CODEWITHSUNDEEP
phpmysqlmysqliwhile-loop
Radu C.
Radu C.

Reputation: 13

mysqli_fetch_array only displays the first result

I struggled all day to display the results of an SQL query using PHP. I have a table in the database named coins with the following columns: - nr_unic (which is the index);
- rank;
- name;
- symbol;
- price_usd;
- price_btc;
What I need to do is to fetch the values of each coin (from the name field) and display the symbol, price_usd, price_btc, and rank. The piece of code which contains the query I am running and trying to display the values is:

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);

    // SQL QUERY
    $sql= "SELECT rank, name, symbol, price_usd, price_btc, 24h_volume_usd FROM coins WHERE rank BETWEEN 1 AND 10 ORDER BY nr_unic DESC LIMIT 10";
    $result = $conn->query($sql);
    $rows = array();
    if ($result) {
    while($row = mysqli_fetch_array($result, MYSQLI_BOTH)) {
    $rows[] =  $row['name'] . " " . $row['price_usd'] . " " . $row['symbol'] . " " . $row['rank'];
    foreach ($rows as $key => $value) {
    echo $value;
    }
    }
    mysqli_free_result ($result);
    }

Thank you!

LATER EDIT Following @Máté Solymosi indications I managed updated the code and to display the results. The problem now is they are getting duplicated: I get the first coin, then the first and the second, then the first, second and third... and so on. The code I use was updated

Upvotes: 1

Views: 1002

Answers (1)

Mate Solymosi
Mate Solymosi

Reputation: 5977

The return statement in your while loop causes the function to terminate immediately, returning just the first row. Instead, you should collect the results in an array and return the array at the end:

$rows = array();

while($row = mysqli_fetch_array($result, MYSQLI_BOTH)) {
  $rows[] = $row['name'] . " " . // ...
}

mysqli_free_result($result);
return $rows;

Upvotes: 2

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