PHP - Mysqli And Php Problem

hi guys

I have two functions and they work well, but now i have a problem in the last if

the email is sent like www.yoursite.com/reset_password.php?userid=0&code=

and not http://your.url/set_new_password.php?userid=564979&code=54c4a2767c2f485185ab72cdcf03ab59

so, the problem is that i can get the value of hash and the userid.
If i do an echo "$hash" in the function code(); it shows the value like 564979&code=54c4a2767c2f485185ab72cdcf03ab59
my question is, how i can do the same in the last if? above two functions. Same occur in the last select select userid from password_reset where code=? none userid is showed, it is always zero.


Code: [Select]
<?
function check($sql, $db, $email) {
if(!empty($_POST['email'])) {

$email = $_POST["email"];
if ($sql = $db->prepare("select email from users where email=?")) {
$sql->bind_param('s', $email);

$sql->execute();

$sql->bind_result($email);

if ($sql->fetch()) {
return true;
}
else {
return false;
}
}
}
}



function code($sql, $db, $hash, $pwdHasher, $userExists, $sendPass) {

if (check($sql, $db, $email)) {

$pwdHasher = new PasswordHash(8, FALSE);

$hash = $pwdHasher->HashPassword($userExists["email"]);

$sendPass=$hash;
($sql = $db->prepare('insert into password_reset (code) values (?)'));
$sql->bind_param('s', $hash);

$sql->execute();
$sql->fetch();
return true;
}
}




if (code($sql, $db, $hash, $pwdHasher, $userExists, $sendPass)) {
($sql = $db->prepare("select userid from password_reset where code=?"));

$sql->bind_param('s', $hash);
$sql->execute();
$sql->bind_result($hash);
$sql->fetch();
echo $hash;
$pwrurl = "www.yoursite.com/reset_password.php?userid=" .$hash . "&code=" . $sendPass;

$mailbody = "Dear user,<br><br>If this e-mail does not apply to you please ignore it. It appears that you have requested a password reset at our website www.yoursitehere.com<br>
To reset your password, please click the link below. If you cannot click it, please paste it into your web browser's address bar.<br> <a href='$pwrurl'>$pwrurl</a>  <br> <br>
Thanks,\nThe Administration";

$mail->MsgHTML($mailbody);

$mail->AddAddress($email,"Membro");

$mail->IsHTML(true);

if(!$mail->Send()) {
echo "Deu erro: " . $mail->ErrorInfo;
} else {
echo "Enviado com sucesso";
}


$sql->close();
$db->close();

}


?>



any help?

thanks

Ok I am trying to use mysqli instead of the usual mysql.  Mysql would be outdated.
With mysqli, sgl-injection is impossible if you use the "?" in those codes.

I would normally use a function but  I've made a simple script to find the error.

I use $parameters and $sql because these are the data I need to give as parameters to the function, so I used it here too
but without the function actually.

Code: [Select]
ini_set('display_errors',1); // 1 == aan , 0 == uit
error_reporting(E_ALL | E_STRICT);
# sql debug
define('DEBUG_MODE',true);  // true == aan, false == uit


$userid = 11;
$lang = 1;
$newLink = "testing123";

$db_host = "localhost";
$db_gebruiker = "root";
$db_wachtwoord = '';
$db_naam = "projecteasywebsite";

$sql= "INSERT tbl_link(userid,linkcat,linksubid,linklang,linkactive,linktitle)  VALUES(?, ?, ?, ?, ?, ?)";
$parameters = '"iiisis", $userid, 1, 0, $lang, 1, $newLink';

echo $parameters;

$mysqli = new mysqli($db_host, $db_gebruiker, $db_wachtwoord, $db_naam);
$stmt = $mysqli->prepare($sql);

$stmt->bind_param($parameters); 

$stmt->execute();

echo "<br><br>". mysqli_connect_errno();
echo "<br><br>". mysqli_report(MYSQLI_REPORT_ERROR);

$stmt->close();
$mysqli->close();


I got Wrong parameter count for mysqli_stmt::bind_param() 

So naturally a problem when we execute : Warning: mysqli_stmt::execute() [mysqli-stmt.execute]: (HY000/2031): No data supplied for parameters in prepared statement    ($stmt->execute()

Is someone using mysqli too ?

