PHP - How To Check If Javascript Is Diabled?

I am unable to execute mysql_num_rows properly in this program.
Code: [Select]
<?php
$link=mysql_connect("localhost","root","");
$db = mysql_select_db("myDB",$link);
$result = mysql_query("SELECT * FROM jobs WHERE jobType=abc"); 
$num = mysql_num_rows($result);
echo "Found $num records";
?>
error is mysql_num_rows() expects parameter 1 to be resource

ok so i can get this to work

so i want to check if there is a video on inside my db if there is no video and nothing happens but if there is i want it to echo video

Code: [Select]

<?php

if($video == ""){

}else{

echo "VIDEO";

}

?>

Hi,

I'd like to know what the most efficient code is for checking the first 2 digits of a string.

eg a strings first 2 digits must  equal '07'

thanks,

hi, im can someone help me out with a file upload?  i want to say if filename=abc.csv then do abc else do nothing

Problem - application produces MD5 PWs which are uppercase, code checks lowercase.
Question- How to use uppercase check?

Code: [Select]
function checkPwd($x,$y)
{
    //Checks if strings are empty

    if(empty($x) || empty($y) )
    {
        //Strings were empty
        return false;
    }
    else if(strlen($x) < 4 || strlen($y) < 4)
    {
        //String length too short
        return false;
    }
    else if(strcmp($x,$y) != 0)
    {
        //Strings do not match
        return false;
    }
    else
    {
        //Password Determined valid
        return true;

    }
}
Question 2: Do I have to change it anywhere else, like PW generation?
Code: [Select]
function GenPwd($length = 7)
{
  $password = "";
  $possible = "0123456789bcdfghjkmnpqrstvwxyz"; //no vowels
 
  $i = 0;
   
  while ($i < $length) {

   
    $char = substr($possible, mt_rand(0, strlen($possible)-1), 1);
       
   
    if (!strstr($password, $char)) {
      $password .= $char;
      $i++;
    }

  }

  return $password;

}

function GenKey($length = 7)
{
  $password = "";
  $possible = "0123456789abcdefghijkmnopqrstuvwxyz";
 
  $i = 0;
   
  while ($i < $length) {

   
    $char = substr($possible, mt_rand(0, strlen($possible)-1), 1);
       
   
    if (!strstr($password, $char)) {
      $password .= $char;
      $i++;
    }

  }

  return $password;

}
Thank you for reading

how do i echo "checked=checked" for a checkbox?  here is what i have:

Code: [Select]
<input type='checkbox' name='ip_automatic_login' id='ip_automatic_login' value='1' if($IPCheck == true) {/"checked=checked/" } />
this code will be echoed!

Hello PHP World,
I don't have access to the database yet for the project I am working on so I created a work around until I get full access to the mysql database.

I Code: [Select]
<?php

$xml_file = "affiliate.xml";

// OPEN XML FILE ######################################################

if(!$xml=simplexml_load_file($xml_file)){
    trigger_error('Error reading XML file',E_affiliate_ERROR);
}
if(!empty($_GET['add'])){

           
$u_name = stripslashes($_POST['name']);
$u_company = stripslashes($_POST['company']);
$u_address = stripslashes($_POST['address']);
$u_city = stripslashes($_POST['city']);
$u_state = stripslashes($_POST['state']);
$u_phone = stripslashes($_POST['phone']);
$u_email = stripslashes($_POST['email']);
$u_desc = stripslashes($_POST['description']);
$u_region = stripslashes($_POST['region']);

// Generate new (appended) ID

foreach($xml as $affiliate){
    $last_id = $affiliate->uid;
}

$id = $last_id+1;

// Add node

$x = $xml->addChild("affiliate");
$x->addChild("uid",$id);
$x->addChild("name",$u_name);
$x->addChild("company",$u_company);
$x->addChild("address",$u_address);
$x->addChild("city",$u_city);
$x->addChild("state",$u_state);
$x->addChild("phone",$u_phone);
$x->addChild("email",$u_email);
$x->addChild("description",$u_desc);
$x->addChild("region",$u_region);

