I'm creating an app that parses json feeds into a listactivity. I've read that new items are added at the bottom of a listview and this is the case in my app. I'm trying to remedy this by sorting the output of my json by id in descending order. Don't know if this will fix my problem or I need to remedy it in my java code. Part one of my question is, will changing the order of output from the database change the order in which the feeds are displayed in the app? If so, I've tried to modify my php script by following the examples on http://php.net/manual/en/function.usort.php but the order doesn't change. Part two, What am I doing wrong in the following php script?
<?php
require("config.inc.php");
$query_params=null;
//initial query
$query = "Select * FROM feeds";
//execute query
try {
$stmt = $db->prepare($query);
$result = $stmt->execute($query_params);
}
catch (PDOException $ex) {
$response["success"] = 0;
$response["message"] = "Database Error!";
die(json_encode($response));
}
// Finally, retrieve all of the found rows into an array using fetchAll
$rows = $stmt->fetchAll();
if ($rows) {
$response["feed"] = array();
foreach ($rows as $row) {
$post = array();
$post["id"] = $row["id"];
$post["name"] = $row["name"];
$post["status"] = $row["status"];
//update our repsonse JSON data
array_push($response["feed"], $post);
}
function cmp($a, $b)
{
if ($a->id == $b->id) {
return 0;
}
return ($a->id > $b->id) ? -1 : 1;
}
$a = array();
usort($a, "cmp");
// echoing JSON response
echo json_encode($response, JSON_NUMERIC_CHECK);
} else {
$response["success"] = 0;
$response["message"] = "No Post Available!";
die(json_encode($response));
}
?>
Looks like you are calling usort on an empty array, instead of the data that you actually want to sort.
Instead of
$a = array();
usort($a, "cmp");
Try
$a = $response['feed'];
usort($a, "cmp");
$response['feed'] = $a;
Blackbelt is right that you can't technically sort a JSON array, at least one with sequential indexes (which results in no indexes at all). In practice, though, when you iterate through the resulting object on the client side, it will be in the order it was defined, so sorting it server-side should not raise any issues.
Related
I'm trying to write a function in php that will allow me to return an array of objects (in my case the objects are the rows in my database(MySQL) from phpMyAdmin) into a Json format.
Here is my function:
public function getAllCases()
{
$stmt = $this->conn->prepare("SELECT * FROM mycase ");
$stmt->execute();
$arr = []
$result = $stmt->get_result();
while($row = $result->fetch_object()) {
$arr[] = $row;
}
return $arr;
}
I couldn't find anything else online and this function is not working.
For example if my table has 4 rows, I want to be able to get 4 objects and each one of those 4 objects should represent a row from the table and it's columns.
Any help would be appreciated.
You can use json_encode and json_decode to convert array to JSON and vice versa.
return json_encode($arr)
To convert a php array into json , Use json_encode() function. And in your code, You mixed up Normal and Prepared Statements.
public function getAllCases()
{
$data = array();
$result = $this->conn->query("SELECT * FROM `mycase`");
$num_rows = $result->num_rows;
if ($num_rows > 0) {
while($row = $result->fetch_assoc()) {
$data[] = $row;
}
}
return json_encode($data);
}
I'm trying to GET a JSON format back when I POST a specific ID to my database. As I get more than one result I have multiple rows, which I want to get back. I do get different arrays back, but it is not a valid JSON Format. Instead of
[{...},{...},{...}]
it comes back as
{...}{...}{...}
Therefore the [...] are missing and the arrays are not separated by commas.
My code is down below. The function "getUserBookingsKl" is defined in a different php.
//get user bookings
public function getUserBookingsKl($id) {
//sql command
$sql = "SELECT * FROM `***` WHERE `hf_id`=$id AND `alloc_to`>DATE(NOW()) AND NOT `confirmation`=0000-00-00 ORDER BY `alloc_from`";
//assign result we got from $sql to $result var
$result = $this->conn->query($sql);
// at least one result
if ($result !=null && (mysqli_num_rows($result) >= 1 ))
{
while ($row = $result->fetch_array())
{
$returArray[] = $row;
}
}
return $returArray;
}
...
...
foreach($userdb as $dataset)
{
$returnArray["group"] = $dataset["kf_id"];
$returnArray["from"] = $dataset["alloc_from"];
$returnArray["to"] = $dataset["alloc_to"];
echo json_encode($returnArray);
# return;
}
// Close connection after registration
$access->disconnect();
It looks like you're sequentially emitting the values, not pushing into an array. You need to make an array, push into it, then call json_encode on the resulting structure:
$final = [ ];
foreach ($userdb as $dataset)
{
$returnArray = [ ];
$returnArray["group"] = $dataset["kf_id"];
$returnArray["from"] = $dataset["alloc_from"];
$returnArray["to"] = $dataset["alloc_to"];
$final[] = $returnArray;
}
echo json_encode($final);
Note that it's important here to not use the same variable inside the loop each time through or you're just pushing the same array in multiple times.
