CODEWITHSUNDEEP
phphtmlmysql
NinePoundHammer
NinePoundHammer

Reputation: 21

Display SQL Query Via HTML Button

My apologies if this is redundant, I searched related topics, but could not find an answer in any of them. I am looking to run SQL queries against a database I have created and display the results on an HTML button click. I have started with a simple scenario where I am just looking select all entities from my "customer" table. I have created the following PHP code to accomplish this:

<?php
define('DB_NAME', 'mis630db');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
define('DB_HOST', 'localhost');

$link = mysql_connect (DB_HOST, DB_USER, DB_PASSWORD);

if (!$link) {
        die('Could not connect:  ' . mysql_error());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
        die('Can\'t use ' . DB_NAME . ':  ' . mysql_error());
}

$resultset = mysql_query("SELECT * FROM Customer");
if($resultset->num_rows !=0) {
    while($rows = $resultset->fetch_assoc())
    {
        $idCustomer = $rows['idCustomer'];
        $CustomerFName = $rows['CustomerFName'];
        $CustomerLName = $rows['CustomerLName'];
        $CustomerAddress = $rows['CustomerAddress'];
        $CustomerPhone = $rows['CustomerPhone'];
        $CustomerEmail = $rows['CustomerEmail'];

        echo"<p> Customer ID: $idCustomer<br />Customer Name:  $CustomerFName $CustomerLName<br />Customer Address:  $CustomerAddress<br />Customer Phone:  $CustomerPhone<br />Customer Email:  $CustomerEmail</p>";
    }
}   
else{
        echo "No results.";
}
?>

When I test that I receive this error:

" Notice: Trying to get property of non-object in C:\xampp\htdocs\report.php on line 20 No results."

I'm not sure why as it should be returning multiple results. Once that is sorted out I want to connect this to an HTML button to display the results, where I would have a few buttons that run various SQL queries I have created :

<html>
<head>
    <title> Reports</title>
</head>
<body>
<div class="header">
    <h1>Reports</h1>
</div>
<form action="report.php" method="post" />
    <table>
    <tr>
        <td><input type="submit" value="Customers" /><td>
    </tr>
</form>
</body>
</html>

Any insight would be much appreciated.

Thanks.

Upvotes: 2

Views: 1622

Answers (2)

Andrew Larsen
Andrew Larsen

Reputation: 1267

Change this line:

if($resultset->num_rows !=0) {

to

if (mysql_num_rows($resultset) != 0) {

Because I don't think you can access a method called num_rows directly from a mysql result object.

And change this line:

while($rows = $resultset->fetch_assoc())

to

while($rows = mysql_fetch_assoc($resultset))

Because I don't think you can access a method called fetch_assoc directly from a mysql result object.

And over to another thing. Mysql_ functions are deprecated I would advice you to start using mysqli_ functions instead. http://php.net/manual/en/book.mysqli.php

Upvotes: 1

jotaelesalinas
jotaelesalinas

Reputation: 1497

Looks like the SQL query SELECT * FROM Customer is failing and thus $resultset is not an object --it probably is null.

SQL tables, columns, etc are case sensitive. Are you sure your table is named Customer and not customer?

Please, note that the extension mysql is deprecated since PHP 5.5 and has been removed in PHP 7. You should be using mysqli or PDO-MySQL instead.

Upvotes: 1

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