$xml->asXML($xml_file);

}
?>


html form

Code: [Select]
<br />
<br />

<h1>Affiliate </h1>

<form name="xml_writing" method="post" action="?add=t">
<table width="393">
<tr><td width="56">
Name:<br /></td><td width="206">
<input type="text" name="name" width="150"/></td>
<td>Region:</td>
<td><select name="region" value="options">
<option value="resellerNE">NorthEast</option>
<option value="resellerSE">SouthEast</option>
<option value="resellerC">Central</option>
<option value="resellerNE">NorthWest</option>
<option value="resellerNE">SouthWest</option>
<option value="resellerNE">International</option>
</select></td>
</tr>
<tr>
<td>
Company:<br /></td><td>
<input type="text" name="company" width="150" /></td>
</tr>
<tr><td>
Address:<br /></td><td>
<input type="text" name="address" width="150"/></td>
</tr>
<tr><td height="45">
City:<br /></td><td>
<input type="text" name="city" width="150"/></td>
<tr><td>
State:<br /></td><td>
<input type="text" name="state" width="150"/></td>
<tr><td>
Phone:<br /></td><td>
<input type="text" name="phone" width="150"/></td>
<tr><td>
Email:<br /></td><td>
<input type="text" name="email" width="150"/></td>
</tr>

<tr><td colspan="2">
Short description of yourself, your practice, your services, and other information of interest to potential clients or collaborators.<br /></td>
<tr><td colspan="2">
<textarea name="description" style="width: 406px; height: 136px;"></textarea></td>
</tr>
<tr><td></td><td>
<input type="submit" value="Submit" /></td>
</tr>
</table></form>

What I am trying to do is have on submit check against the email node in the xml file to see if email exist.
if exist string already a member else insert.

if you can help with code or point to something that will help.


"Everyone needs a mentor to lead them into success."
Quote by Shelby Poston

Hi,

I have no knowledge about coding.
But I have very little knowledge in creating or editing simple web pages.
How do you create a web page that checks user account name and password of a website in batch mode to see if they are valid.
For example:
I can put account details in an empty form in a "username:password" format and submit to check if they are valid.
The "username:password" will be checked one by one in a fashion of "http://username:password@www.(name of site).com".
It basically connects to the site's URL with a "username:password@" attached to it and see if it redirects to either "Members Area" or "Access Denied".
Example:
I put TestUser:TestPass in the form and submit.
Now it will connect to "http://TestUser:TestPass@www.(name of site).com"

When the checking is done a result page will come showing the valid ones.

How can all this be done?

Thanks

Hey guys, I've set up a database with a login and logout script for my site..

There is a TINYINT value called admin and it either equals 1 or 0 depending on whether the user is an admin or not..

The registration script works perfectly to create the table value and the login script works fine for the site..

The question I had was if I wanted to add a link to the bottom of every page that said:

Go to Administration Panel and make it only viewable by ADMINS I figured this little script would work..


Here would be the end of the page:
Code: [Select]
  <br />
  <center>Copyright &copy 2010 <a href="http://www.website.com">www.WEBSITE.com</a></center>

  <?php include('includes/start_admincheck.php'); ?>
    <center><a href="<?php echo $homedir .'admin.php'; ?>">Go to Administration Panel</a></center>
  <?php include('includes/end_admincheck.php'); ?>

</body>

</html>
Inside start_admincheck.php I have:

(NOTE: $cUsername refers to a setcookie and $cAdmin does as well.. They are defined on my Variable page included at the top.)
Code: [Select]
<?php include('variables/variables.php'); ?>

<?php
  mysql_connect("$mysql_hostname", "$mysql_username", "$mysql_password") or die(mysql_error());
  mysql_select_db("$mysql_database") or die(mysql_error());

  if(isset($cUsername))
  {
    $check = mysql_query("SELECT * FROM users WHERE username = '$cUsername'")or die(mysql_error());

    while($info = mysql_fetch_array( $check ))
    {
      if (($cAdmin == 1) && ($info['admin'] == 1))
      {
?>
And this is the end_admincheck.php
Code: [Select]
<?php include('variables/variables.php'); ?>

<?php
      }

      else
        die(); 
    } 
  }
  else
    die();
?>
?>
I get this Parse error thrown at the bottom of the page:

Code: [Select]
Parse error: syntax error, unexpected $end in /*******/includes/start_admincheck.php on line 15
Naturally I would checkout line 15 in start_admincheck.php, but normally when I get an $end error it is the last line of the code and leaves me lost..