Hello everyone,

For two weeks now, I'm trying to get this database connection in my query. Can someone give me a solution and tell me what I've done wrong? Am I overlooking something?

<?php

class Mysql{
public function connect(){
$mysqli = new mysqli('localhost','root','','login');
}
}

class Query extends Mysql{
public function runQuery(){
$this->result = parent::connect()->query("select bla bla from bla bla");
}
}

$query = new Query;
$query->runQuery();

?>

hello , I'm starting to use mysqli and i have few questions.

is there a guide for mysqli?

and how do i use this functions at mysqli ?
mysql_num_rows
mysql_query
mysql_fetch_assoc
mysql_fetch_array

thanks , Mor.

Hi, The following code is what I want in that it creates a menu and I can select and display a table row. I still need to use that selection to update the "lastused". I really appreciate your help.

 

<!DOCTYPE><html><head><title>email menu</title></head>     
    <body><center>
    <form name="form" method="post" action="">
    <?php
    $con=mysqli_connect("localhost","root","cookie","homedb");
    //============== check connection
    if(mysqli_errno($con))
    {echo "Can't Connect to mySQL:".mysqli_connect_error();}
    else
    {echo "Connected to mySQL</br>";}
       //This creates the drop down box
    echo "<select name= 'target'>";
    echo '<option value="">'.'--- Select email account ---'.'</option>';
    $query = mysqli_query($con,"SELECT target FROM emailtbl");
    $query_display = mysqli_query($con,"SELECT * FROM emailtbl");
    while($row=mysqli_fetch_array($query))
    {echo "<option value='". $row['target']."'>".$row['target']
    .'</option>';}
    echo '</select>';
    ?>
    <input type="submit" name="submit" value="Submit"/><!-- update "lastused" using selected "target"-->
    </form></body></html>

<!DOCTYPE><html><head><title>email menu</title></head>    
    <body><center>
    <?php
    $con=mysqli_connect("localhost","root","cookie","homedb");
    if(mysqli_errno($con))
    {echo "Can't Connect to mySQL:".mysqli_connect_error();}
        if(isset($_POST['target']))
      {
    $name = $_POST['target'];
    $fetch="SELECT target,username,password,emailused,lastused, purpose, saved FROM emailtbl WHERE target = '".$name."'";
    $result = mysqli_query($con,$fetch);
    if(!$result)
    {echo "Error:".(mysqli_error($con));}
   $lastused = "CURDATE()"; // update "lastused" using selected "target"
    //display the table
    echo '<table border="1">'.'<tr>'.'<td bgcolor="#ccffff align="center">'. 'Email menu'. '</td>'.'</tr>';
    echo '<tr>'.'<td>'.'<table border="1">'.'<tr>'.'<td bgcolor="#ccffff align="center">'.'target'.'</td>'.'<td bgcolor="#ccffff align="center">'.'username'.'</td>'.'<td bgcolor="#ccffff align="center">'.'password'.'</td>'.'<td bgcolor="#ccffff align="center">'.'emailused'.'</td>'.'<td bgcolor="#ccffff align="center">'.'lastused'.'</td>'.'<td bgcolor="#ccffff align="center">'.'purpose'. '</td>'.'<td bgcolor="#ccffff align="center">'. 'saved' .'</td>'.'</tr>';
    while($data=mysqli_fetch_row($result))
    {echo ("<tr><td>$data[0]</td><td>$data[1]</td><td>$data[2]</td><td>$data[3]</td><td>$data[4]</td><td>$data[5]</td><td>$data[6]</td></tr>");}
    echo '</table>'.'</td>'.'</tr>'.'</table>';
      }
    ?>
       </body></html>



 

I have just started using MySQLi and am clueless it is giving me the follow errors in which i do not understand Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:\xampp\htdocs\Login\connect.php on line 23

