I am kinda new to SQL and I am having a problem. I am following the tutorial from Here to try to get data from my database into my android app.
I have it working, but it only gets one line. I know it should be possible to iterate through the database and echo out every line that matches the ID to JSON, but I do not know how. How would I do this?
My current code:
<?php
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
require_once('dbConnect.php');
$sql = "SELECT * FROM LATLON WHERE DeviceID='".$id."'";
$r = mysqli_query($con,$sql);
$res = mysqli_fetch_array($r);
$result = array();
array_push($result,array(
"INDEX"=>$res['INDEX'],
"DeviceID"=>$res['DeviceID'],
"LAT"=>$res['LAT'],
"LON"=>$res['LON']
)
);
echo json_encode(array("result"=>$result));
mysqli_close($con);
}
?>
EDIT
I edited my code to this, but I am getting nothing. Still not quite understanding what is happening. Thanks for all the help!
<?php
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
require_once('dbConnect.php');
$sql = "SELECT * FROM LATLON WHERE DeviceID='".$id."'";
$r = mysqli_query($con,$sql);
while($row = $result->mysqli_fetch_array($r, MYSQLI_ASSOC)){
$array[] = $row;
}
echo json_encode($array);
mysqli_close($con);
}
?>
You can iterate through mysqli_fetch_array();
while($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
$array[] = $row;
}
echo json_encode($array);
When you use the OO interface, the method names don't begin with mysqli_, that's only used for procedural calls. You also have no variable named $result, the result is in $r.
So it should be:
while ($row = $r->fetch_array(MYSQLI_ASSOC)) {
or:
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
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 need to print multiple database queries using json_encode(). I have this code below that outputs database records using json_encode. This works fine.
<?php
error_reporting(0);
include('db.php');
$result = $db->prepare('SELECT username,firstname,lastname FROM user');
$result->execute(array());
$data = array();
while ($row = $result->fetch()) {
$data[] = $row;
}
echo json_encode($data);
?>
I need to add another database query so that I can print them together using the same json_encode. Here is the database query that I want to add so that I can print them together using json_encode:
<?php
include('db.php');
$result = $db->prepare('SELECT price AS price1,goods AS product FROM provision_table');
$result->execute(array());
$data_pro = array();
while ($row = $result->fetch()) {
$data_pro[] = $row;
}
echo json_encode($data_pro);
?>
How can I accomplish this?
I think this might hit what you're looking for.
<?php
error_reporting(0);
include('db.php');
$result = $db->prepare('select username,firstname,lastname from user');
$result->execute(array());
$data = array();
while ($row = $result->fetch()) {
$data[] = $row;
}
$result = $db->prepare('select price as price1,goods as product from provision_table');
$result->execute(array());
$data_pro = array();
while ($row = $result->fetch()) {
$data_pro[] = $row;
}
// Combine both arrays in a new variable
$all_data['user'] = $data;
$all_data['pro'] = $data_pro;
echo json_encode($all_data);
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);
?>
in my last project I used foreach loop to assign to every mysqli result a variable, like $r->mydata, but I formatted my pc accidentally, so I lost my core file and I can't remember how exactly I did that. I remember that I did something like this
$result = $db->query("SELECT * FROM data");
if($result->num_rows){
while ($row = $result->fetch_object()) {
foreach ($row as $r){
$row[] = $r;
}
}
}
And I can access the result from outside the while loop like this:
<?php echo $r->mydata ?>
Can anyone edit my code so it will work like before?
it would be easier to use
$rows=$result->fetch_all(MYSQLI_ASSOC);
rather than looping through all the rows and building an array.
Maybe you don't remember how you did it, but if you did it once you should know at least which approach you followed to solve this, let help you to understand the code first:
$result = $db->query("SELECT * FROM data");
if($result->num_rows>0){
//iterating only if the table is not empty
while ($row = $result->fetch_object()) {
//Here you are iterating each row of the database
foreach ($row as $r){
//Here you are iterating each column as $r
//and (trying) adding it again to the $row array
$row[] = $r;
}
}
}
Why would you like to access to the $row outside the loop, if what you want to do is print each row, you can do it inside the loop:
$result = $db->query("SELECT * FROM data");
if($result->num_rows>0){
while ($row = $result->fetch_object()) {
foreach ($row as $r){
echo $r.'<br>';
}
}
}
If you do something like what follows, what part of the 'printed' data do you need? You may have been 'specializing' the result set by eliminating the rows to only use a single row...
$result = $db->query("SELECT * FROM data");
$r = new stdClass();
// Only loop if there is a result.
if ($result)
{
$r->myData = []; // You aren't exactly clear on 'myData'. In this instance, I am setting the rows inside of it.
while ($row = $result->fetch_object())
{
$r->myData[] = $row;
}
$result->close(); // Free result set
}
print_r($r->myData);
$records = array();
$result = $db->query("SELECT * FROM data");
if($result->num_rows){
while ($row = $result->fetch_object()) {
$records[] = $row;
foreach($records as $r);
}
}
and now you can access the result from any place in the page,example echo inside html h1:
<h1>My name is: <?php echo $r->name ?></h1>
I have a strange question. I need to send some code to a client without having access to the server to test my code. In addition, it's using postgreSQL which I've never used, and I've not done PHP for a while!
In order to save some time, I'd really appreciate if someone could tell me if this code will do what I want?
example feed
<?
$sql = "SELECT * FROM V_SIDE_MENU_E";
include 'db.inc.php';
?>
db.inc.php
$connectString = 'host=localhost dbname=myDatabase user=foo password=bar';
$link = pg_connect($connectString);
if (!$link) {
echo "error";
} else {
$result = pg_query($link, $sql);
$rows = array();
while($r = pg_fetch_assoc($result)) {
$rows[] = $r;
}
print json_encode($rows);
}
I would change
$rows = array();
while($r = pg_fetch_assoc($result)) {
$rows[] = $r;
}
print json_encode($rows);
into
print json_encode(array_values(pg_fetch_all($result)));
But that's just a style thing -- your code should work.
Tested on your mysql ( it looks like it will work ). Your SELECT will work same in PostgreSQL like mySQL