php insert to mysql issue [duplicate] - php

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"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 5 years ago.
I have an issue inserting data to table in php,the problem I am facing maybe it is not difficult to solve but Im new in php development and could not solve it, here is my code
<table class="table-fill">
<tbody class="table-hover">
<tr>
<td class="text-left"><input type = "text" class = "form-control"
name = "name"></td>
<td class="text-left">name</td>
</tr>
<tr>
<td class="text-left"><input type = "text" class = "form-control"
name = "Quantity" id="QTY"></td>
<td class="text-left">QTY</td>
</tr>
<tr>
<td class="text-left"><input type = "text" class = "form-control"
name = "Type" id="type"></td>
<td class="text-left">type</td>
</tr>
<tr>
<td class="text-left"><input type = "text" class = "form-control"
name = "price" id="prs"></td>
<td class="text-left">price</td>
</tr>
<tr>
<td class="text-left"><input type="file" name="fileToUpload" id="fileToUpload"></td>
<td class="text-left">image</td>
</tr>
<tr>
<td class="text-left"> <input type="submit" value="Submit1" name="Submit1"></td>
<td class="text-left">Register</td>
</tr>
</tbody>
</table>
</div>
<div id="footer">
<div>
</div>
</div>
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn ) {
die('Could not connect: ' . mysql_error());
}
if (isset($_post('Submit1')))
{
$name=filter_input(INPUT_POST,'name');
$QTY=filter_input(INPUT_POST,'Quantity');
$type=filter_input(INPUT_POST,'Type');
$sql = "INSERT INTO products (ProductName, Quantity,Price)
VALUES ('$name','$QTY','$type')";
//('$_POST['name']','$_POST['Quantity']','$_POST['Type']')";
mysql_select_db('pharmacynew');
$retval = mysql_query( $sql, $conn );
if(! $retval ) {
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($conn);
} // put your code here
?>
</body>
</html>
the problem Iam facing when I run the page ,it shows me this error
"Fatal error: Cannot use isset() on the result of a function call (you can use "null !== func()" instead) "
Can any body tell me why is this problem appearing and how to overcome it,
many thanks

Answer 1

The problem is in this line:
if (isset($_post('Submit1')))
By using regular brackets ( and ) you're trying to use $_post as a function instead of a variable. That's not allowed within isset().
You need to use angled brackets [ and ] to read values from the $_post array (which should be $_POST in uppercase, by the way).
Simply change that line to
if (isset($_POST["Submit1"]))