mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result

General Tech Bugs & Fixes 2 years ago

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Posted on 16 Aug 2022, this text provides information on Bugs & Fixes related to General Tech. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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manpreet Tuteehub forum best answer Best Answer 2 years ago

 

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

or

Call to a member function fetch_array() on boolean / non-object

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, 'SELECT ...');
// mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
while( $row=mysqli_fetch_array($result) ) {
    ...

and

$result = $mysqli->query($mysqli, 'SELECT ...');
// Call to a member function fetch_assoc() on a non-object
while( $row=$result->fetch_assoc($result) ) {
    ...

and

$result = $pdo->query('SELECT ...', PDO::FETCH_ASSOC);
// Invalid argument supplied for foreach()
foreach( $result as $row ) {
    ...

and

$stmt = $mysqli->prepare('SELECT ...');
// Call to a member function bind_param() on a non-object
$stmt->bind_param(...);

and

$stmt = $pdo->prepare('SELECT ...');
// Call to a member function bindParam() on a non-object
$stmt->bindParam(...);
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manpreet 2 years ago

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

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