CODEWITHSUNDEEP
phpmysql
Pawan Kalyan
Pawan Kalyan

Reputation: 51

PHP how to fetch row value by unique value is passed

For my website I am generating unique value email and random code as login, so when the user login with the above email and code, so I need to echo the details of that user only.

Example: In my database im having first name, last name, mobile number, email, code. So if the user login with email and code, I have to Echo all details as first name, last name, mobile number

In my case how to use exact query to echo all the details by email and code.

$sql = 'select firstname, lastname, mobileno from clients where code=; & email=
If (is_array($row = $db->queryRow($sql))) 
Echo($firstname, $lastname, $mobileno) = $row;

Please help.

Upvotes: 1

Views: 122

Answers (2)

Junius L
Junius L

Reputation: 16132

 $code = $_POST['code'];
 $email = $_POST['email'];

$sql = $db->prepare("SELECT firstname, lastname, mobileno FROM `clients` WHERE code = ? AND email = ?");
$sql->execute(array($code, $email));
 $results = $sql->fetchAll(PDO::FETCH_ASSOC);

 foreach ($results as $row) {

  echo $row['firstname'].'<br>';
  echo $row['lastname'].'<br>';
 echo $row['mobileno'].'<br>';

 }

Upvotes: 2

Jeroen Bellemans
Jeroen Bellemans

Reputation: 2035

This is the html to login:

<form action="" method="POST">
<input name="email" type="email"><br>
<input name="code" type="text"><br>
<input name="submit" type="submit" value="login">
</form>

This will be different for each user as he has his own details to login...

Submitting the form:

if (isset($_POST['submit'])) {

$email = $_POST['email'];
$code = $_POST['code'];

    $sql = mysql_query("SELECT firstname, lastname, mobileno FROM `clients` WHERE code = '$code' AND email = '$email'");

    while ($row = mysql_fetch_array($sql)) {

    echo $row['firstname'].'<br>';
    echo $row['lastname'].'<br>';
    echo $row['mobileno'].'<br>';

    }
}

Upvotes: 0

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