Mysqli fetch_object is giving me hassles - php

All this section is supposed to do is collect an int from active and use it in the if statement to continue with the code. Can somebody please tell me why this is not working?
$act_qry = "Select active FROM user_m WHERE username = '$username' and password = '$Menrypted_password'";
$result_act = mysqli_query ( $connMS, $act_qry );
$value_act = mysqli_fetch_object($result_act);
if($value_act == 1)
{
//Do php stuff.
}

I am having this table in my database:-http://prntscr.com/9tnalx
Check this code:-
<?php
error_reporting(E_ALL);
ini_set('display_errors',1);
//connect to database
$link = mysqli_connect('localhost','root','','stack');
$act_qry = "Select product_id FROM bom WHERE bom_description = 'Table Tops'";
$result_act = mysqli_query ( $link, $act_qry ) or die(mysqli_error($link));
$value_act = mysqli_fetch_object($result_act);
if($value_act->product_id == 1)
{
echo $value_act->product_id;
}
// or you can do this
$value_act = mysqli_fetch_assoc($result_act);
if($value_act['product_id'] == 1)
{
echo $value_act->product_id;
}
mysqli_close($link);
?>
Output on my browser:- http://prntscr.com/9tnazj
Note:- I hope you can understand the code by checking my screenshot of table.Thanks

Related

Fail to update the data in database

I try to update the data in the database, but when I run the code, there is no error message appear, looks like its a logical error but I still don't have any clue about what is happening with my code.
Here is the code
<?php
include("conn.php");
SESSION_START();
if($_SESSION["loggedin"]!="true"&& $_SESSION['login'] != '')
header("location:login.php");
$aid = $_SESSION["usr"];
$result = mysql_query("select r.CustomerID from customer c inner join results r on r.CustomerID = c.CustomerID where c.Username = '".$aid."' ");
if (false === $result) {
echo mysql_error();
}
$row = mysql_fetch_assoc($result);
?>
<?php
if (isset($_POST["submitbtn"]))
{
$bookid = $_POST["bookid"];
$LP = $_POST["LP"];
$budget = $_POST["budget"];
$smokep = $_POST["SmokeP"];
$spreq = $_POST["sp_req"];
$query = mysql_query("UPDATE `results` SET LP = '$LP', budget = '$budget', SmokeP = '$smokep', sp_req = '$spreq'
WHERE results.BookID = '".$bookid."' and results.CustomerID = '".$result."'");
if (false === $query)
{
echo mysql_error();
}
?>
<script type = "text/javascript">
alert("Amendment Saved!!");
</script>
<?php
}
?>
Is the error coming from the select query? Or the if statement for the submitbtn went wrong?
First of all you cant put session start here
You must put it on the first line after open php tag
Second
update res='$ new_value' where ...
Tell me if it's not usefull to try another solution

Array objects not being printed in input fields and sql query not receiving the id value

I am getting the id from another page but i am not being able to pass it to the sql query. If i define any value to $id instead of 0 then the query works but otherwise it fails.
Secondly, i would like to display the values of the array in respective input fields. I tried using
<?php
echo $result_array['institutename'][0];
?>
in the body part but it didnt work out.
My rest code is as follows:
(I know the mysql functions are deprecated but i would move on to mysqli as soon as i have solved this problem)
<?php
include 'connect.php';
$id=0;
$result_array=array();
if(isset($_REQUEST['id'])){
$id=(int)$_REQUEST['id'];
//$uid=$id;
if(!empty($id)){
$sql = "SELECT * FROM institute WHERE id =$id";
$result = mysql_query($sql);
while($row = mysql_fetch_assoc($result)){
$result_array[]=$row;
}
}
}
if ($_SERVER['REQUEST_METHOD'] == 'POST' && $_POST['form_institutedetails'] == 'saveinstitutedetails')
{
$mysql_table='institute';
$institutename = $_POST['institutename'];
$established = $_POST['established'];
$regno = $_POST['reg_no'];
$branch = $_POST['branch'];
$initials = $_POST['initials'];
$address=$_POST['address'];
$pin=$_POST['pin'];
$contact1=$_POST['contact1'];
$contact2=$_POST['contact2'];
$contact3=$_POST['contact3'];
$fax1=$_POST['fax1'];
$fax2=$_POST['fax2'];
$email=$_POST['email'];
$website=$_POST['website'];
if(isset($_POST['head_office'])){
$head_office=$_POST['head_office'];
}
else{
$head_office="Branch";
}
if (!preg_match("/^.+#.+\..+$/", $email))
{
$error_message = 'Email is not a valid email address. Please check and try again.';
}
if (empty($error_message))
{
$newinstitutename = mysql_real_escape_string($institutename);
$newestablished = mysql_real_escape_string($established);
$newregno = mysql_real_escape_string($regno);
$newbranch = mysql_real_escape_string($branch);
$newaddress = mysql_real_escape_string($address);
$newpin = mysql_real_escape_string($pin);
$newemail = mysql_real_escape_string($email);
$newwebsite = mysql_real_escape_string($website);
$ho = mysql_real_escape_string($head_office);
include 'connect.php';
$sql = "UPDATE `".$mysql_table."` SET `institutename`='$newinstitutename', `established`='$newestablished', `regno`='$newregno', `branch`='$newbranch', `initials`='$initials', `address`='$newaddress', `pin`='$newpin', `contact1`='$contact1', `contact2`='$contact2', `contact3`='$contact3', `fax1`='$fax1', `fax2`='$fax2', `email`='$newemail', `website`='$newwebsite', `head_office`='$ho' WHERE `id`=$id";
$result = mysql_query($sql, $db);
mysql_close($db);
$error_message='Updated Successfully!.';
}
}
?>
When you are unsure about the structure of an array, you can always do a print_r during development.
print_r($result_array);
In this case, it is an index array of associative arrays.
To access the first record's institutename (and probably the only record since it looks like you used an unique key in your query), you can use
echo $result_array[0]['institutename'];

