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']),
);
}
}
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 making a script to monitorize data from some VPSs, which is being stored in a MySQL database.
$result = mysql_query("SELECT * FROM data");
$data = array();
while ($row = mysql_fetch_array($result)) {
$data[] = array(
'id' => $row['id'],
'hostname' => $row['hostname'],
'loadavrg' => $row['load average']
);
}
I would like to separase data by hostnames so I can read it like the following example:
foreach($data['sitename.com'] as $d)
echo $d['loadavrg'];
I already tried with the following code (just for testing), but didn't work:
$result = mysql_query("SELECT * FROM data WHERE hostname='sitename.com'");
$data = array();
while ($row = mysql_fetch_array($result)) {
$data[] = array(
'sitename.com' => array(
'id' => $row['id'],
'loadavrg' => $row['load average']
)
);
}
I'm just missing the right way and syntax to achieve it :X
Every element should be added as subarray of 'sitename.com':
while ($row = mysql_fetch_array($result)) {
$data['sitename.com'][] = array(
'id' => $row['id'],
'loadavrg' => $row['load average']
);
}
Try this,
while ($row = mysql_fetch_array($result)) {
$hostname = $row['hostname'];
$data[$hostname][] = array(
'id' => $row['id'],
'loadavrg' => $row['load average']
);
}
I'm trying to fetch events from database to calendar but at this stage it just outputs the first event it gets from the database and the json when i echo it it returns just the first event
<?php
include("dbcon.php");
$events1 = array();
$sql345="select * from takimet";
$result = mysql_query($sql345)or die(mysql_error());
while ($row=mysql_fetch_array($result)){
$id = $row['agent_id'];
$title = $row['ezitimi'];
$start = $row['datestamp'];
$events1 = array(
'id' => "$id",
'title' => "$title",
'start' => "$start"
);
}
echo json_encode($events1);
?>
You are over writing the same array occurance each time round you loop.
Instead
$events1 = array();
while ($row=mysql_fetch_array($result)){
$events1[] = array( //<--- changed
'id' => $row['agent_id'],
'title' => $row['ezitimi'],
'start' => $row['datestamp']
);
}
I want to get all locations in table travel_location in mysql. This is my code
$select = $this->_db_table->select()->from(travel_location, array('*'));
$result = $this->_db_table->fetchAll($select);
if(count($result) == 0) {
throw new Exception('not found',404);
}
while ($row1 = mysql_fetch_array($result)){
$user_object = new Api_Model_User($row1);
$count = 1;
$json = array($json[$count] = array(
'travel_location_id' => $user_object->travel_location_id,
'city_id' => $user_object->city_id,
'user_id' => $user_object->user_id,
'location_name' => $user_object->location_name,
'description' => $user_object->description,
'longitude' => $user_object->longitude,
'latitude' => $user_object->latitude,
'created_time' => $user_object->created_time,
'updated_time' => $user_object->updated_time));
$count++;
}
It doesn't work. I print $row1 = mysql_fetch_array($result) and it returns false, so I think it's wrong because of this line. How can I fix it?
If you use fetchAll from Zend_Db_Table you get a Zend_Db_Table_Rowset as result.
Try this:
foreach ($result as $row) {
// $row should be a Zend_Db_Table_Row object
// you can cast to array
$rowArray = $row->toArray();
$user_object = new Api_Model_User($rowArray);
}
Read more about here and here
$result = mysql_query($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"]);