i have a small problem with my php code.
i've created a class for Clients, to return me a list with all clients. The problem is when i call the function from outsite, it dont show me the clietsList.
<?php
$returnData = array();
class Clients
{
public function readAll(){
$query = "SELECT * FROM clients";
$result = $this->con->query($query);
if ($result->num_rows > 0){
while ($row = $result->fetch_assoc()){
$returnData['clientsList'][] = $row;
}
return $returnData;
}
else{
echo "No found records";
}
}
}
if ($auth->isAuth()){
$returnData['message'] = "you are loggedin, so is ok";
$clientsObj = new Clients();
$clientsObj->readAll();
}
echo json_encode($returnData);
?>
and the result is like that, but without clietslist
{
"message": "you are loggedin, so is ok"
}
Where im doing wrong? Thanks for your answer. i want to learn php, im begginer. thanks in advance.
You need to get return value of function in a variable:
$returnData['message'] = "you are loggedin, so is ok";
$clientsObj = new Clients();
$returnData['clients'] = $clientsObj->readAll();
$returnData
on you first line will not be same as $returnData
in your function. $returnData
in readAll function will be a new variable.
I suggest you to read about variable scope in php. https://www.php.net/manual/en/language.variables.scope.php
Now, You will need to store returned value from function in a variable like this
$returnData['message'] = "you are loggedin, so is ok";
$clientsObj = new Clients();
$clientListArray = $clientsObj->readAll(); //store returned value in a variable
$returnData['clientsList'] = $clientListArray['clientsList'];
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.