CODEWITHSUNDEEP
phphtmlsqloracle-database
BrownEye
BrownEye

Reputation: 979

Compare form data against database and perform an action

I have a website where the user must already exist in the database to be able to proceed to the next page. I have made a query that checks their first name, last name, and email address against those stored in the DB and I can get it to print the result if they match, but I can't get it to return a true or something I can work with in order to perform an action. I need to send an email to the user then redirect them to another page, could you help out with this too?

Here's my code

<?php

include("dbconnect.php");

$firstname=$_POST["firstname"];
$lastname=$_POST["lastname"];
$email=$_POST["email"];

$connect = $conn; 
if (!$connect) {
    $e = oci_error();
    trigger_error(htmlentities($e['message'], ENT_QUOTES), E_USER_ERROR);
}

// Prepare the statement
$stid = oci_parse($connect, "SELECT * FROM USERS WHERE firstname = '$firstname' AND lastname = '$lastname' AND email = '$email'");

if (!$stid) {
    $e = oci_error($connect);
    trigger_error(htmlentities($e['message'], ENT_QUOTES), E_USER_ERROR);
}

// Perform the logic of the query
$r = oci_execute($stid);
if (!$r) {
    $e = oci_error($stid);
    trigger_error(htmlentities($e['message'], ENT_QUOTES), E_USER_ERROR);
}

// Fetch the results of the query
print "<table border='1'>\n";
while ($row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS)) {
    print "<tr>\n";
    foreach ($row as $item) {
        print "    <td>" . ($item !== null ? htmlentities($item, ENT_QUOTES) : "&nbsp;") . "</td>\n";
    }
    print "</tr>\n";
}
print "</table>\n";

oci_free_statement($stid);
oci_close($connect);

?>

If you could please help me to run an action if true (send an email and redirect) and display an error if the user does not exist in the database. Thanks - I'm very new to PHP

Upvotes: 0

Views: 82

Answers (1)

Marcus Rommel
Marcus Rommel

Reputation: 1294

You can solve your problem with multiple solutions. I will give you two recommendations. The first one is to use a SQL select that counts the returning rows of an executed query. Try this:

SELECT * FROM YOUR_TABLE WHERE SURNAME LIKE '%surname%' AND NAME LIKE '%name%' AND EMAIL like '%email%'

Execute this and after the execute command insert this: echo oci_num_rows($stid) . " people with this data.\n";

If it is 1 you have somebody with this data in your database.

You can also use JavaScript. If you want I can explain that also. EDIT: here you go with pure javascript:

A short example of comparasion in JS
http://jsfiddle.net/marcusrommel/tcm4ak1p/

Upvotes: 1

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