Jump to content

sungpeng

Members
 View Profile  See their activity

  • Posts

    290

  • Joined

  • Last visited

 Content Type 

Posts posted by sungpeng

    [SOLVED] simple script cannot work

    in PHP Coding Help

    Posted 

    
<?php

if($_POST[action]=="Submit")
{




include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/config.php"); 






$loginname=$_POST[loginname];
echo "$loginname";
$insert_student="insert into users ('llid','loginname') values ('3','$loginname')"; 
mysql_query($insert_student) or die ("cannot insert");

   





?>



<?php }else{ ?>



                 
<?php include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/header.php");  ?>

<html>

<head>

<title>Agents Register</title>

</head>

<body>

                 

                     <table width="400" align="center">
                    <br> <form name="form" action="<?php print $PHP_SELF ?>" onSubmit="return validate(this);" method="post">
                      

                         
                        <tr> 
                          <td width="35%">Loginname</td>
                          <td><input type="text" name="loginname" size="25"></td>
                        </tr>
                        
                         

                        <tr> 
                          <td align="center" colspan="2"><input type="submit" name="action" value="Submit"></td>
                        </tr>
                      </table>
                    </form>




          
</body>

</html>
<?php 
}
?>
   

     

    Can anyone help, I don't know why simple script why cannot work? always show cannot insert

    [SOLVED] error on script

    in PHP Coding Help

    Posted 

    <?php
include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/config.php");
$uid='sungpeng1';
$query = "SELECT * from users where loginname='$uid'";
while($resu=mysql_query($query)){
echo "$resu[llid]";
}
?>

     

    Hi I see nothing wrong with my script, but it give me an error FATAL ERROR maximum execution time is 30 seconds on line 6. Why is that so? Can anyone help?

    [SOLVED] error message mysql num rows

    in PHP Coding Help

    Posted 

    
<?php


if($_POST[action]=="Update")
{	
include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/config.php");
$indexnumber=$_GET[llid];		
$user=$_POST[user];
$passwd=$_POST[pass];
echo $user;
echo $passwd;
echo $indexnumber;

$check_result=0;
$check_login=mysql_query("select * from users where email=$user and passwd=$passwd and llid=$indexnumber");
$check_result=mysql_num_rows($check_login);
echo $check_result;



}else{ ?>






<html>

<title>Upload an image to a database</title>


<body>




<h2>Update with new information</h2>

<form name="form" enctype="multipart/form-data" method="post" action="<?php echo"$PHP_SELF?llid=$_GET[llid]"; ?>">
Please insert the images
<input type='file' name='imagefile'><br>


<br>
USER : <input name="user"><br>
PASSWORD : <input name="pass"><br>


<input type="submit" name="action" value="Update">
</form>

</body>
</html>

<?php
} ?>

     

    Can help I keep having error message warning: mysql_num_row(): supplied argument is not a valid MySql result resource in insertphoto.php on line 17

    [SOLVED] Upload images error

    in PHP Coding Help

    Posted 

    <?php

include 'config2.php';
if($_POST[action]=="Update")
{	

$check_result=0;
$check_login=mysql_query("select * from users where email='$_POST[user]' and passwd='$_POST[pass]' and pid='$_GET[pid]'");
$check_result=mysql_num_rows($check_login);		


}

if($check_result>0)
{

$indexnumber=$_GET[pid];
$folder_path="photo/$indexnumber/";


if (file_exists("photo/$indexnumber"))
{
}
else
{
mkdir($folder_path,0777);
}

echo $indexnumber;


		$file_upload_err=0;
		$max_file_size="10000000"; 
		$max_file_size_kb="500";
		$file_upload_err_msg="";

		if($_FILES['imagefile']['size']>$max_file_size)
		{
		echo "image file too big";	
            

            }elseif
($imagefile && ($_FILES['imagefile']['type']!="image/jpeg" && $_FILES['imagefile']['type']!="image/pjpeg"))
		{
			$type=$_FILES['imagefile']['type'];
				echo "image file type wrong";
                
		}





}else{ ?>






<html>

<title>Upload an image to a database</title>


<body>




<h2>Update with new information</h2>

<form name="form" enctype="multipart/form-data" method="post" action="<?php echo"$PHP_SELF?pid=$_GET[pid]"; ?>">
Please insert the images
<input type='file' name='imagefile'><br>


<br>
USER : <input name="user"><br>
PASSWORD : <input name="pass"><br>


<input type="submit" name="action" value="Update">
</form>

</body>
</html>

<?php
} ?>

     

    Hi I keep editing the above code but it seem not working. Can anyone help? Parse error: syntax error?