Something I'm missing guys?

As always, thanks in advance 

I am new to PHP.

I developed this class, I wonder if there's anything wrong or that I can improve. I could not test it because I'm in school.

Thanks in advance.

Code: [Select]
<?



class user {



var $userID,

$schoolID,

$userName,

$userPass,

$dbHost,

$dbUser,

$dbName,

$dbPass,

$dbUserTable;

$dbSchoolTable;



function dbInfo() {

$this->dbHost = 'localhost';

$this->dbUser = '';

$this->dbName = '';

$this->dbPass = '';

$this->dbUserTable = '';

$this->dbSchoolTable = '';

}



function registerUser($userName, $userPass) {

$dbLink = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass);

if(!$dbLink) die("Could not connect to database: " . mysql_error());



mysql_select_db($this->dbName);



$query = "INSERT INTO $this->dbUserTable VALUES (NULL, \"$userName\", \"$userPass\")";

$result = mysql_query($query);



if(!$result) {
echo "Fail.";
} else {
$this->userID = mysql_insert_id();
}



mysql_close($dbLink);



$this->userName = $userName;

$this->userPass = $userPass;

}



function registerSchool($schoolName) {

$dbLink = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass);

if(!$dbLink) die("Could not connect to database: " . mysql_error());



mysql_select_db($this->dbName);



$query = "INSERT INTO $this->dbSchoolTable VALUES (NULL, \"$schoolName\")";

$result = mysql_query($query);



if(!$result) {
echo "Fail.";
} else {
$this->schoolID = mysql_insert_id();
}



mysql_close($dbLink);



$this->schoolName = $schoolName;

}


function userLogin() {

$dbLink = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass);

if(!$dbLink) die("Could not connect to database: " . mysql_error());



mysql_select_db($this->dbName);



$query = "SELECT * FROM $this->dbUserTable WHERE userName = \"$this->userName\" AND userPass = \"$this->userPass\" LIMIT 1";

$result = mysql_query($query);



if(!$result) {
echo "Fail.";
} else {
while($row = mysql_fetch_array($result)) {
session_start();
$_SESSION['userID'] = $row['userID'];
session_write_close();
}
}



mysql_close($dbLink);

}


function changePass($newPass) {

$dbLink = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass);

if(!$dbLink) die("Could not connect to database: " . mysql_error());



mysql_select_db($this->dbName);

$query = "SELECT * FROM $this->dbUserTable WHERE userName = \"$this->userName\" LIMIT 1";

$result = mysql_query($query);

if(!$result) {
echo "Fail.";
} else {

$query = "UPDATE $this->dbUserTable SET userPass = \"$newPass\" WHERE userName = \"$this->userName\"";

$result = mysql_query($query);



if(!$result) {
echo "Fail";
} else {
$this->userPass = $newPass;
}
}



mysql_close($dbLink);

}



}

?>

I am writing a script, but keep getting a 500 error page.

I've tried various tricks to isolate the problem to no avail.

My old system would always send me an email to report the error, but since my new box, it isn't working

The major change from my old setup is that I am now using fast CGI

I rememeber that there is a log that records errors but I havn't been able to find it if there is one... Or I am looking in the wrong places.

Can anyone nudge me in the right direction on this?

