Php – Warning: thesqli_error() expects exactly 1 parameter, 0 given error

mysqliPHP

I get the following error

Warning: mysqli_error() expects exactly 1 parameter, 0 given

The problem is with this line of the code:

$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error());

The whole code is

session_start();

require_once "scripts/connect_to_mysql2.php";

//Build Main Navigation menu and gather page data here

$sqlCommand = "SELECT id, linklabel FROM pages ORDER BY pageorder ASC";

$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error()); 

$menuDisplay = '';
while ($row = mysqli_fetch_array($query)) { 
    $pid = $row["id"];
    $linklabel = $row["linklabel"];

    $menuDisplay .= '<a href="index.php?pid=' . $pid . '">' . $linklabel . '</a><br />';

} 
mysqli_free_result($query);

The included file has the following line

$myConnection = mysqli_connect("$db_host","$db_username","$db_pass","$db_name") or die ("could not connect to mysql"); with reference to $myConnection, why do I get this error?

Best Answer

mysqli_error() needs you to pass the connection to the database as a parameter. Documentation here has some helpful examples:

http://php.net/manual/en/mysqli.error.php

Try altering your problem line like so and you should be in good shape:

$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error($myConnection));

Related Solutions

Php – thesql_fetch_array()/thesql_fetch_assoc()/thesql_fetch_row()/thesql_num_rows etc… expects parameter 1 to be resource

A query may fail for various reasons in which case both the mysql_* and the mysqli extension will return false from their respective query functions/methods. You need to test for that error condition and handle it accordingly.

mysql_* extension:

NOTE The mysql_ functions are deprecated and have been removed in php version 7.

Check $result before passing it to mysql_fetch_array. You'll find that it's false because the query failed. See the mysql_query documentation for possible return values and suggestions for how to deal with them.

$username = mysql_real_escape_string($_POST['username']);
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '$username'");

if($result === FALSE) { 
    die(mysql_error()); // TODO: better error handling
}

while($row = mysql_fetch_array($result))
{
    echo $row['FirstName'];
}

mysqli extension
procedural style:

$username = mysqli_real_escape_string($mysqli, $_POST['username']);
$result = mysqli_query($mysqli, "SELECT * FROM Users WHERE UserName LIKE '$username'");

// mysqli_query returns false if something went wrong with the query
if($result === FALSE) { 
    yourErrorHandler(mysqli_error($mysqli));
}
else {
    // as of php 5.4 mysqli_result implements Traversable, so you can use it with foreach
    foreach( $result as $row ) {
        ...

oo-style:

$username = $mysqli->escape_string($_POST['username']);
$result = $mysqli->query("SELECT * FROM Users WHERE UserName LIKE '$username'");

if($result === FALSE) { 
    yourErrorHandler($mysqli->error); // or $mysqli->error_list
}
else {
    // as of php 5.4 mysqli_result implements Traversable, so you can use it with foreach
    foreach( $result as $row ) {
      ...

using a prepared statement:

$stmt = $mysqli->prepare('SELECT * FROM Users WHERE UserName LIKE ?');
if ( !$stmt ) {
    yourErrorHandler($mysqli->error); // or $mysqli->error_list
}
else if ( !$stmt->bind_param('s', $_POST['username']) ) {
    yourErrorHandler($stmt->error); // or $stmt->error_list
}
else if ( !$stmt->execute() ) {
    yourErrorHandler($stmt->error); // or $stmt->error_list
}
else {
    $result = $stmt->get_result();
    // as of php 5.4 mysqli_result implements Traversable, so you can use it with foreach
    foreach( $result as $row ) {
      ...

These examples only illustrate what should be done (error handling), not how to do it. Production code shouldn't use or die when outputting HTML, else it will (at the very least) generate invalid HTML. Also, database error messages shouldn't be displayed to non-admin users, as it discloses too much information.

Php – Warning: thesqli_error() expects parameter 1 to be thesqli, string given in

There are three main problems with your code:

Wrong SQL. You have to learn basic SQL syntax prior trying to use SQL from PHP.
Wrong mysqli syntax. You have to use prepared statements instead of adding variables right in the query.
Wrong error reporting. You have to set mysqli in exception mode instead of checking every query result manually.

To solve them one by one, read these three helpful answers:

Best Answer

Related Solutions

Php – thesql_fetch_array()/thesql_fetch_assoc()/thesql_fetch_row()/thesql_num_rows etc… expects parameter 1 to be resource

Php – Warning: thesqli_error() expects parameter 1 to be thesqli, string given in

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