I would like to randomize the order of my json objects. Here is the output:
And here is my code so far:
// check for empty result
if (mysql_num_rows($result1) > 0) {
// looping through all results
// products node
while ($row = mysql_fetch_array($result1)) {
// temp user array
$feedMain["user_id"] = $row["user_id"];
// push single product into final response array
array_push($response["feedMain"], $feedMain);
}
// success
$response["success"] = 1;
}
// check for empty result
if (mysql_num_rows($result2) > 0) {
// looping through all results
// products node
while ($row = mysql_fetch_array($result2)) {
// temp user array
$feedMain["user_id"] = $row["user_id"];
// push single product into final response array
array_push($response["feedMain"], $feedMain);
}
// success
$response["success"] = 1;
}
// check for empty result
if (mysql_num_rows($result3) > 0) {
// looping through all results
// products node
while ($row = mysql_fetch_array($result3)) {
// temp user array
$feedMain["user_id"] = $row["user_id"];
// push single product into final response array
array_push($response["feedMain"], $feedMain);
}
// success
$response["success"] = 1;
// echoing JSON response
echo json_encode($response, JSON_UNESCAPED_UNICODE);
}
I tried something like :
echo json_encode(shuffle($response), JSON_UNESCAPED_UNICODE);
and other code snippets but nothing won't work for me..
I just want to randomize the order of the json objects. Thanks.
result1,2 and 3 are mysql statements :)
shuffle will not walk the array deeply. What you should shuffle is $response["feedMain"]:
shuffle($response["feedMain"]);
echo json_encode($response, JSON_UNESCAPED_UNICODE);
Why is my method only returning the first row of my table? I can't understand why and it's driving me nuts. I'm sure it's something very simple.
public function getTitlesForRegistrationForm() {
$result = $this->_db->query("SELECT UserTitleID, UserTitleName FROM UserTitles");
$i=0;
$array[0] = "No result";
foreach($result->fetch(PDO::FETCH_ASSOC) as $row){
$array[$i] = $row;
$i++;
}
return $array;
}
Thanks.
It may be because your $result->fetch() call doesn't return an iterable value, but either a result row or FALSE. PHP's foreach only works on iterable values. Updating your code to something like this should do the trick:
public function getTitlesForRegistrationForm() {
$result = $this->_db->query("SELECT UserTitleID, UserTitleName FROM UserTitles");
$i=0;
$array[0] = "No result";
while (($row = $result->fetch(PDO::FETCH_ASSOC)) !== FALSE) {
$array[$i] = $row;
$i++;
}
return $array;
}
You need to fetch() inside a while loop. It will return only one row each time it is called.
I've also taken the liberty of refactoring away your $i counter. Instead the No result is appended onto the array in the first position (for whatever purpose you planned to use it), and subsequent rows are appended on with [].
public function getTitlesForRegistrationForm() {
$array = array();
$result = $this->_db->query("SELECT UserTitleID, UserTitleName FROM UserTitles");
// Why are you putting No Result onto the array?
// I've left it in, but it doesn't make sense to me.
$array[] = "No result";
while ($row = $result->fetch(PDO::FETCH_ASSOC)){
$array[] = $row;
}
return $array;
}
It's because you are using fetch(). You need fetchall().
http://www.php.net/manual/en/pdostatement.fetchall.php
[edit]Wow. I need to learn to type faster.
Try:
while($row = $result->fetch(PDO::FETCH_ASSOC)){
foo();
}
Try while instead of foreach. Foreach only iterates over a single row that is returned.
code:
while($row = fetch(PDO::FETCH_ASSOC)) {
$array[$i] = $row;
$i++;
}
Just replace the fetch method with fetchAll
This question already has answers here:
JSON encode MySQL results
(16 answers)
Closed 1 year ago.
I've spent a couple of hours looking through several the similar answers before posting my problem.
I'm retrieving data from a table in my database, and I want to encode it into a JSON. However, the output of json_encode() is only valid when the table has one single row. If there is more than one row, the test at http://jsonlint.com/ returns an error.
This is my query:
$result = mysql_query($query);
$rows = array();
//retrieve and print every record
while($r = mysql_fetch_assoc($result)){
$rows['data'] = $r;
//echo result as json
echo json_encode($rows);
}
That gets me the following JSON:
{
"data":
{
"entry_id":"2",
"entry_type":"Information Relevant to the Subject",
"entry":"This is my second entry."
