CODEWITHSUNDEEP
phpmysql
Juliver Galleto
Juliver Galleto

Reputation: 9037

display contents from mysql return only one record

Im using this code to retrieve all records associated with the usertype="user" row from mysql

class user{
public function getnews(){
    $sql5 = "SELECT * FROM news WHERE usertype = 'user'";
    $result = mysqli_query($this->db,$sql5);
    $data = '';
    while($row = mysqli_fetch_assoc($result)){
        $data = '<article>';
        $data .= '<h3>' . $row['title'] . '</h3>';
        $data .= '<p>' . $row['content'] . '</p>';
        $data .= '<footer>By: ' . $row['whoposted'] . ' / ' . $row['dateposted'] . '</footer>'; 
        $data .= '</article>';
    }
    echo $data;
}
}

and then display the records with.

$user = new user();

$user->getnews()

and the structure of my database is.

id, title, content, whoposted, dateposted, usertype.

and assume i have one records that have usertype="admin"

id=autoincrement, title="this is the title", content="this is the content", dateposted="2015/03/08", usertype="admin"

and 5 records that have usertype="user"

id=autoincrement, title="this is the title", content="this is the content", dateposted="2015/03/08", usertype="user"

now, as you can see from my mysql query, i suppose to get all the records that has usertype="user" and it suppose to display all records except for one record which has usertype="admin" but it only display one record.

anyone has idea whats wrong with my code?

any help would be greatly appreciated. Thank you!

Upvotes: 0

Views: 132

Answers (1)

Abhik Chakraborty
Abhik Chakraborty

Reputation: 44844

You need to declare $data outside the loop, you are doing inside the loop and hence inside each loop its getting over-written and you are getting the last value.

$data = '';
while($row = mysqli_fetch_assoc($result)){
        $data .= '<article>';
        $data .= '<h3>' . $row['title'] . '</h3>';
        $data .= '<p>' . $row['content'] . '</p>';
        $data .= '<footer>By: ' . $row['whoposted'] . ' / ' . $row['dateposted'] . '</footer>'; 
        $data .= '</article>';
    }

Upvotes: 3

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