How to save table data in session

I have problem in little project,
how can I save table data in session?
<?php
session_start();
include 'connect.php';
if (isset($_POST["email"]))
{
$email = $_POST["email"];
$password = $_POST["password"];
$r=mysql_query("SELECT * FROM user_login WHERE `uemail` ='".$email."' AND `upass` = '".$password."'");
$s = $_POST["userid"];
$n=mysql_query("SELECT * FROM user_data WHERE `userid` ='".$s."'");
$q=mysql_fetch_assoc($n);
$_SESSION["name"]=$q["nfname"];
$k=mysql_num_rows($r);
if ($k>0)
{
header("location:user/index.php");
}
else
header("location:login.php");
}
?>
this code not working !! :(
please help !
You probably just missed the
session_start();
But here is the dildo (deal tho) xD
Your Login script is not secure, try this at the top of your index.php or whatever rootfile you have.
<?php
session_start();
function _login($email, $password) {
$sql = "SELECT * FROM user_login
WHERE MD5(uemail) ='".md5(mysql_real_escape_string($email))."'
AND MD5(upass) = '".md5(mysql_real_escape_string($password))."'";
$qry = mysql_query($sql);
if(mysql_num_rows($qry) > 0) {
// user with that login found!
$sql = "UPDATE user_login SET uip = '".$_SERVER['REMOTE_ADDR']."', usession = '".session_id()."'";
mysql_query($sql);
return true;
} else {
return false;
}
}
function _loginCheck() {
$sql = "SELECT * FROM user_login WHERE uip = '".$_SERVER['REMOTE_ADDR']."' AND MD5(usession) = '".md5(session_id())."'";
$qry = mysql_query($sql);
if(mysql_num_rows($qry) > 0) {
// user is logged in
$GLOBALS['user'] = mysql_fetch_object($qry);
$GLOBALS['user']->login = true;
} else {
// user is not logged in
$GLOBALS['user'] = (object) array('login' => false);
}
}
if(isset($_POST['login'])) {
if(_login($_POST["email"], $_POST["password"])) {
// login was successfull
} else {
// login failed
}
}
_loginCheck(); // checkes every Page, if the user is logged in or if not
if($GLOBALS['user']->login === true) {
// this user is logged in :D
}
?>
Ok, I'll bite. First 13ruce1337, and Marc B are right. There is a lot more wrong with this than not being able to get your data into your session.
Using PDO ( as 13ruce1337 links you too ) is a must. If you want to keep using the same style of mysql functions start reading up on how. Marc B points out that session_start(); before any html output is required for sessions to work.
As for your code, you got along ways to go before it is ready for use but here is an example to get you started
if (isset($_POST["email"])) {
//mysql_ functions are being deprecated you can instead use
//mysqli_ functions read up at http://se1.php.net/mysqli
/* Manage your post data. Clean it up, etc dont just use $_POST data */
foreach($_POST as $key =>$val) {
$$key = mysqli_real_escape_string($link,$val);
/* ... filter your data ... */
}
if ($_POST["select"] == "user"){
$r = mysqli_query($link,"SELECT * FROM user_login WHERE `uemail` ='$email' AND `upass` = '$password'");
/* you probably meant to do something with this query? so do it*/
$n = mysqli_query($link,"SELECT * FROM user_data WHERE userid ='$userid'");
//$r=mysql_fetch_assoc($n); <- this overrides your user_login query
$t = mysqli_fetch_array($n);
$_SESSION["name"] = $t['nfname'];
/* ... whatever else you have going on */

