res2);
$arr[$id_binatang] = array();
array_push($arr[$id_binatang], $row1['nama_binatang'], $row1['id_user'], $row1['jenis_binatang'], $row1['ras_binatang'], $row1['foto_binatang'], $row2['nama_vaksin'], $row2['id_data_waktu_vaksinasi'], $row2['status_vaksin'], $row2['tanggal_vaksin'], $row2['tanggal_datang']);
}
echo "RESULT:";
echo "<table border=1><tr><th>id binatang</th><th>nama binatang</th><th>id user</th><th>jenis binatang</th><th>ras binatang</th><th>foto binatang</th><th>nama vaksin</th><th>id data waktu vaksin</th><th>status vaksin</th><th>tanggal vaksin</th><th>tanggal datang</th></tr>";
foreach($arr as $key => $val){
echo "<tr><td>$key</td><td>".implode("</td><td>", $val)."</td></tr><br>";
}
?>
and here's the result
now i want to generate the table into json, but i don't know what to put inside the json encode, i tried:
echo '{"data_vaksinasi_menu":'.json_encode($arr[$id_binatang]).'}';
but instead it gave me null
Try this:
echo json_encode(array('data_vaksinasi_menu' => $arr));
i have a trouble, when select table user in yii2 , data no show up
$sqlGetuser = "select * from user ";
$sqlquery = Yii::$app->db->createCommand($sqlGetuser)->query();
foreach($sqlquery as $row){
// echo $row;
// echo "saya";
echo $row['username'];
}
The error is in this line:
$sqlquery = Yii::$app->db->createCommand($sqlGetuser)->query();
query() returns yii\db\DataReader. To return array of rows use queryAll():
$rows = Yii::$app->db->createCommand($sqlGetuser)->queryAll();
If you want to use query(), you need to read data differently:
$command = $connection->createCommand('SELECT * FROM "user"');
$reader = $command->query();
while ($row = $reader->read()) {
$rows[] = $row;
}
// equivalent to:
foreach ($reader as $row) {
$rows[] = $row;
}
// equivalent to:
$rows = $reader->readAll();
See more in official docs.
Also quote table name as adviced since it's reserved word.
I get an array of values returned from the following function:
function get_subscribitions($user)
{
$user = mysql_real_escape_string ($user);
$sql = "SELECT 'user_id' FROM `subscribe` WHERE subscriber = '$user'";
$result = mysql_query($sql);
$rows = array();
while ($row = mysql_fetch_assoc($result)) {
$rows[] = $row;
}
mysql_free_result($result);
return $rows;
I now want to use these values in a new function, where each "user_id" is used to collect text from the database through this function:
function get_text($writer) {
$writer = mysql_real_escape_string ($writer);
$sql = "SELECT * FROM `text` WHERE user_id='$writer' ORDER BY timestamp desc";
$result = mysql_query($sql);
$rows = array();
while ($row = mysql_fetch_assoc($result)) {
$rows[] = $row;
}
mysql_free_result($result);
return $rows;
However the returned value from the first function is an array, and as I've learnt the hard way, arrays cannot be treated by "mysql_real_escape_string".
How can I make the second function handle the values that I got from the first function?
Any responses appreciated.
Thank you in advance.
Your first mistake is to use mysql_fetch_assoc when only selecting one column. You should use mysql_fetch_row for this. This is likely going to fix your primary problem.
Could look like this:
$subs = get_subscribitions($whateverId);
$texts = get_text($subs);
function get_subscribitions($user)
{
$user = mysql_real_escape_string ($user);
$sql = "SELECT 'user_id' FROM `subscribe` WHERE subscriber = '$user'";
$result = mysql_query($sql);
$rows = array();
while ($row = mysql_fetch_row($result)) {
$user_id = $row['user_id'];
$rows[$user_id] = $user_id;
}
mysql_free_result($result);
return $rows;
}
function get_text($writer) {
$writers = implode(",", $writer);
$sql = "SELECT * FROM `text` WHERE user_id IN ({$writers}) ORDER BY timestamp DESC";
$result = mysql_query($sql);
$rows = array();
while ($row = mysql_fetch_array($result)) {
$rows[] = $row;
}
mysql_free_result($result);
return $rows;
}
This will save you a lot of time, because you can get all data from 'text' in one statement.