Notice: Trying to get property of non-object in C:\xampp\htdocs\Login\connect.php on line 25

Notice: Use of undefined constant mysqli - assumed 'mysqli' in C:\xampp\htdocs\Login\connect.php on line 32

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\Login\connect.php on line 32

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in C:\xampp\htdocs\Login\connect.php on line 33    can someone please explain to me why i am getting these?   and my code is

$mysqli_db = mysqli_select_db("$db_name");

if($mysqli_db->connect_errno) {

	printf("Database not found: %s\n", $mysql->connect_error);
	exit();
}

$sql = "SELECT * FROM $tbl_name WHERE username='$username' AND password='$password'";
$result = mysqli_query($sql);
$row = mysqli_fetch_assoc($result);

I just got rid off most the errors the only ones left are  Warning: mysqli_query() expects at least 2 parameters, 1 given in C:\xampp\htdocs\Login\connect.php on line 32

Fatal error: Call to undefined function mysqli_result() in C:\xampp\htdocs\Login\connect.php on line 33   Code Updated: 

$mysqli_db = mysqli_select_db($mysqli_connect, $db_name);

if(!$mysqli_db) {

	printf("Database not found: %s\n", $mysqli->connect_error);
	exit();
}

$sql = "SELECT * FROM $tbl_name WHERE username='$username' AND password='$password'";
$query = mysqli_query($sql);
$result = mysqli_result($query);
$row = mysqli_fetch_assoc($result);

Edited by Tom8001, 30 November 2014 - 12:43 PM.

When running the following code i get the error:

Call to undefined method mysqli::errno() 

the code:
$conn = new mysqli(HOST, USER, PASSWORD, DATABASE);
if ($conn->errno() !== 0)
{
$msg = $conn->error();
throw new connErrorException($msg, 'Connect');
}

I am fairly new to classes but as i understand it this should be correct. I am using mysql 5.1 so mysqli is on by default. I have even checked the php ini and everything looks fine there in respect to this. Any advice?

I am using mysqli, OO, to connect to MySQL.

I have only today started looking at this and am used to:
Code: [Select]
<?php
$con = mysql_connect();//etc
mysql_close($connection);
?>

Am I right that with mysqli (OO) that I don't need to set a connection variable wither when connecting or closing??
Code: [Select]
<?php
mysqli::connect();//etc
mysqli::close();
?>

What about with multiple databases, does mysqli keep track for me, as I am used to this:
Code: [Select]
<?php
$con1 = mysql_connect();//db1
$con2 = mysql_connect();//db2
?>
//etc

The below code produces a dropdown and when a selection is made and submitted produces
a table row with a link to the record but the link doesn't work.  suggestions?

---------------------------------------------------------------------------

<!DOCTYPE><html><head>
<title>lookup menu</title>
</head>
<body><center><b>
<form name="form" method="post" action="">

<?php
// error_reporting(0);
error_reporting(E_ALL ^ E_NOTICE);
include 'homedb-connect.php';

//This creates the drop down box
echo "<select name= 'target'>";
echo '<option value="">'.'--- Select account ---'.'</option>';
$query = mysqli_query($con,"SELECT target FROM lookuptbl");
$query_display = mysqli_query($con,"SELECT * FROM lookuptbl");
while($row=mysqli_fetch_array($query))
{echo "<option value='". $row['target']."'>".$row['target']
.'</option>';}
echo '</select>';
?>
<input type="submit" name="submit" value="Submit"/>
</form><center>

<?php
// error_reporting(0);
error_reporting(E_ALL ^ E_NOTICE);
include 'homedb-connect.php';

if(isset($_POST['target']))
{
$name = $_POST['target'];
$fetch="SELECT target, purpose, user, password, email, visits, date,
 saved FROM lookuptbl WHERE target = '".$name."'";
$result = mysqli_query($con,$fetch);
if(!$result)
{echo "Error:".(mysqli_error($con));}

