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


LearningScholar
Go to solution Solved by QuickOldCar,

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Hey guys,

 

Earlier I mentioned making search criteria in my database. I also had to put in something to make it add to the database. I made some code like this:

<form id="form" name="form" method="post" action="Website2.php">
  <p>
    <label for="Genre">Genre</label>
    <input type="text" name="Genre" id="Genre" />
  </p>
  <p>
    <label for="Naam">Naam</label>
    <input type="text" name="Naam" id="Naam" />
  </p>
  <p>
    <label for="Jaar">Jaar</label>
    <input type="text" name="Jaar" id="Jaar" />
  </p>
  <p>
    <label for="Regisseur">Regisseur</label>
    <input type="text" name="Regisseur" id="Regisseur" />
  </p>
  <p>
    <input type="submit" name="Verzenden" id="Verzenden" value="Verzenden" />
  </p>
</form>
<?php
	$Genre = $_POST["Genre"];
	$Naam = $_POST["Naam"];
	$Jaar = $_POST["Jaar"];
	$Regisseur = $_POST["Regisseur"];
		/*Hier hoeft geen verbinden meer gemaakt te worden met de database*/
				if (isset($_POST['Verzenden']) && trim($_POST['Verzenden'])!=''){
                                $sql = "INSERT INTO 
				Movies (Genre, Naam, Jaar, Regisseur)
				VALUES ('$Genre', '$Naam', '$Jaar', '$Regisseur')";
					$resultaat = mysqli_query($db, $sql);
					$verbreken = mysqli_close($db);
					echo "De gegevens van $Naam zijn opgeslagen in de database.";}
                                else echo 'Hier kunt u iets toevoegen';
?>

However it didn't work.

I got the problem  "Warning: mysqli_query() expects parameter 1 to be mysqli, null given in...  on line 119

Can you guys help please?

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  • Solution

Along with what cyberRobot said...

 

For your own good, take a look at mysqli_real_escape_string() and protect those $_POST variables before using them in your query.

 

Furthermore you should check all of those $_POST values are set and not empty.

Some changes.

<form id="form" name="form" method="post" action="Website2.php">
<p>
<label for="Genre">Genre</label>
<input type="text" name="Genre" id="Genre" />
</p>
<p>
<label for="Naam">Naam</label>
<input type="text" name="Naam" id="Naam" />
</p>
<p>
<label for="Jaar">Jaar</label>
<input type="text" name="Jaar" id="Jaar" />
</p>
<p>
<label for="Regisseur">Regisseur</label>
<input type="text" name="Regisseur" id="Regisseur" />
</p>
<p>
<input type="submit" name="Verzenden" id="Verzenden" value="Verzenden" />
</p>
</form>

<?php
if (isset($_POST['Verzenden']){
include("/path/to/database/connection");//<--database connection http://php.net/manual/en/function.mysqli-connect.php
$errors = array();
if(isset($_POST['Genre']) && trim($_POST['Genre']) != ''){

    $Genre = mysqli_real_escape_string($db, trim($_POST['Genre']));

} else {
    $errors[] = 'No Genre';
}

if(isset($_POST['Naam']) && trim($_POST['Naam']) != ''){

    $Naam = mysqli_real_escape_string($db, trim($_POST['Naam']));

} else {
    $errors[] = 'No Naam';
}


if(isset($_POST['Jaar']) && trim($_POST['Jaar']) != ''){

    $Genre = mysqli_real_escape_string($db, trim($_POST['Jaar']));

} else {
    $errors[] = 'No Jaar';
}

if(isset($_POST['Regisseur']) && trim($_POST['Regisseur']) != ''){

    $Regisseur = mysqli_real_escape_string($db, trim($_POST['Regisseur']));

} else {
    $errors[] = 'Regisseur';
}
	
    /*Hier hoeft geen verbinden meer gemaakt te worden met de database*/
if (empty($errors)) {
    $sql = "INSERT INTO
    Movies (Genre, Naam, Jaar, Regisseur)
    VALUES ('$Genre', '$Naam', '$Jaar', '$Regisseur')";
    $resultaat = mysqli_query($db, $sql);

    if(!$resultaat){
        echo 'No query resultaat';  
    } else {
        echo "De gegevens van $Naam zijn opgeslagen in de database.";
    }
    $verbreken = mysqli_close($db);
} else {
	foreach($errors as $error){
	    echo "$error <br />";
}
} else {
    echo "Hier kunt u iets toevoegen <br />";
}
?>
Edited by QuickOldCar
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using 'mysqli_real_escape_string' worked

 

 

Using that just escaped the variables

 

but are the '$errors' and '$error' necessary?

 

Because I wrote to only execute the queries if no errors...yes

 

If you wanted to remove them for some reason you can, but then you need something like this

 

instead of

if (empty($errors)) {

would do

if($Genre && $Naam && $Jaar && $Regisseur){

or

if(isset($Genre) && isset($Naam) && isset($Jaar) && isset($Regisseur)){

I doubt you want any empty columns in your database.

Otherwise you should set a default in your database or the query fails

Could be a custom default, NULL

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