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I'm new to PHP and developing a login form. Please find below the code I used. When I tried it gave me the following error:
Fatal error: Can't use function return value in write context in C:\xampp\htdocs\forsiteSystem\login.php on line 3
Please help me to fix the issue.
Source code for thems/login.html:
<html>
<head>
<title></title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
</head>
<body>
<form action=".\login.php" method="get">
Name: <input type="text" name="name"><br>
E-mail: <input type="text" name="email"><br>
<input type="submit" id="Submit_button">
</form>
</body>
</html>
Source code for index.php:
<?php
// venushka thisara dharmasiri
require 'config.php';
require 'thems\login.html';
?>
Source code for login.php:
<?php
if(isset($_POST("Submit_button"))==true)
print("Submit button pressed");
else
print("submit button sorry");
?>
Source code for config.php:
<?php
$dbUser="root";
$dbPassword="";
$dbName="forsitelogin";
$dbHost="localHost";
$dbConnection= mysql_connect($dbHost, $dbUser, $dbPassword);
if($dbConnection)
{
mysql_select_db($dbName);
//print("Sucessfully connected to database");
}
else
die("<strong>Cound not connect to database </strong> ");
?>
798
Should be $_POST["Submit_button"] instead of $_POST("Submit_button")
133
The error the script returns explains it:
Fatal error: Can't use function return value in write context in C:\xampp\htdocs\forsiteSystem\login.php on line 3
If you don’t understand the meaning of the error—and believe me, most error messages are bizarre even to experienced programmers—look at the line number referenced. And looking at line 3 in login.php shows me the error; $_POST("Submit_button") is invalid:
if(isset($_POST("Submit_button"))==true)
print("Submit button pressed");
else
print("submit button sorry");
It should be $_POST["Submit_button"]:
if(isset($_POST["Submit_button"])==true)
print("Submit button pressed");
else
print("submit button sorry");
But looking at it further, why is there an ==true? It can simply be like this:
if(isset($_POST["Submit_button"]))
print("Submit button pressed");
else
print("submit button sorry");
But I would recommend doing a better check on that value like this:
if(array_key_exists("Submit_button", $_POST) && !empty(trim($_POST["Submit_button"])))
print("Submit button pressed");
else
print("submit button sorry");
I find that using array_key_exists and a combination of !empty with trim works better for basic user submitted data verification.
44
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