//display the table
        echo '<table border="1"><tr><td bgcolor="#ccffff" align="center">lookup menu</td></tr>
        <tr><td>
        <table border="1">
        <tr>
        <td> Target </td>
        <td> Purpose </td>
        <td> User </td>
        <td> Password </td>
        <td> Email </td>
        <td> Visits </td>
        <td> Date </td>
        <td> Saved </td>
        </tr>';

        while($data=mysqli_fetch_row($result))                              
          {
$url= "http://localhost/home/crud-link.php?target=". $data[0];
$link= '<a href="'.$url.'">'. $data[0]. '</a>';

echo ("<tr><td> $link </td><td>$data[1]</td><td>$data[2]</td><td>$data[3]</td>
<td>$data[4]</td><td>$data[5]</td><td>$data[6]</td><td>$data[7]</td></tr>");
}     
        echo '</table>
        </td></tr></table>';
  }
?>
</body></html>

 

I dont know whether the statement is correct.....i just tried it.....and it didn't work.

   $stmt->bind_param('ssiiiss',$_POST['name'],$_POST['email'],$_POST['d'],$_POST['m'],$_POST['y'],$_POST['add'],$_POST['phone']);

here my first two values are strings and next 2 tiny int's next is int and last 2 again strings.

The following code is supposed to submit everything in the database on a table, Currently its doing the first row in the database. Not sure whats wrong..


function getNews(){
$query = $this->con->query("SELECT * FROM `". $this->prefix ."news`");
while ($result = $query->fetch_assoc()) {
if($result['serveradded'] == 1){
$queryy = $this->con->query("SELECT * FROM `". $this->prefix ."news` WHERE serveradded=1");
while ($resultt = $queryy->fetch_assoc()) {
return "<tr><td>The Advertisement ". ucfirst($resultt['name']) ." was created</td><td>". $resultt['date'] ."</td></tr>";
}
} elseif($result['servupdate'] == 1){
$queryyy = $this->con->query("SELECT * FROM `". $this->prefix ."news` WHERE servupdate=1");
while ($resulttt = $queryyy->fetch_assoc()) {
return "<tr><td>".  $result['type'] ." ". ucfirst($resulttt['name']) ." was Updated by ". ucfirst($resulttt['updatedby']) ." </td><td>". $resulttt['date'] ." </td></tr>";
}
} elseif($result['newuser'] == 1){
$queryyyy = $this->con->query("SELECT * FROM `". $this->prefix ."news` WHERE newuser=1");
while ($resultttt = $queryyyy->fetch_assoc()) {
return "<tr><td>Account Created: ". ucfirst($resultttt['name']) ." was created</td><td>". $resultttt['date'] ."</td></tr>";
}

}

}

}

I am trying to extend my knowledge from using basic mysql_connect etc. into something better.

I am searching on Google, and looking up as much info as possible but find that I learn best with a brief overview that allows me to channel down my searches.

(1) I know PDO v mysql_connect v mysqli isn;t right - I think PDO is a class, mysql_connect is a command and mysqli is something else. Am I on the right track?

(2) What is the best solution? I am using a full OOP coding style and want to learn the best - I am a novice at this but would consider myself an 8/10 programmer at other parts of PHP.

An very brief overview to get me started and channel my Google searches would be much appreciated.

- I have heard PDO is slower.
- I don't need support for different databases (only MySQL).

Thanks in advance.

Greeting all.

I am brand new to the forum and I would like to ask a question regarding mysqli.

I am not certain why the following code works:
Code: [Select]
...
$sqlCmd = "select * from usrdata";
$rs = $mysqli->query($sqlCmd);
...

and the following code gives me a 500 internal error:
Code: [Select]
...
$sqlCmd = "select * from usrdata";
$result= dbQuery($sqlCmd);
...
function dbQuery($sqlCmd){
   $rt = array();
   $rs = $mysqli->query($sqlCmd);
   while($row = $rs->fetch_assoc ()){
    $rt[] = $row;
   }
   return 'records: ' . json_encode($rt);
}

The error occurs on the mysqli->query.  Thanks for any help. Cheers!

