警告：mysqli_stmt :: bind_param（）变量数与预准备语句中的参数数量不匹配

Question

Bear with me I realize there are many posts about this error. But I have looked through them and can't seem to find the solution for my particular problem.

I have tried "isi", "sss", and "sii" doesn't work. Not sure what to do any help would be great.

Not too sure where to use var_dump(); to find out more information about my problem. I'm pretty new to PHP that would help a lot too, to know.

Thanks for bearing with me! Sorry about a repeat topic.

code:

<tr>
             <form action="Voting_action.php" method="post">
            <td><br />
                <input type="submit" class="buttontable1" value="<?php echo $random; ?>" name="name"/>
            </td>
                </form>
                <form action="Voting_action.php" method="post">
            <td><br />
                <input type="submit" class="buttontable1" value="<?php echo $random3; ?>" name="name"/>
            </td>
                </form>
         </tr>
         <tr>
             <form action="Voting_action.php" method="post">
            <td><br />
                <input type="submit" class="buttontable1" value="<?php echo $random6; ?>" name="name"/>
            </td>
                </form>
                <form action="Voting_action.php" method="post">
            <td><br />
                <input type="submit" class="buttontable1" value="<?php echo $random4; ?>" name="name"/>
            </td>
                </form>
         </tr>
         <tr>
             <form action="Voting_action.php" method="post">
            <td><br />
                <input type="submit" class="buttontable1" value="<?php echo $random5; ?>" name="name"/>
            </td>
                </form>
                <form action="Voting_action.php" method="post">
            <td><br />
                <input type="submit" class="buttontable1" value="<?php echo $random2; ?>" name="name"/>
            </td>
</tr>

<?php
include ('login-home.php');
$mysqli = new mysqli("", "", "", "");
if ($mysqli->connect_error) {
    echo "Failed to connect to MySQL: (" . $mysqli->connect_error . ") " . $mysqli->connect_error;
}

if (!($stmt = $mysqli->prepare("INSERT INTO table(id, name, votes) VALUES (id, '".$_POST['name']."', '".$votes."')"))) {
    echo "Prepare failed: (" . $mysqli->error . ") " . $mysqli->error;
}
$id = 1;

This line:

if (!$stmt->bind_param("isi",$id, $_POST['name'], $votes)) {
    echo "Binding parameters failed: (" . $stmt->error . ") " . $stmt->error;
}

if (!$stmt->execute()) {
    echo "Execute failed: (" . $stmt->error . ") " . $stmt->error;
}
$stmt->close();
?>

Answer 1

Your query preparation contains zero parameters, since you just dumped the values in there, completely defeating the purpose of prepared statements. Instead, try this:

 if (!($stmt = $mysqli->prepare("INSERT INTO table(id, name, votes) VALUES (?,?,?)"))) {

Then continue as you are.