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Basic PHP Question/Help

cdecker15

By cdecker15
in PHP Coding Help

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cdecker15 Newbie

cdecker15

Posted
Posted

The below code works and all, but I would like to figure out how to get rid of the global variable. I tried without declaring it as a global variable and it did not work. Any other way to do that? Also, how would I go about returning two variables(right now I am returning $id), how can I return both $id and $first_name from the database.

<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="css/reset.css">
<link rel="stylesheet" type="text/css" href="css/structure.css">
<script type="text/javascript">
window.onload = function () {
    var inputs = document.getElementsByTagName('input');
    for (i = 0; i < inputs.length; i++) {
        inputs[i].onkeyup = function () {
            if (this.value.match(/^([0-9][0-9][0-9][0-9][0-9])$/)) {
                document.forms["punch"].submit();
            }
        };
    }
};
$("form").bind("keypress", function (e) {
    if (e.keyCode == 13) {
        $("#btnSearch").attr('value');
        //add more buttons here
        return false;
    }
});
</script>
</head>
<?php
$id = 0;
function getID($staffid){
	$servername = "localhost";
	$username = "root";
	$password = "x";
	$dbname = "hr";
	$conn1 = new mysqli($servername, $username, $password, $dbname);
		if ($conn1->connect_error) {
			die("Connection failed: " . $conn1->connect_error);
		} 
	
		$sql = "SELECT id, first_name, employee_id FROM Employees WHERE employee_id='".$staffid."'";
		$result = $conn1->query($sql);
		global $id;
		if ($result->num_rows > 0) {
		while($row = $result->fetch_assoc()) {
			$id = $row["id"];
		}
		return $id;
		} else {
			return false;
		}
		$conn1->close();
}
if (isset($_POST['staffid'])){
	$staffid = $_REQUEST['staffid'];
	
	if(getID($staffid)) {
		$status = "<font color='green'>Punched</font>";
		echo $id;
	} else {
	$status = "<font color='red'>Invalid ID, try again</font>";
	}
}


?>
<body>

<form class="box login" action="" id="punch" method="post">
<fieldset class="boxBody">
	  <label>Staff ID</label>
	<p><input id="staffid" type="text" autofocus="autofocus" name="staffid"></p>
</fieldset>	
<fieldset class="boxBody2">
	  <center><img src="clock.png" /><br><?php if ($status) {echo $status;} else {echo 'Please enter your Staff ID';}?></center>
		</fieldset>	
</form>

</body>
</html>
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https://forums.phpfreaks.com/topic/298356-basic-php-questionhelp/
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cdecker15 Newbie

cdecker15

Posted
  • Author
Posted

I changed it to the following, and it is now returning 0, not 1 like it should. Any ideas?

<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="css/reset.css">
<link rel="stylesheet" type="text/css" href="css/structure.css">
<script type="text/javascript">
window.onload = function () {
    var inputs = document.getElementsByTagName('input');
    for (i = 0; i < inputs.length; i++) {
        inputs[i].onkeyup = function () {
            if (this.value.match(/^([0-9][0-9][0-9][0-9][0-9])$/)) {
                document.forms["punch"].submit();
            }
        };
    }
};
$("form").bind("keypress", function (e) {
    if (e.keyCode == 13) {
        $("#btnSearch").attr('value');
        //add more buttons here
        return false;
    }
});
</script>
</head>
<?php
$id = 0;
function getID($staffid, $id, $first_name){
	$servername = "localhost";
	$username = "root";
	$password = "x";
	$dbname = "hr";
	$conn1 = new mysqli($servername, $username, $password, $dbname);
		if ($conn1->connect_error) {
			die("Connection failed: " . $conn1->connect_error);
		} 
	
		$sql = "SELECT id, first_name, employee_id FROM Employees WHERE employee_id='".$staffid."'";
		$result = $conn1->query($sql);
		if ($result->num_rows > 0) {
		while($row = $result->fetch_assoc()) {
			$id = $row["id"];
			$first_name = $row["first_name"];
		}
		return array($id, $first_name);
		} else {
			return false;
		}
		$conn1->close();
}
if (isset($_POST['staffid'])){
	$staffid = $_REQUEST['staffid'];
	
	if(getID($staffid)) {
		$status = "<font color='green'>Punched</font>";
		echo $id;
		echo $first_name;
	} else {
	$status = "<font color='red'>Invalid ID, try again</font>";
	}
}


?>
<body>

<form class="box login" action="" id="punch" method="post">
<fieldset class="boxBody">
	  <label>Staff ID</label>
	<p><input id="staffid" type="text" autofocus="autofocus" name="staffid"></p>
</fieldset>	
<fieldset class="boxBody2">
	  <center><img src="clock.png" /><br><?php if ($status) {echo $status;} else {echo 'Please enter your Staff ID';}?></center>
		</fieldset>	
</form>

</body>
</html>
gizmola Prolific Member

gizmola

Posted
Posted

This line:

 

if(getID($staffid)) {
--- depends on getID returning a boolean equivalent value. You have changed the function so that it no longer works that way. You will need to rethink the way that code is written if you want to change the way getID() works. Returning false or an array is no longer compatible with returning false OR some positive integer that will evaluate to true.
hansford Member

hansford

Posted
Posted

I don't understand why you need the $id variable as you don't seem to use it. The only id variable I see you use is the $staffid

$arg = getID($staffid);

if(is_array($arg)) 
{
  $status = "<font color='green'>Punched</font>";
  echo $arg[0]; //$id
  echo $arg[1]; //$first_name
} 
else 
{
  $status = "<font color='red'>Invalid ID, try again</font>";
}

Change the arguments in this function to this:

function getID($staffid){
gizmola Prolific Member

gizmola

Posted
Posted

I plan on using the ID later in the file, for a different MySQL statement. So what should I be using for this?

This is how you create spaghetti.

 

functions should be discrete and do one thing. Planning to "add a different MySQL statement later" isn't the right solution. You should write a different function for that.

 

It's also clear that getID() is a bad name for this function. It really should be named something like:

 

function getEmployee($id) {
}
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