     

    [SOLVED] display only "registration done" after form processed

    in PHP Coding Help

    Posted 

    
<?php

if($_POST[action]=="Submit")
{




include ("".$_SERVER['DOCUMENT_ROOT']."/includes/config.php"); 

$loginname=$_POST["loginname"];
echo "registration done";

}
?>


                 
<?php 
if(!$action) 
{
?>


<html>

<head>

<title>Register</title>

</head>

<body>       

                     <table width="<?php print $number1 ?>" align="center">
                    <br> <form name="form" action="<?php print $PHP_SELF ?>" onSubmit="return validate(this);" method="post">
                      

                       
                        <tr> 
                          <td width="35%">Login name</td>
                          <td><input type="text" name="loginname" size="25"></td>
                        </tr>
                       
                        <tr> 
                          <td align="center" colspan="2"><input type="submit" name="action" value="Submit"></td>
                        </tr>
                      </table>
                    </form>    
</body>

</html>
<?php 
}
?>
   

     

    Hi good day, Can I check with you how to display only "registration done" after the form processed. Currently after the form processed, it still show the bottom form.

    [SOLVED] error when dealing with session

    in PHP Coding Help

    Posted

    housing accesscontrol.php

    <?php 
include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/config.php");

session_start();

if (isset($_POST['uid'])) {
$uid = $_POST['uid'];
} else {
$uid = $_SESSION['uid'];
}
if (isset($_POST['pwd'])) {
$pwd = md5($_POST['pwd']);
} else {
$pwd = $_SESSION['pwd'];
}


if(!isset($uid) || !isset($pwd) )
{
  ?>
  <html>
  <head>
  <title> Please Log In for Access </title>
  </head>
<body>
  <table align=center width=300 border=0 cellspacing=0 cellpadding=0 bgcolor="#2f4f4f">
  <tr><td>
   <table border=0 width=100% cellspacing=1 cellpadding=1>
    <form action="<?=$_SERVER['PHP_SELF']?>" method=POST>
    <tr><td BGCOLOR="#2f4f4f"><FONT SIZE="-1" FACE="Verdana,Tahoma,Arial,Helvetica,sans-serif" COLOR="#FFFFFF">
    <B>Please Log In For Access:</B>
    </td></tr>
    <tr><td BGCOLOR="#c7c7c7"><FONT SIZE="-1" FACE="Verdana,Tahoma,Arial,Helvetica,sans-serif">
You must log in to access this area of the site.
     </td></tr>
    <tr>
     <td BGCOLOR="#fffff0">
      <table width=100% border=0 cellspacing=0 cellpadding=0>
    <tr>
     <td><FONT SIZE="-1" FACE="Verdana,Tahoma,Arial,Helvetica,sans-serif">Email Address:</td>
     <td><input type=text name="uid" size="20" value=""></td>
    </tr>
        <tr>
     <td><FONT SIZE="-1" FACE="Verdana,Tahoma,Arial,Helvetica,sans-serif">Password:</td>
     <td><input type=password name="pwd" size="20"></td>
    </tr>
    <tr>
     <td colspan=2 align=center>
      <input type=submit name="Login" value="Login">
     </td>
    </tr>
    </form>
      </table>
     </td>
    </tr>
   </table>
  </td></tr>
</table>
  </body>
  </html>
  <?php
  exit;
}
//Clean the input submitted to mysql
$uid=addslashes($uid);
$pwd=addslashes($pwd);

//this puts the variable into the session

$_SESSION['uid'] = $uid; 
$_SESSION['pwd'] = $pwd;

$sql = "SELECT * FROM users WHERE email = '$uid' AND passwd = '$pwd' ";

