mysqli_fetch_array()/mysqli_fetch_assoc()/mysqli_fetch_row() expects parameter 1 to be resource or mysqli_result, boolean given
I am trying to select data from a MySQL table, but I get one of the following error messages:
mysql_fetch_array() expects parameter 1 to be resource, boolean given
or
mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
This is my code:
$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');
while($row = mysql_fetch_array($result)) {
echo $row['FirstName'];
}
The same applies to code like
$result = mysqli_query($mysqli, 'SLECT ...');
// mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
while( $row=mysqli_fetch_array($result) ) {
...
and
$result = $mysqli->query($mysqli, 'SELCT ...');
// Call to a member function fetch_assoc() on a non-object
while( $row=$result->fetch_assoc($result) ) {
...
and
$result = $pdo->query('SLECT ...', PDO::FETCH_ASSOC);
// Invalid argument supplied for foreach()
foreach( $result as $row ) {
...
and
$stmt = $mysqli->prepare('SLECT ...');
// Call to a member function bind_param() on a non-object
$stmt->bind_param(...)
and
$stmt = $pdo->prepare('SLECT ...');
// Call to a member function bindParam() on a non-object
$stmt->bindParam(...)Answer:
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.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:
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.
http://stackoverflow.com/questions/2973202/mysqli-fetch-array-mysqli-fetch-assoc-mysqli-fetch-row-expects-parameter-1