Thanks,

David

could anyone please help me with the code which is i have already displayed data from two table as a drop-down list and a multi select list but now i need to select one from  drop-down list and one or more from multi select list  the insert into another database table.
here is the coding. it shows no error message( blank page after submit) and nothing inserted on the take able
<?php 

$con = mysql_connect("localhost","root",""); 
if (!$con) 
  { 
  die('Could not connect: ' . mysql_error()); 
  } 
 error_reporting(E_ALL); 
ini_set('display_errors', 1);  

  mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); 



$result  = mysql_query("SELECT * FROM course") or trigger_error('MySQL error: ' . mysql_error()); 


echo '<select name ="cid[]" multiple="multiple" size="10">'; 

while($row = mysql_fetch_array($result)) 
{ 
    echo '<option value="' . $row['CourseID'] . '">' . $row['CourseName'] . '</option>'; 
} 

echo '</select>'; 

// ---------------- 

$result  = mysql_query("SELECT * FROM student") or trigger_error('MySQL error: ' . mysql_error()); 


echo '<select name="sid">'; 

while($row = mysql_fetch_array($result)) 
{ 

    echo '<option value="' . $row['StudentID'] . '">' . $row['StudentName'] . '</option>'; 
} 

echo '</select>'; 


mysql_close($con); 

?> 
<form id="form1" name="form1" method="post" action="update_result.php"> 
  <label> 
  <input type="submit" name="Submit" value="Submit" /> 
  </label> 
</form> 
</body> 
</html>


update_result.php
<?php 

$con = mysql_connect("localhost","root",""); 
if (!$con) 
  { 
  die('Could not connect: ' . mysql_error()); 
  } 
  
  mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); 
   
  

if (!empty($_POST['sid']) && !empty($_POST['cid'])) 
{ 
    foreach ($_POST['cid'] as $key => $course) 
    { 
     $courses .= $course."-"; 
     } 
    $courses = rtrim($courses,"-"); 
    $student = $_POST['sid']; 
    $sql = "INSERT INTO test (StudentID, CourseID) VALUES('".mysql_real_escape_string($student)."','".mysql_real_escape_string($courses)."')"; 
    $query = mysql_query($sql) or trigger_error('MySQL error: ' . mysql_error()); 

    if (mysql_affected_rows() > 0) 
    { 
        echo mysql_affected_rows() . ' rows added.'; 
    } 
}   
mysql_close($con); 

?> 
</body> 
</html>

Code: [Select]

<table border="1">
<tr>
<th>ID</th>
<th>Image</th>
</tr>

<?php

for($i=0; $i<=100; $i++)
{
$url = 'http://localhost/images/icon_'.$i.'_64x64.jpg';
echo "<tr><td>".$i."</td><td><img src='".$url."' /></td></tr>";

}

?>

</table>


So after testing this script, I get empty cells in the table if the image doesn't exist. Can someone show me how to make it display only the images that exists in that folder ?

basically this is a search function


<form  method="post" action="test1.php">
<input type = "hidden" name="submitted" value ="true" />
  <label>TYPE:
  <select name="field">
    <option value = "sid">StudentID</option>
    <option value = "sname">StudentName</option>
  </select>
  </label>
  <label>WORD:
  <input type="text" name="searchword" />
  </label>
<input type="submit"  />
</form>


-------------
<?php
if (isset($_POST['submitted'])){

$con = mysql_connect("localhost","root","");

  mysql_select_db("uni", $con);
  

$field= $_POST['field'];
$searchword = $_POST['searchword'];
$query = "SELECT* FROM student WHERE $field = '$searchword'";
$result = mysqli_query($con,$query) or die ('error data'); //-----------> error line

echo"<table>";
echo "<tr><th>StudentID</th><th>StudentName</th></tr>";
while($row = mysqli_fetch_array($result)){

echo "<tr><td>";
echo "$row ['sid']";
echo "</td><td>";
echo "$row ['sname']";
echo "</td></tr>";
}

echo"</table>";

}

mysql_close($con);

?>


NB: the error message "Warning: mysqli_query() expects parameter 1 to be mysqli, resource given in C:\wamp\www\test1.php on line 21
error data"