I have a script I wrote to return records for cases out of the database. I am returning one record for my mysql query when there are actually two records. This is what I am returning:
{ "cases": [ {"name":"Test Case for App","number":"3846"}] }
I should see:
{ "cases": [ {"name":"Test Case for App","number": "2903"}, {"name":"Test Case 2","number": "2856"} ] }
Here is my source:
$sql = "select * from cases as c join contacts_cases as conc on c.id = conc.case_id where conc.contact_id = '1b360507'";
$query = mysql_query($sql);
// If we find a match, create an array of data, json_encode it and echo it out
if (mysql_num_rows($query) > 0)
{
$row = mysql_fetch_assoc($query);
$response = array(
'name' => $row['name'],
'number' => $row['case_number']
);
echo '{ "cases": [ ', json_encode($response), "] }";
If you are expecting more than one result you should try
if (mysql_num_rows($query) > 0)
{
$responses = array();
while($row = mysql_fetch_assoc($query)) {
$responses[] = array(
'name' => $row['name'],
'number' => $row['case_number']
);
}
echo '{"cases": ' . json_encode($responses) . '}';
}
You need to loop through all the rows, you're just getting one.
Also, don't try to build the JSON yourself. Make the array how you want then json_encode the entire thing.
$cases = array();
while ($row = mysql_fetch_assoc($query)) {
$cases[] = array(
'name' => $row['name'],
'number' => $row['case_number']
);
}
echo json_encode(array('cases' => $cases));
Related
I'm working on my REST API responses with PHP.
Here is my backend that generates the responses:
$query = $this->db->query("SELECT some_params from some_table ");
$result = $query->result();
foreach($result as $row){
$obj[] = array(
'title' => $row['title'],
'price' => $row['price'],
);
}
print_r(json_encode($obj));
and with that I have the following response: an array of json objects
[
{
"title":"Marketing",
"price":"0"
},
{
"title":"SAP B1",
"price":"10"
}
]
What I would like to do is return a new object, something like this:
{
"apiVersion": "2.0",
"data": {
{
"title":"Marketing",
"price":"0"
},
{
"title":"SAP B1",
"price":"10"
}
}
}
Does anyone have any idea how I could implement this? thanks!
You can easily have a class or function to do that for you. Some thing like below should help,
// provided you are using php > 7 . otherwise parmas
// can become $data, $apiversion ( without typecasting )
function api_reponse(array $data, string $apiVersion)
{
return json_encode([
'apiVersion' => $apiVersion,
'data' => $data
])
}
You can then use,
$query = $this->db->query("SELECT some_params from some_table ");
$result = $query->result();
foreach($result as $row){
$obj[] = array(
'title' => $row['title'],
'price' => $row['price'],
);
}
print_r( api_response($obj, '2.0') );
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 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']),
);
}
}
Hi I have an error when I call a function.
"Warning: Illegal string offset 'id' in C:\xampp\htdocs\blog\posts.php
on line 28
2"
function:
function get_short_posts() {
$sql = "SELECT * FROM posts ORDER by id DESC LIMIT 0, 5";
$result = mysql_query($sql);
while($row = mysql_fetch_assoc($result)) {
$timestamp = new DateTime($row['date']);
return array (
"id" => $row['id'],
"title" => $row['title'],
"content" => $row['content'],
"author" => $row['author'],
"date" => $timestamp->format('d-m-Y'),
"time" => $timestamp->format('H:i')
);
}
Call:
require_once "functions.php";
$_posts = get_short_posts();
foreach($_posts as $_post) {
echo $_post['id'];
}
You know you return after the first iteration in the get_short_posts function right, so the foreach will not work as expected.
Try:
<?php
function get_short_posts() {
$sql = "SELECT * FROM posts ORDER by id DESC LIMIT 0, 5";
$result = mysql_query($sql);
$return = array();
while($row = mysql_fetch_assoc($result)) {
$timestamp = new DateTime($row['date']);
$return[] = array (
"id" => $row['id'],
"title" => $row['title'],
"content" => $row['content'],
"author" => $row['author'],
"date" => $timestamp->format('d-m-Y'),
"time" => $timestamp->format('H:i')
);
}
return $return;
}
?>
<?php
require_once "functions.php";
foreach(get_short_posts() as $_post) {
echo $_post['id'];
}
?>
Also, Don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO, or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.
function get_short_posts() {
$sql = "SELECT * FROM posts ORDER by id DESC LIMIT 0, 5";
$result = mysql_query($sql);
while($row = mysql_fetch_assoc($result)) {
$timestamp = new DateTime($row['date']);
$data [] = array (
"id" => $row['id'],
"title" => $row['title'],
"content" => $row['content'],
"author" => $row['author'],
"date" => $timestamp->format('d-m-Y'),
"time" => $timestamp->format('H:i')
);
}
return $data;
}
you return data, so the loop stops, save your data in a array and return that array like abive code
Your code is wrong it should be as below, assuming the query is returning data as mentioned.
function get_short_posts() {
$sql = "SELECT * FROM posts ORDER by id DESC LIMIT 0, 5";
$result = mysql_query($sql);
$rows = array();
$return_data = array();
while($row = mysql_fetch_assoc($result)) {
$timestamp = new DateTime($row['date']);
$data = array (
"id" => $row['id'],
"title" => $row['title'],
"content" => $row['content'],
"author" => $row['author'],
"date" => $timestamp->format('d-m-Y'),
"time" => $timestamp->format('H:i')
);
$return_data[] = $data;
}
return $return_data ;
}
require_once "functions.php";
$posts = get_short_posts();
foreach($posts as $key=>$val) {
echo $val['id'];
}
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);
}
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