sultankhan

how can my admin see all user and maonitor them please any one help me

what can i should do with this code that it show me my user profile which is login <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("test", $con); $result = mysql_query("SELECT * FROM users where id='GET'"); ?> <div class= "nine"><table border="0" cellpadding="1" cellspacing="0" id="tblsubmit" …

i have store an image into my database. but i am unable to display the image what i have done till now is under.... any help will be appreciated. thanks in advance! <? $query="SELECT * from testimonial"; $ret = mysqli_query($mysql,$query); if (isset($ret) && $ret->num_rows>0) { while($row=mysqli_fetch_array($ret)) { $body=$row['body']; $name=$row['name']; $image=$row['img']; …

hi evry one may be this a tip task or easy but for me it is not easy i have show.php and showpost.php page in show.php i am showing my picture while in showpost.php i am showing post now if a user add picture it should be on of my …

i have image table with the follwing column `ano(my image id) aphote adteail exactar ` 6 image` and i have an comments table comeents `cid(comt id) user1(userid),ano(iamge id) comments` ` 1 10 here is problem hiiiiiiiii` i want if a user comments on picture then the picture id should add …

hi every one i have create a table with name upload id int notnull name varchar notnull type varchar notnull size varchar notnull content varchar not null and my code is the follwing but instead of iamge it show garbage value <form method="post" enctype="multipart/form-data" action="image.php"> <table width="350" border="0" cellpadding="1" cellspacing="1" …

i have the follwing table name email pass name1 email1 pass1 name2 eamil2 pass2 how can i cahnge name1 from my admin page please any one help me

hiii! Can any one help me to add super user to my login script that can veiw and edit all user, and evry user can veiw his/her profile like facebook. hope you people will help me.