Hey.
I was writing my class object for a database connection and while I was writing the query part, I came to wonder whether I should use mysql_real_escape_string or mysqli_real_escape_string to make the query mysql friendly, what's the difference?

Dear Sir/Madame

I am making a website where user can insert data and wait for the admin to approve/reject the form. Now i am stuck with the update status where an admin can submit with a click pending to approval or reject with comments. I am new to PHP programming. Can somebody help me with the issue. Part 1 is inserting the data and part two is fetching the data but i am unable to solve the status approve/reject and comment at the same time on the view.php? page. Kindly help. Thank you. 

<?php


$host="localhost";
$username="root";
$pass="";
$db="ems1";

$conn=mysqli_connect($host,$username,$pass,$db);
if(!$conn){
	die("Database connection error");
}

// insert query for register page
if(isset($_REQUEST['proposal']))
{

$details=$_POST['details'];
$location=$_POST['location'];
$date=$_POST['date'];
$time=$_POST['time'];

$status="Pending";
$comment=$_POST['comment'];
	
	
	 $query="INSERT INTO `proposal` (`details`,`location`,`date`,`time`,`status`,`comment`) VALUES ('$details','$location','$date','$time','$status','$comment')";
	
	$res=mysqli_query($conn,$query);
	if($res){
		$_SESSION['success']="Not Inserted successfully!";
		header('Location:');
	}else{
		echo "Leave not Applied, please try again!";
	}
	}
?>

<div class="col-xs-6 col-xs-push-3 well">
<form class="form-horizontal" method="post" action="" >
<input type="hidden" name="proposal" value="">
  <fieldset>
    <legend>New Proposals  </legend>
	        <!----left box----------->
		<!----right box----------->
	<div class="col-xs-9">
<div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b>Details:</b></label>
      <div class="col-lg-9">
        <input type="text" name="details" class="form-control">
      </div>
    </div>
   <div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b>Location:</b></label>
      <div class="col-lg-9">
        <input type="text" name="location" class="form-control" >
      </div>
    </div>
	<div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b>Date:</b></label>
      <div class="col-lg-9">
        <input type="date" name="date"  class="form-control">
      </div>
    </div>
	<div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b>Time:</b></label>
      <div class="col-lg-9">
        <input type="time" name="time"  class="form-control" >
      </div>
    </div>
	
      <div class="col-lg-9">
          <input type="hidden" name="status" class="form-control" >
      </div>
    </div>
	<div class="form-group">
      <label for="inputEmail" class="col-lg-3"><b></b></label>
      <div class="col-lg-9">
          <input type="hidden" name="comment"  class="form-control">
      </div>
    </div>
	</div>
	
    
    <div class="form-group">
      <div class="col-lg-12">
        <button type="reset" class="btn btn-default">Cancel</button>
        <button type="submit" class="btn btn-primary">Submit</button>
      </div>
    </div>
  </fieldset>
</form>
</div>
</div>	

<body>
    
<h2 style="text-align:center; color:orangered;">
DASHBOARD
</h2>
<table>   
<h3>
                    <tr style="background-color:#E4EBC5; color:orangered;">

                    <th>ID</th>
                    <th>Details</th>
                    <th>Location</th>
                    <th>Status</th>
                    <th>Comment</th>
                    

                    </tr>
                    </h3>
</table>
    
  
<?Php
////////////////////////////////////////////
require "dbconfig.php"; // MySQL connection string

$count="SELECT  id,details,location,time,status,comment FROM proposal";

    
if($stmt = $connection->query($count)){


while ($nt = $stmt->fetch_assoc()) {
echo "
  

<body>
<table>

<tr>   
   
  <td><a href=view.php?id=$nt[id]>$nt[id]</a></td>
  <td>$nt[details]</td>
  <td>$nt[location]</td>
  <td>$nt[status]</td>
  <td>$nt[comment]</td>
  

     </tr>
     
      </table>
  </body>
  ";
   
}

}else{
echo $connection->error;
}
?>