$result = mysql_query($query);
$leaderboard = array();
while($row = mysql_fetch_assoc($result)) {
$leaderboard[$row["username"]] = $row["score"];
}
$output = array
(
'status' => 1,
'content' =>$leaderboard
);
print_r(json_encode($output));
right now the $output array is such JSON:
{"tim":"120","john":"45","larry":"56"}
but I want to have them as key-value pair so instead I want to be like:
{"name":"tim","score":120","name":"john","score="45", etc.}
and if I need that way, how do I modify the $leaderboard array so the output would be like that?
$leaderboard[] = Array('name' => $row["username"], 'score' => $row["score"]);
Related
I'm using Google Orgchart in my project. In that I'm returning JSON OBJECT from PHP file.
Problem
My Problem is when I hardcode the value, It works fine. When I return data from PHP file. It did not work. I guess the data format which is returning from PHP file is not correct. File below.
$result = mysql_query("SELECT * FROM emp");
while($row = mysql_fetch_array( $result )) {
$arr1 = array(
'v' => $row['name'],
'f' => $row['name']+'<div style="color:red; font-style:italic">President</div>',
'' => $row['rep'],
'' => $row['des'],
);
array_push($dataarray, $arr1);
}
echo json_encode($dataarray);
which returns object like below
How it Should be
My hardcorded JSON OBJECT below
[
[{v:'Prabhkar', f:'Prabhkar<div style="color:red; font-style:italic">President</div>'},
'', 'The President'],
[{v:'Raguram', f:'Raguram<div style="color:red; font-style:italic">GM</div>'},
'Prabhkar', 'GM']
]
Console Screenshot below:
Do I need to create a one more array in PHP file. How I suppose to change the PHP array according to above screenshot. sorry for my english. Thank you.
You need to wrap 'v' and 'f' in a array, then push other values to parent array.
$result = mysql_query("SELECT * FROM emp");
while($row = mysql_fetch_array( $result )) {
$arr1 = array(
array(
'v' => $row['name'],
'f' => $row['name'] . '<div style="color:red; font-style:italic">President</div>'
),
$row['rep'],
$row['des']
);
array_push($dataarray, $arr1);
}
echo json_encode($dataarray);
In your hard coded array the first key has an array inside so you must change your code like this
$result = mysql_query("SELECT * FROM emp");
$dataarray = [];
while($row = mysql_fetch_array( $result )) {
$arr1 = array(
array(
'v'=> $row['name'],
'f' => $row['name'].'<div style="color:red; font-style:italic">President</div>',),
$row['rep'],
$row['des'],
);
array_push($dataarray, $arr1);
}
echo json_encode($dataarray);
Your internal structure is wrong. Your internal structure is an array, with the first being a map, followed by two values. Your current implementation is an array, with only a map.
$result = mysql_query("SELECT * FROM emp");
while($row = mysql_fetch_array( $result )) {
$arr1 = array(
array(
'v' => $row['name'],
'f' => $row['name'] . '<div style="color:red; font-style:italic">President</div>',
),
$row['rep'],
$row['des']);
array_push($dataarray, $arr1);
}
echo json_encode($dataarray);
I am trying to extract data from mysql database into a datatable using ajax, and php.
The code for my response.php file is below:
<?php
$result = mysql_query("select * from orders");
while ($row = mysql_fetch_array($result)) {
$data = array(
array(
'Name' => $row['jobnumber'],
'Empid' => $row['ID'],
'Salary' => $row['product']
)
);
}
$results = array(
"sEcho" => 1,
"iTotalRecords" => count($data),
"iTotalDisplayRecords" => count($data),
"aaData" => $data
);
/*while($row = $result->fetch_array(MYSQLI_ASSOC)){
$results["data"][] = $row ;
}*/
echo json_encode($results);
?>
Why is this only returning one result in my front end table?
http://orca.awaluminium.com/test.php
link above shows table.
You're replacing value of $data instead of pushing new rows in an array.
Change the following line.
$data = array(
array(
'Name'=>$row['jobnumber'],
'Empid'=>$row['ID'], 'Salary'=>$row['product']
)
);
To
$data[] = array(
'Name'=>$row['jobnumber'],
'Empid'=>$row['ID'], 'Salary'=>$row['product']
);
Also put $data=array(); before string while() looop.
You have to do foreach
while ($row = mysql_fetch_array($result)){
foreach($row as $a)
{$data[] = array(
array('Name'=>$a['jobnumber'], 'Empid'=>$a['ID'], 'Salary'=>$a['product']),
);
}
}
Im generating Chart4PHP.
In sample it takes data like this
$p->data = array(array(array("2010/10",-48),array("2011/01",238),array("2011/02",395)));
I have array "rows" constructed of row[date][units].
Im storing it in this way:
$rows = array();
for(...)
{
$row[date] = $mydate;
$row[units]= $myunits;
$rows[]=$row;
}
What I should make additionally to be able to use it as $p->data = $rows;
To add the extra array container, call array() with the rows array as the argument.
$data = array(array('date' => "2010/10", 'units' => -48),
array('date' => "2011/01", 'units' => 238),
array('date' => "2011/02", 'units' => 395));
foreach ($data as $d) {
$mydate = $d['date'];
$myunits = $d['units'];
$rows[] = array($mydate, $myunits);
}
$p->data = array($rows);
I am trying to make a dynamic JSON array in PHP, however when I try to do so it returns "Array". Here is the code I am currently using:
<?php
require '../../scripts/connect.php';
$array = '';
if($result = $db->query("SELECT * FROM art") or die ($db->error)){
if($count = $result->num_rows) {
while($row = $result->fetch_object()){
$array .= array(
'title' => $row->title,
'image' => "http://www.thewebsite.com/img/2.jpg",
'rating' => 7.7,
'releaseYear' => 2003,
'genre' => array(
'0' => $row->category,
'1' => $row->subcategory
)
);
}
}
}
echo json_encode($array);
?>
Can anyone suggest how I might go about fixing this?
And if anyone has suggestions about creating a dynamic JSON array, some help would be much appreciated.
Change your declaration of $array to be an array:
$array = array();
Then in your while loop, when you add the new array to $array, push it like this:
$array[] = array('title'=>$row->title, etc...)
I'm building an angular application, so I have a PHP script extracting the database data into a JSON when requested.
This is what I'm using to extract the data into an array:
$values = array();
$query = "SELECT * FROM photos ORDER BY id";
$result = $mysqli->query($query);
while ($row = $result->fetch_array(MYSQL_ASSOC)) {
array_push($values, $row);
}
json_encode($values);
But it gives no result.
After some testing, if I change json_encode($values); to print_r($values); it actually print the table array, here an extract:
Array (
[0] => Array (
[id] => 4
[title] => Feel the Moment
[views] => 6
)
It seems the script is not creating a valid array.
$query = "SELECT * FROM photos ORDER BY id";
if ($result = $this->mysqli->query($query)) {
$json = array();
while ($row = $result->fetch_assoc()) {
$json[] = $row;
}
echo json_encode($json);
}
Your while is not executing, probably because you don't have any returned values from database.
I tested this code and it works well:
$json = array();
$json[] = array(
"id" => 1,
"category" => 'a',
"tags" => 't');
$json[] = array(
"id" => 2,
"category" => 'b',
"tags" => 'tt');
echo json_encode($json);
The result is:
[{"id":1,"category":"a","tags":"t"},{"id":2,"category":"b","tags":"tt"}]
And you'd better use mysqli_fetch_row or mysqli_fetch_assoc.
Try to debug your code. You may use var_dump($result->fetch_assoc()) to see if the query has some results.