$result = mysql_query($sql);

if (!$result) {
echo "A database error occurred while checking your login details";
}
//if bad user/pass combo access denied
if (mysql_num_rows($result) == 0) {

  unset($_SESSION['uid']);
  unset($_SESSION['pwd']);
  ?>
  <html>
  <head>
  <title> Access Denied </title>
  </head>
  <body>
  <h1> Access Denied </h1>
  <p>There are several reasons this may be happening:<BR>
  <UL><LI>Your username or password is incorrect</LI>
  <LI>You have forgotten your login information. <a href="password_reset.php">Lost Password</a></LI></UL>
  To return to our login page, <a href="index.php">click here</a>.</p>
  </body>
  </html>
    <?php
  exit;
}

?>

    [SOLVED] error when dealing with session

    in PHP Coding Help

    Posted

     

    accesscontrol.php

    <?php 
include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/config.php");

session_start();

if (isset($_POST['uid'])) {
$uid = $_POST['uid'];
} else {
$uid = $_SESSION['uid'];
}
if (isset($_POST['pwd'])) {
$pwd = md5($_POST['pwd']);
} else {
$pwd = $_SESSION['pwd'];
}


if(!isset($uid) || !isset($pwd) )
{
  ?>
  <html>
  <head>
  <title> Please Log In for Access </title>
  </head>
<body>
  <table align=center width=300 border=0 cellspacing=0 cellpadding=0 bgcolor="#2f4f4f">
  <tr><td>
   <table border=0 width=100% cellspacing=1 cellpadding=1>
    <form action="<?=$_SERVER['PHP_SELF']?>" method=POST>
    <tr><td BGCOLOR="#2f4f4f"><FONT SIZE="-1" FACE="Verdana,Tahoma,Arial,Helvetica,sans-serif" COLOR="#FFFFFF">
    <B>Please Log In For Access:</B>
    </td></tr>
    <tr><td BGCOLOR="#c7c7c7"><FONT SIZE="-1" FACE="Verdana,Tahoma,Arial,Helvetica,sans-serif">
You must log in to access this area of the site.
     </td></tr>
    <tr>
     <td BGCOLOR="#fffff0">
      <table width=100% border=0 cellspacing=0 cellpadding=0>
    <tr>
     <td><FONT SIZE="-1" FACE="Verdana,Tahoma,Arial,Helvetica,sans-serif">Email Address:</td>
     <td><input type=text name="uid" size="20" value=""></td>
    </tr>
        <tr>
     <td><FONT SIZE="-1" FACE="Verdana,Tahoma,Arial,Helvetica,sans-serif">Password:</td>
     <td><input type=password name="pwd" size="20"></td>
    </tr>
    <tr>
     <td colspan=2 align=center>
      <input type=submit name="Login" value="Login">
     </td>
    </tr>
    </form>
      </table>
     </td>
    </tr>
   </table>
  </td></tr>
</table>
  </body>
  </html>
  <?php
  exit;
}
//Clean the input submitted to mysql
$uid=addslashes($uid);
$pwd=addslashes($pwd);

//this puts the variable into the session

$_SESSION['uid'] = $uid; 
$_SESSION['pwd'] = $pwd;

$sql = "SELECT * FROM users WHERE email = '$uid' AND passwd = '$pwd' ";

$result = mysql_query($sql);

if (!$result) {
echo "A database error occurred while checking your login details";
}
//if bad user/pass combo access denied
if (mysql_num_rows($result) == 0) {

  unset($_SESSION['uid']);
  unset($_SESSION['pwd']);
  ?>
  <html>
  <head>
  <title> Access Denied </title>
  </head>
  <body>
  <h1> Access Denied </h1>
  <p>There are several reasons this may be happening:<BR>
  <UL><LI>Your username or password is incorrect</LI>
  <LI>You have forgotten your login information. <a href="password_reset.php">Lost Password</a></LI></UL>
  To return to our login page, <a href="index.php">click here</a>.</p>
  </body>
  </html>
    <?php
  exit;
}

?>

     

     

     

    test2.php

    


<?php 

if($_POST[action]=="Update")
{

echo "hello";

    
}

?>

<html>

<title>Upload an image to a database</title>


<body>


<h2>Update with new information</h2>
<form name="form" enctype="multipart/form-data" method="post" action="<?php echo"$PHP_SELF?rid=$_GET[rid]"; ?>">
<?php 
include ("".$_SERVER['DOCUMENT_ROOT']."/housing/accesscontrol.php");
?>
<input type=text name='images' class='bginput'>


<input type="submit" name="action" value="Update">
</form>

</body>
</html>

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.