PHP - Mysqli Select Statement

I wrote a code with prepared statements MySQLi: Code: [Select]
<?php 
include_once 'dbinfo.php';
if(isset($_POST['kuldes']))
{
$name = trim($_POST['nev']);
$username = $_POST['felh_nev'];
$password = $_POST['jelszo'];
$email = $_POST['email'];
$phone = $_POST['telefon'];
$gender = $_POST['sex'];
$hobby = $_POST['hobby'];
$regfelt = $_POST['regfelt'];
$name = strip_tags($name);
$name = stripslashes($name);
$username = strip_tags($username);
$email = strip_tags($email);
$phone = strip_tags($phone);
$date = date("d-m-Y");
if($name == NULL || $username == NULL || $password == NULL || $email == NULL || $phone == NULL || $gender == NULL)
{
    echo "Please complete the form below or one of the boxes is empty.";
}
elseif(strlen($username) <= 3 || strlen($username) >= 30){
    $final_report.="Your username must be between 3 and 30 characters..";
    }
elseif($stmt = $connect->prepare('SELECT * FROM users WHERE username=?'))
{    $stmt->bind_param('s', $username);
    $stmt->execute();
    $stmt->bind_result($username);
    while ($stmt->fetch())
    {
        printf("Name: %s\n", $name);
        $final_report.="The username is already in use!";
    }
    $stmt->close();
}elseif(strlen($password) <= 6 || strlen($password) >= 12){
    $final_report.="Your password must be between 6 and 12 digits and characters..";
    }
elseif(!eregi("^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$", $email)){ 
    $final_report.="Your email address was not valid..";
    }
elseif(!eregi("^[0-9]{1,3}-[0-9]{1,3}-[0-9]{1,10}$",$phone)){
    $final_report.="Phone number is invalid. Only numbers with hyphen. Allowed format: countrycode-areacode-phonenumber";
    }
elseif(!isset($hobby)){
    $final_report.="Youd didn't select any hobbies";
    }
elseif(!isset($regfelt)){
    $final_report.="You didn't accept the terms";
    }
else
    {
if ($stmt = $connection->prepare('INSERT INTO users(name,sex,email,phone_number,username,password,hobby) VALUES(?, ?, ?, ?, ?, ?, ?)'))
{
    $stmt->bind_param('sssssss', $name, $sex, $email, $phone_number, $username, $password, $hobby);
    $stmt->execute();
    $stmt->close();
}  
}}?>
<h1>Registration Form</h1>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" name="registration_form" method="POST">
   <p>Name: <input type="text" name="nev" value="<?php echo (isset($name) ? $name : '');  ''?>" size=25></p>
   <p>Username: <input type="text"  name="felh_nev" value="<?php echo (isset($username) ? $username : ''); ?>" size=10></p>
  <p>Password: <input type="password" name="jelszo" size=10></p>
  <!--<p>Password again:<input type="password" name="password_confirmation"></p>-->
    <p>E-mail: <input type="text" name="email" value="<?php echo (isset($email) ? $email : ''); ?>"/></p>
    <p>Phone number: <input type="text" name="telefon" value="<?php echo (isset($phone) ? $phone : ''); ?>"/></p>
   <p>Sex:
       <label><input type="radio" name="sex" value="no" >Female</label>
       <label><input type="radio" name="sex" value="ferfi" >Male</label></p>   
   <p>Favorite hobbies (Using CTRL you can select more than one):</p>
      <select name="hobby[]" size="4" multiple>
        <option value="sport">Sport</option>
        <option value="mozi">Movies</option>
        <option value="kirandulas">Hiking</option>
        <option value="olvasas">Reading</option>
  </select>
  <p><input name="regfelt" type="checkbox" value="elfogad">I accept the terms!</p>
   <p><input name="kuldes" type="submit" value="Submit form">
  <input name="reset" type="reset" value="delete"></p>
   <table width="501" border="1">
      <tr>
        <td><?php echo $final_report; ?></td>
      </tr>
    </table>
   <p>&nbsp;</p>
</form>And I get the following error message: Warning: mysqli_stmt::bind_result() [mysqli-stmt.bind-result]: Number of bind variables doesn't match number of fields in prepared statement in
I don't understand what's the problem is, many people can't give solution for this? Anyone who can help me? It's brain racking.

Hi,

I want to restructure my database tables so that I can have one table (t_incidents) to hold foreign keys instead of holding foreign keys in the table "t_persons" because one person can commit more than one offense.

