Getting all EmployeeID's in dropdown list from mysql database? - php

Question

In my case, I am getting only the EmployeeId which i have registered latest and not all the employeeId's from the table in database.
Code :
public function getEmployeeId() {
if (!isset($_SESSION["email"]) || !isset($_SESSION["passwrd"])) {
header("Location:index.php");
// Cannot Access this page without Login.
}
if (empty($_POST)) {
$query = mysqli_query($this->connection, "SELECT EmployeeId FROM employeeprofile") or die("Query execution failed: " . mysqli_error());
**while ($row = $query->fetch_assoc()) {**
$id = $row["EmployeeId"];
$_SESSION["id"] = $id;
}
}
}
i am using this session value in tag in my html form.
<select name="EmployeeId" required autofocus class='form-control'>
<option value=""> ----Select----</option>
<?php
if (isset($_SESSION["id"])) {
echo "<option value = " . $_SESSION["id"] . ">" . $_SESSION["id"] . "</option>";
}
?>
</select>
Someone suggested me to check the array , but i am confused.

Answer 1

You're currently overwriting the ID on each iteration. You need to store them in an array in which you append each ID instead:
// Make this session item an array
$_SESSION['id'] = [];
$query = mysqli_query($this->connection, "SELECT EmployeeId FROM employeeprofile") or die("Query execution failed: " . mysqli_error());
while ($row = $query->fetch_assoc()) {
// Push the id to the array
$_SESSION['id'][] = $row["EmployeeId"];
}
Now you need to iterate through the array with the ID's when printing the options:
if (isset($_SESSION["id"])) {
foreach ($_SESSION['id'] as $id) {
echo "<option value='$id'>$id</option>";
}
}
This should work, as long as you have session_start() in the top of your scripts.
I would probably not use sessions to store results like this, but I don't know enough of your code to be able to give you a good solution.

Answer 2

i thik the problem is that you should retrieve data in array from query and you can apply foreach loop to print the employee.
first get all the employee
$query = mysqli_query($this->connection, "SELECT EmployeeId FROM employeeprofile") or die("Query execution failed: " . mysqli_error());
$result = query($query);
now print the retrieved employee in option field
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<option value=" . $row["id"] . ">" . $row["id"] . "</option>"
}
} else {
echo "<option value="">No result found.</option>"
}

Answer 3

For me wouldn't advise to use sessioning, then to populate your select input, you do the below code, this is just a method of doing this anyway
public function getEmployeeId() {
if (!isset($_SESSION["email"]) || !isset($_SESSION["passwrd"])) {
header("Location:index.php");
// Cannot Access this page without Login.
}
$ids = array();
if (empty($_POST)) {
$query = mysqli_query($this->connection, "SELECT EmployeeId FROM employeeprofile") or die("Query execution failed: " . mysqli_error());
while($row = mysqli_fetch_array($query)){
$ids[] = $row['EmployeeId']; //we populate our array here
}
}
return $ids; //return the array here
}
from your html you do this
<select>
<option value="">Select</option>
<?php
$ids = getEmployeeId();
for($i=0;$i<count($ids); $i++){
?>
<option value="<?php echo $ids[$i];?>"><?php echo $ids[$i];?></option>
<?php }?>
</select>

Answer 4

It is because, you are putting only last value of array in $_SESSION["id"] variable
Change $_SESSION["id"] = $id; to $_SESSION["id"][] = $id;
and to print
if (isset($_SESSION["id"])) {
foreach($_SESSION["id"] as $v){
echo "<option value ='" . $v . "'>" . $v."</option>";
}
}