The solution is to avoid placing arrays in your $rows array in the first function. Instead of:
while ($row = mysql_fetch_assoc($result)) {
$rows[] = $row;
}
try:
while ($row = mysql_fetch_assoc($result)) {
$rows[] = $row['user_id'];
}
This will place only the value from column 'user_id' in the $rows array.
In order to use the second function you must iterate over the array returned from the first one. Something like this could work for you:
$user_subscriptions = get_subscribitions($user);
foreach($user_subscriptions as $subscription) {
$texts = get_text($subscription['user_id']);
foreach($texts as $text) {
// do something with the fetched text
}
}
As George Cummins says,
while ($row = mysql_fetch_assoc($result)) {
$rows[] = $row['user_id'];
}
and, to speed up the second function:
function get_text($writer)
{
$sql = "SELECT * FROM `text` WHERE user_id in (".implode(',',$writer).") ORDER BY timestamp desc";
$rows = array();
if ($result = mysql_query($sql))
{
while ($row = mysql_fetch_assoc($result))
{
$rows[] = $row;
}
mysql_free_result($result);
}
return $rows;
}
The change to the query means that you only do one in total rather than one for each ID thus removing the time taken to send the query to the server and get a response multiple times. Also, if the query fails, the function returns an empty array
Use :
string implode ( string $glue , array $pieces )
// example, elements separated by a comma :
$arrayasstring = impode(",", $myarray);
function get_subscribitions($user)
{
$user = mysql_real_escape_string ($user);
$sql = "SELECT 'user_id' FROM `subscribe` WHERE subscriber = '$user'";
$result = mysql_query($sql);
$row = mysql_fetch_row($results);
mysql_free_result($result);
return intval($row['user_id']);
}
it return only the user id you can used it in the 2nd function
My mind got stuck when i try to develope this:
i have a table in my database called "article" whit two column, "name" and "price".
How can i extract all rows from my table and echo all column in JSON?
i really can't understand how to convert result in JSON. My mind it's stuck like never before. i need to echo something like this:
{"items": {
"items":[
{"name": "firstitemname",
"price": "5"
},
{"name": "secondone",
"years": "3"
}],
}}
Please help me fixing my buggy code!
<?php
$query = mysql_query("SELECT * FROM itemlist");
$nameitem = array();
$itemprice = array();
while($row = mysql_fetch_array($query)){
array_push($nameitem , $row['nome']);
array_push($itemprice, $row['pix']);
}
?>
You would simply edit your PHP as follows.
<?php
$query = mysql_query("SELECT * FROM itemlist");
$items = array();
while($row = mysql_fetch_array($query)){
$items[] = array('name' => $row['nome'], 'price' => $row['pix']);
}
echo json_encode(array('items'=>$items));
?>
http://php.net/manual/en/function.json-encode.php
JSON is super easy to deal with in PHP.
If you're using PHP 5.2 or greater, you can use the json_encode function to do exactly what you're trying to do: http://www.php.net/manual/en/function.json-encode.php
For your code, you should be able to do something like this:
$query = mysql_query("SELECT * FROM itemlist");
$json_output = array();
while($row = mysql_fetch_assoc($query)){
$json_output[] = json_encode($row);
}
Here, $json_output will contain an array of strings with the json encoded string of each row as each array element. You can output these as you please.
<?php
$result = mysql_query("select * from item_list");
$rows = array();
while($r = mysql_fetch_assoc($result))
{
$rows[] = $r;
}
print json_encode($rows);
?>
Convert Data table to json with following code :
echo(json_encode($array));
for example ( select data from mysql and convert to json ):
public function SELECT($tableName,$conditions){
$connection = mysqli_connect($hostname, $userName, $password,$dbName);
try {
if (!$connection)
die("Connection failed: " . $connection->connect_error);
else
{
$qry = "";
if(!$this->IsNullOrEmptyString($conditions))
$qry = "SELECT * FROM `".$tableName."` WHERE ".$conditions;
else
$qry = "SELECT * FROM `".$tableName."`";
$result = mysqli_query( $connection, $qry);
if($result) {
$emparray = array();
while($row =mysqli_fetch_assoc($result))
$emparray[] = $row;
echo(json_encode($emparray));
}
else
echo(mysqli_error($connection));
}
mysqli_close($connection);
} catch(Exception $ex) {
mysqli_close($connection);
echo($ex->getMessage());
}
}