How to use PHP to connect to a MySQL database and run a query against it

Question

I am able to connect to my database, as it says Connection Successful when I go to the page. The problem is that I'm still not able to run a query against my database and have the results display.

I've tried following the instructions that were given to me in my class and kept getting HTTP 500 Error. So, I started to search the internet and have my config file set up differently than what my instructor said and was able to get it to say that the connection is successful.

Here is the main file:

<?php

    $configd = include('config2.php');
    $conn = mysqli_connect($configd['dbhost'], $configd['dbuser'], $configd['dbpass'], $configd['dbname']);

    if (!$conn) {
      die("Connection failed: " . mysqli_connect_error());
    }
      echo "Connected successfully";

    $sql = "SELECT Last_Name, First_Name FROM sample";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
      while($row = $result->fetch_assoc()) {
        echo "Last Name: " . $row["Last_Name"]. "<br>";
        echo "First Name: " . $row["First_Name"]. "<br><hr>"
      }
    } else {
      echo "0 results";
    }
    $conn->close();
?>

Here is the config file:

<?php
return array(
 dbhost => 'localhost',
 dbuser => 'cdefauw',
 dbpass => 'sherbert00',
 dbname => 'sample',
);

?>

When this is run, it's supposed to show a query of all of the first and last names that are within the sample database I made on phpmyadmin. When I run it right now, all I get is HTTP Error 500. If I remove the if ($result->num_rows >0) statement, then it says connection successful.

You aren't checking for error status returned by `query()`. This is a pretty basic mistake, even though it's common. Check for errors every time you use `query()`. See https://www.php.net/manual/en/mysqli.error.php for an example. — Bill Karwin, Jul 07 '19 at 18:03

Mike Foxtech · Answer 1 · 2019-07-07T19:28:12.723

-1

check like this

if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

or

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

edited Jul 07 '19 at 19:28

answered Jul 07 '19 at 18:08

Mike Foxtech

1,633
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your assumption (as well as one of code examples) are wrong – Your Common Sense Jul 07 '19 at 19:10
@YourCommonSense which exactly? – Mike Foxtech Jul 07 '19 at 19:12
`it is wrong` one – Your Common Sense Jul 07 '19 at 19:12
I wrote that this code is incorrect – Mike Foxtech Jul 07 '19 at 19:13
1

**checking `if (!$conn) {`is not wrong**. mysqli_connect returns false on error – Your Common Sense Jul 07 '19 at 19:17
yes, but in this case it is false. Because the object will be created in any case. And there will never be fulfilled !$conn – Mike Foxtech Jul 07 '19 at 19:20
it returns FALSE. try it yourself and see – Your Common Sense Jul 07 '19 at 19:23
Yes you are right – Mike Foxtech Jul 07 '19 at 19:29
it is mysqli to blame. Although it is stated in the manual that new mysqli and mysqli_connect are synonyms, but new mysqli indeed always returns an object but mysql_connect returns an object on success but false on error, for backward compatibility I believe – Your Common Sense Jul 07 '19 at 19:33
On top of it all, I see no need to check for connection errors this way. Enable exceptions and you can catch them if needed. `die/exit` is a very bad habit. – Dharman Jul 07 '19 at 22:55

How to use PHP to connect to a MySQL database and run a query against it

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