SQL / PHP if NULL show else

We are trying to complete / fix the below code, We cant seem to do the following.
Check if 'Check_Activation' is set to 'NULL' within the Database
IF value is NULL direct the user to one of the forms (1,2,3)
And finally if the 'Check_Activation' has already been activated and isn't 'NULL' prevent user from accessing one of the 3 forms.
I know its basicly there but we can't seem to figure out the final bug.
Please have a quick look at the code below and if anyone notices anything that isn't right please advice us.
Paste Bucket / Version
Website URL
<?php
$username = $_POST['username'];
$activation_code = $_POST['activation_code'];
$activation_codeurl = $activation_code;
$usernameurl = $username;
$db_host = "localhost";
$db_name = "aardvark";
$db_use = "aardvark";
$db_pass = "aardvark";
$con = mysql_connect("localhost", $db_use, $db_pass);
if (!$con){
die('Could not connect: ' . mysql_error());
}
mysql_select_db($db_name, $con);
$checkcustomer = mysql_query("SELECT `Check_Activation` FROM `members` WHERE `Username` = '".$username."' & `Activation` = '".$activation_code."'; ");
$array = mysql_fetch_array($checkcustomer);
if (is_null($array['Check_Activation'])) {
$username = substr($username, 0, 1);
if($username == '1') {
$redirect_url='form-one.php?membernumber='.$usernameurl.'&activation_code='.$activation_codeurl;
} elseif($username == '2') {
$redirect_url='form-two.php?membernumber='.$usernameurl.'&activation_code='.$activation_codeurl;
} elseif($username == '3') {
$redirect_url='form-three.php?membernumber='.$usernameurl.'&activation_code='.$activation_codeurl;
}
header("Location:". $redirect_url);
}
else
{
?>
Try this, You need to fetch the row from table and then you can check the values,
$val = mysql_fetch_array($checkcustomer);
if (is_null($val['Check_Activation']))
instead of
$val = mysql_query($checkcustomer);
if ($val == 'NULL')
NOTE: Use mysqli_* functions or PDO instead of using mysql_* functions(deprecated)
before I get into the technicality of what your are trying to accomplish, I have some advice for your code in general. You should avoid using the mysql api as it is deprecated, and use the mysqli api instead. I think you will also find that it is easier to use.
Now for the code:
You have this line in your code which seems to be incorrect, $checkcustomer is a result set from your previous query, so why are you running it as a query again?
$val = mysql_query($checkcustomer);
You already have the result set so do this:
$array = mysql_fetch_array($checkcustomer);
And then take the value of Check_Aviation;
if (is_null($array['Check_Aviation'])) {
//Do Something
}
Should solve your issue

problem with fetching results from mysql via php function

can some one point out the problem with this code? It supposed to fetch data from mysql but it returns blank. Here is the full code.
<ul class="list">
<?php
require("object/db.class.php");
error_reporting(0);
function entry_tickets() {
if($_SESSION['dojopress_global:root'] == false) {
$entry_ticket .= <<<ENTRY_TICKET
<li><p><img src="http://cdn1.iconfinder.com/data/icons/humano2/32x32/apps/gnome-keyring-manager.png" />Access denied</p></li>
ENTRY_TICKET;
} elseif($_SESSION['dojopress_global:root'] == true) {
$q = "SELECT * FROM `notice` ORDER BY nid LIMIT 12 DESC";
$r = mysql_query($q);
if ( mysql_num_rows($r) == 0 ) {
$entry_ticket .= <<<ENTRY_TICKET
<li><p><img src="http://cdn1.iconfinder.com/data/icons/humano2/32x32/status/dialog-information.png" /> Nothing to display</p></li>
ENTRY_TICKET;
} elseif ( $r !== false && mysql_num_rows($r) > 0 ) {
while ( $a = mysql_fetch_array($r) ) {
$nid = stripslashes($a['nid']);
$note = stripslashes($a['note']);
$type = stripslashes($a['type']);
$private = stripslashes($a['private']);
$date = stripslashes($a['date']);
$author = stripslashes($a['author']);
function note_type($type) {
if($type == 1) { $type = "posted a comment!"; } elseif($type == 2) { $type = "raised a support ticket!"; } else { }
return ($type);
}
$entry_ticket .= <<<ENTRY_TICKET
<li><p> $athor, note_type($type)</p></li>
ENTRY_TICKET;
return $entry_ticket;
}
}
}
}
echo entry_tickets();
?>
</ul>
<div style="clear:both;height:10px;"></div>
sorry forgot db.class.php
<?php
session_start();
//connect.php
$host = "localhost";
$db_user = "root";
$db_pass = "";
$db_name = "au";
$connectClass = mysql_connect("$host", "$db_user", "$db_pass") or die ("Couldn't establish connection to database server.");
$dbObject = mysql_select_db("$db_name", $connectClass) or die ("Couldn't select database.");
?>
error reporting disabled error code
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\wamp\www\ageis\note.php on line 22
Your mysql syntax looks bad. You have:
SELECT * FROM `notice` ORDER BY nid LIMIT 12 DESC
try
SELECT * FROM `notice` ORDER BY nid DESC LIMIT 12
Erik's comments should have pointed you in the right direction, however:
Figure out where PHP logs errors. Either turn up php's error reporting level so you see more errors, or start tailing your apache error_log
You should always check for errors after running mysql_query and do some sort of logging on failure. This is probably best accomplished by writing a wrapper function for mysql_query that does this, or using a 3rd party db wrapper that has already solved this problem.
You're redefining function note_type every time through your while loop. You have a return call outside the function, below it. Really I can't see how this is even syntactically correct. It looks like you have a large problem with mismatched brackets.
As for an actual database issue, you should check mysql_error() if no rows return from the call because it's likely something went wrong.
Also I recommend using PDO which is a true database class instead of the native function based ones.

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