Warning: mysqli_query() expects parameter 1 to be mysqli, null given in

Question

I am trying to build a simple custom CMS, but I'm getting an error:

Warning: mysqli_query() expects parameter 1 to be MySQLi, null given in

Why am I getting this error? All my code is already MySQLi and I am using two parameters, not one.

$con=mysqli_connect("localhost","xxxx","xxxx","xxxxx");

//check connection
if (mysqli_connect_errno($con))
{
    echo "Failed to connect to MySQL:" . mysqli_connect_error();
}

function getPosts() {
    $query = mysqli_query($con,"SELECT * FROM Blog");
    while($row = mysqli_fetch_array($query))
    {
        echo "<div class=\"blogsnippet\">";
        echo "<h4>" . $row['Title'] . "</h4>" . $row['SubHeading'];
        echo "</div>";
    }
}

This is a scoping issue. You need to pass `$con` to `getPosts()` — Phil, Sep 18 '13 at 02:06

Phil · Accepted Answer · 2019-11-04T02:57:45.303

30

As mentioned in comments, this is a scoping issue. Specifically, $con is not in scope within your getPosts function.

You should pass your connection object in as a dependency, eg

function getPosts(mysqli $con) {
    // etc

I would also highly recommend halting execution if your connection fails or if errors occur. Something like this should suffice

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); // throw exceptions
$con=mysqli_connect("localhost","xxxx","xxxx","xxxxx");

getPosts($con);

edited Nov 04 '19 at 02:57

answered Sep 18 '13 at 02:10

Phil

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1

so this is unneed in mysql? for some reason this only happens with mysqli? – Philip Sep 18 '13 at 02:12
4

The (deprecated) MySQL extension would use the last known connection if the `$link_identifier` argument was not passed in to functions like `mysql_query`. The MySQLi extension offers no such convenience. – Phil Sep 18 '13 at 02:16

score 11 · Answer 2 · answered Jul 03 '14 at 06:07

use global scope on your $con and put it inside your getPosts() function like so.

function getPosts() {
global $con;
$query = mysqli_query($con,"SELECT * FROM Blog");
while($row = mysqli_fetch_array($query))
    {
        echo "<div class=\"blogsnippet\">";
        echo "<h4>" . $row['Title'] . "</h4>" . $row['SubHeading'];
        echo "</div>";
    }
}

score 0 · Answer 3 · answered Sep 18 '13 at 02:08

The getPosts() function seems to be expecting $con to be global, but you're not declaring it as such.

A lot of programmers regard bald global variables as a "code smell". The alternative at the other end of the scale is to always pass around the connection resource. Partway between the two is a singleton call that always returns the same resource handle.

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in

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