}
}
{
"data":{
"entry_id":"1",
"entry_type":"My Opinion About What Happened",
"entry":"This is my first entry."
}
}
When I run the test at http://jsonlint.com/, it returns this error:
Parse error on line 29:
..."No comment" }}{ "data": {
---------------------^
Expecting 'EOF', '}', ',', ']'
However, if I only use this first half of the JSON...
{
"data":
{
"entry_id":"2",
"entry_type":"Information Relevant to the Subject",
"entry":"This is my second entry."
}
}
... or if I only test the second half...
{
"data":{
"entry_id":"1",
"entry_type":"My Opinion About What Happened",
"entry":"This is my first entry."
}
}
... the same test will return "Valid JSON".
What I want is to be able to output in one single [valid] JSON every row in the table.
Any suggestion will be very much appreciated.
The problem is you're spitting out separate JSON for each row, as opposed to doing it all at once.
$result = mysql_query($query);
$rows = array();
//retrieve and print every record
while($r = mysql_fetch_assoc($result)){
// $rows[] = $r; has the same effect, without the superfluous data attribute
$rows[] = array('data' => $r);
}
// now all the rows have been fetched, it can be encoded
echo json_encode($rows);
The minor change I've made is to store each row of the database as a new value in the $rows array. This means that when it's done, your $rows array contains all of the rows from your query, and thus you can get the correct result once it's finished.
The problem with your solution is that you're echoing valid JSON for one row of the database, but json_encode() doesn't know about all the other rows, so you're getting a succession of individual JSON objects, as opposed to a single one containing an array.
You need to change your PHP code into something like this:
$result = mysql_query($query);
$rows = array();
//retrieve every record and put it into an array that we can later turn into JSON
while($r = mysql_fetch_assoc($result)){
$rows[]['data'] = $r;
}
//echo result as json
echo json_encode($rows);
I think you should do
$rows = array();
while($r = mysql_fetch_assoc($result)){
$rows[]['data'] = $r;
}
echo json_encode($rows);
echo should be placed outside of the loop.
I was trying the same in my PHP, so I came whit this...
$find = mysql_query("SELECT Id,nombre, appaterno, apmaterno, semestre, seccion, carrera FROM Alumno");
//check that records exist
if(mysql_num_rows($find)>0) {
$response= array();
$response["success"] = 1;
while($line = mysql_fetch_assoc($find)){}
$response[] = $line; //This worked for me
}
echo json_encode($response);
} else {
//Return error
$response["success"] = 0;
$response["error"] = 1;
$response["error_msg"] = "Alumno could not be found";
echo json_encode($response);
}
And, in my Android Class...
if (Integer.parseInt(json.getString("success")) == 1) {
Iterator<String> iter = json.keys();
while (iter.hasNext()) {
String key = iter.next();
try {
Object value = json.get(key);
if (!value.equals(1)) {
JSONObject jsonArray = (JSONObject) value;
int id = jsonArray.getInt("Id");
if (!db.ExisteAlumo(id)) {
Log.e("DB EXISTE:","INN");
Alumno a = new Alumno();
int carrera=0;
a.setId_alumno(id);
a.setNombre(jsonArray.getString("nombre"));
a.setAp_paterno(jsonArray.getString("appaterno"));
a.setAp_materno(jsonArray.getString("apmaterno"));
a.setSemestre(Integer.valueOf(jsonArray.getString("semestre")));
a.setSeccion(jsonArray.getString("seccion"));
if(jsonArray.getString("carrera").equals("C"))
carrera=1;
if(jsonArray.getString("carrera").equals("E"))
carrera=2;
if(jsonArray.getString("carrera").equals("M"))
carrera=3;
if(jsonArray.getString("carrera").equals("S"))
carrera=4;
a.setCarrera(carrera);
db.addAlumno(a);
}
}
} catch (JSONException e) {
// Something went wrong!
}
}
I must have spent 15 hours on this issue. Every variation discussed above was tried. Finally I was able to get the 'standard solution' working. The issue, very oddly, appears to be this:
When the interval is set beyond 14 hours, json appears to be unable to parse it. There must be a limit to JSON.
$sql= "SELECT cpu_name, used, timestamp FROM tbl_cpu_use WHERE timestamp>(NOW() - INTERVAL 14 HOUR) ORDER BY id";
$result=mysql_query($sql);
if ($result){
$i=0;
$return =[];
while($row = mysql_fetch_array($result, MYSQL_NUM)){
$rows[] = $row;
}
echo json_encode($rows);
}else{
echo "ERROR";
}