Proper mysqli_query and mysqli_error configuration

Question

When launching my code I get a failed query and the following errors:

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

mysqli_error() expects parameter 1 to be mysqli, string given in

<?php
include('mysql_config.php');

function mysqlConnect()
{
    global $mysql_hostname, $mysql_username, $mysql_password, $mysql_database;
    $link = mysqli_connect($mysql_hostname, $mysql_username, $mysql_password) 
    or die('Could not connect: ' . mysqli_error());
    mysqli_select_db($link,$mysql_database) or die('Could not select database');
    return $link;
}

function mysqliClose($link)
{
    mysqli_close($link);
}

function sendQuery($query)
{
    $result = mysqli_query($link, $query) or die('Query failed: ' . mysqli_error("could not query"));
    return $result;
}

?>

How do I properly format the mysqli_query and mysqli_error functions?

Where did `$link` come from? Check your `sendQuery` function.. Error is clear.. `$link` is not defined.. — Bhuvan Rikka, Dec 22 '12 at 12:53
You have a scope problem: http://php.net/manual/en/language.variables.scope.php — Pekka, Dec 22 '12 at 12:53
You better consider to change `or die(...)` approach to error handling. See http://www.phpfreaks.com/blog/or-die-must-die for starters. — peterm, Dec 22 '12 at 20:51
Does this answer your question? [Warning: mysqli\_query() expects parameter 1 to be mysqli, null given in](https://stackoverflow.com/questions/18862743/warning-mysqli-query-expects-parameter-1-to-be-mysqli-null-given-in) — Dharman, Jan 04 '20 at 20:40

hek2mgl · Accepted Answer · 2012-12-22T20:48:35.343

2

There are two errors in the code above:

You missed to declare $link global as $mysql_hostname etc.
You passed the wrong argument type to mysqli_error() it expects mysqli and you passed a string

I have changed your example:

<?php

include('mysql_config.php');

// declaring an additional global var.
$link = NULL;

function mysqlConnect()
{
    global $link; // using the global $link
    global $mysql_hostname, $mysql_username, $mysql_password, $mysql_database;
    $link = mysqli_connect($mysql_hostname, $mysql_username, $mysql_password) 
    or die('Could not connect: ' . mysqli_connect_error());
    mysqli_select_db($link,$mysql_database) or die('Could not select database');
    return $link;
}

function mysqliClose($link)
{
    mysqli_close($link);
}

function sendQuery($query)
{
    global $link; // using the global $link
    $result = mysqli_query($link, $query) or die('Query failed: '
      . mysqli_error($link)); // note $link is the param
    return $result;
}

edited Dec 22 '12 at 20:48

answered Dec 22 '12 at 13:02

hek2mgl

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In a statement `die('Could not connect: ' . mysqli_error())` you better use `mysqli_connect_error()` instead of `mysqli_error()`. See [mysqli_connect](http://php.net/manual/en/mysqli.construct.php) – peterm Dec 22 '12 at 20:42
1

It is a very bad idea to use `die(mysqli_error($conn));` in your code, because it could potentially leak sensitive information. See this post for more explanation: [mysqli or die, does it have to die?](https://stackoverflow.com/a/15320411/1839439) – Dharman Jan 04 '20 at 20:39
@Dharman I agree on that. I just took the code from the OP. Not sure if it's worth to change the answer 8 years later, especially because I'm not coding PHP since years :) ... Feel free to edit it! – hek2mgl Jan 04 '20 at 21:26

Proper mysqli_query and mysqli_error configuration

1 Answers1