However, I need to know how the join should be implemented with MySQLi prepared statement.

I need some one to review the following statement for me and advise:

if ($stmt = $mysqli->prepare("Select t_persons.PersonID    
                                                                    
        ,t_persons.FamilyName
        ,t_persons.FirstName
        ,t_persons.OtherNames
        ,t_persons.Gender
        ,t_persons.CountryID
        ,t_persons.ImagePath    

        ,t_incidents.IncidentID
        ,t_incidents.Incident
        ,t_incidents.IncidentDate
        ,t_incidents.PersonID
        ,t_incidents.CountryID
        ,t_incidents.OffenceKeywordID
        ,t_incidents.StatusID
        ,t_incidents.AgencyID

        ,t_status.StatusID
        ,t_status.Status

        ,t_offencekeyword.KeywordID
        ,t_offencekeyword.Keyword

        ,t_countries.CountryID
        ,t_countries.Country

        ,t_agencies.AgencyID
        ,t_agencies.Agency

From t_persons
    Inner Join t_incidents On t_persons.PersonID = t_incidents.PersonID
    Inner Join t_incidents On t_countries.CountryID = t_incidents.CountryID
    Inner Join t_incidents On t_status.StatusID = t_incidents.StatusID
    Inner Join t_incidents On t_offenceskeyword.KeywordID = t_incidents.KeywordID
    Inner Join t_incidents On t_agencies.AgencyID = t_incidents.AgencyID

Where t_persons.PersonID = '$PersonID'")) {

Regards.

josephbupe

I have two tables...one is scheduled_umps and one is games.  Both tables have a column 'game_id'. 

I need to select how many games a person is scheduled for in a given time frame.

Here's what I have for the select...and it's not working (You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE scheduled_umps.ump_id = '34' AND games.game_id = scheduled_umps.game_id AN' at line 1)

Code: [Select]
<?php
$today_is_this_date = date('Y-m-d');
$four_months_ago = date('Y-m-d', strtotime('-4 months', strtotime($today_is_this_date)));
$four_months_from_now = date('Y-m-d', strtotime('+4 months', strtotime($today_is_this_date)));

/// How many games have you worked? 
//$query = "SELECT COUNT(ump_id) FROM scheduled_umps WHERE `ump_id` = '$_SESSION[ump_id]'";   ///this one selects how many games you've worked NO MATTER WHAT THE DATE

$query = "SELECT scheduled_umps.game_id, games.game_id, COUNT(scheduled_umps.ump_id) WHERE scheduled_umps.ump_id = '$_SESSION[ump_id]' AND games.game_id = scheduled_umps.game_id AND games.game_id BETWEEN $four_months_ago AND $four_months_from_now";

$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
echo "You have been scheduled on <b>". $row['COUNT(ump_id)'] ."</b> dates for the $_SESSION[association_name].";
}
?>
[\code]

Any ideas where I'm going wrong?  Thanks for the help...I HATE multi-table selects!

I have a Select statement that relies on two session variables to complete the statement. One variable, $egroup, supplies a column name. I cannot make the SELECT work. It works fine if I substitute the real name of one of the columns  such as "egroup1", or "egroup6", but that's what the variable $egroup is for. Can anyone tell me where I'm wrong?

I guess the answer is in the syntax, but I'm not very good at this and I've been on it for too long. Any help gratefully taken.

Code: [Select]
<?php                 
$query1 = mysql_query("SELECT * FROM topics WHERE managerId='".$managerId."' AND "$egroup"= 1 ORDER BY title ASC");
?>

$sql=mysql_query("SELECT * FROM `buds` WHERE `level`<='$user_level' UNION SELECT * FROM `buds`, `unlocked_buds` WHERE buds.`id` = unlocked_buds.`bud_id` ORDER BY buds.`id` ASC") or die("A MySQL error has occurred.<br />Your Query: " . $sql . "<br /> Error: (" . mysql_errno() . ") " . mysql_error());



I have been trying to learn about UNION select statements. I ran the query above and got this response:

Quote

Error: (1222) The used SELECT statements have a different number of columns



I think I know what the problem is, but not sure how to fix it.

There is two columns, one is "buds" which holds the flowers seeds info. The seconds is "unlocked_buds" which just links the "id" from buds to "user_id" to the user table.

Both buds and unlocked_buds are both 8 columns.

What do I need to learn?