Using a variable in php function returning value

Question

I have a function that is supposed to give me a staff members name from his ID number so my php is like this

$staffId = $_GET['staff_id'];
$staff = staff_load($staffId);

and my function is like this

function staff_load()
{
$dbh = dbh_get(); //connects to database

$sql = 'select user_name from user_staff where user_id = ?';
$stmt = $dbh->prepare($sql);
$stmt->execute();
$staff = $stmt->fetch();

dbh_free($dbh); //disconnects from database
return $staff;
}

But when I try and use the $staff variable, it shows nothing. I can't work out what I'm doing wrong. I've tried a bunch of variants and gotten nowhere except frustrated.

<td>Book for ' . $staff . '</td>

Your function definition (function staff_load()) needs to accept the value passed to it (function staff_load($id)), and bind it in the SQL query — Mark Baker
– Mark Baker, Commented Jan 23, 2017 at 9:24
function staff_load() takes no parameter, it should be function staff_load($id) and then inside your execute() add execute([$id]) — Jaquarh
– Jaquarh, Commented Jan 23, 2017 at 9:24
You should be getting errors from this code. Look at your error log — RiggsFolly
– RiggsFolly, Commented Jan 23, 2017 at 9:25
You ->prepare() a query, but you dont bind the parameter to it — RiggsFolly
– RiggsFolly, Commented Jan 23, 2017 at 9:26
Add error reporting to the top of your file(s) while testing right after your opening PHP tag for example <?php error_reporting(E_ALL); ini_set('display_errors', 1); to see if it yields anything. — RiggsFolly
– RiggsFolly, Commented Jan 23, 2017 at 9:26

Barmar · Accepted Answer · 2017-01-23 20:18:58Z

3

First, you forgot to include the parameter in your function definition:

function staff_load($id) {

Then you need to bind the parameter of the prepared statement.

$sql = 'select user_name from user_staff where user_id = $1';
$stmt = $dbh->prepare($sql);
$stmt->execute(array($id));

Next, fetch() returns an array, you need to extract the user_name element from the array:

$row = $stmt->fetch();
if ($row) {
    $staff = $row['user_name'];
} else {
    $staff = false;
}

edited Jan 23, 2017 at 20:18

answered Jan 23, 2017 at 9:29

Barmar

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9 Comments

Kilisi Over a year ago

changed function to this but still nothing function staff_load($staffId) { $dbh = dbh_get(); $sql = 'select user_name from user_staff where user_id = :staffId'; $stmt = $dbh->prepare($sql); $stmt->execute(); $stmt->bindParam(':staffId', $staffId); $row = $stmt->fetch(PDO::FETCH_ASSOC); if ($row) { $staff = $row['user_name']; } else { $staff = false; } dbh_free($dbh); return $staff; }

Barmar Over a year ago

Are there any errors? Have you enable PDO's error signalling? Can you confirm that you're using PDO as I assumed?

Barmar Over a year ago

I assumed it was PDO because $stmt->fetch() is the syntax for that API. What API are you using?

Kilisi Over a year ago

can it be written without api? I'm not a programmer, but doesn't have pdo in other functions. It's just php to postgresql database

Kilisi Over a year ago

this is a statement that works function user_options() { $dbh = dbh_get(); $options = ''; $sql = 'select user_id, user_name from user_staff where user_id>7 ORDER BY user_name'; $stmt = $dbh->prepare($sql); $stmt->execute(); while (true) { $r = $stmt->fetch(); if (is_bool($r)) break; $options .= '<option value="' . $r['user_id'] . '">' . $r['user_name'] . '</option>'; } dbh_free($dbh); return $options; }

|

Dima · Accepted Answer · 2017-01-23 09:29:54Z

0

first of all you have to pass parameter to staff_load function

function staff_load($staffId)

secondly you should bind parameter

$sql = 'select user_name from user_staff where user_id = :user_id';
$stmt = $dbh->prepare($sql);
$stmt->bindParam(':user_id', $staffId);

edited Jan 23, 2017 at 9:29

answered Jan 23, 2017 at 9:27

Dima

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1 Comment

Kilisi Over a year ago

user_id should equal the $staffId value

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