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craygo

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Posts posted by craygo

    [SOLVED] PHP mySql session Login help please

    in PHP Coding Help

    Posted

    Change the login.php to this

    <?PHP
session_start();
//check that the user is calling the page from the login form and not accessing it directly
//and redirect back to the login form if necessary
if (!isset($_POST['username']) || !isset($_POST['password'])) {
header( "Location: http://localhost/test2/login.htm" );
}
//check that the form fields are not empty, and redirect back to the login page if they are
elseif (empty($_POST['username']) || empty($_POST['password'])) {
header( "Location: http://localhost/test2/login.htm" );
}
else{

//convert the field values to simple variables

//add slashes to the username and md5() the password
$user = addslashes($_POST['username']);
$pass = md5($_POST['password']);


//set the database connection variables

$dbHost = "localhost";
$dbUser = "root";
$dbPass = "";
$dbDatabase = "randv";

//connet to the database

$db = mysql_connect("$dbHost", "$dbUser", "$dbPass") or die ("Error connecting to database.");

mysql_select_db("$dbDatabase", $db) or die ("Couldn't select the database.");

$result=mysql_query("select * from users where username='$user' AND password='$pass'", $db);

//check that at least one row was returned

$rowCheck = mysql_num_rows($result);
if($rowCheck > 0){
while($row = mysql_fetch_array($result)){

  //egister a variable
  $_SESSION['username'] = $row['username'];
  $_SESSION['logged_in'] = 1;

  //successful login code will go here...
  echo 'Success!';

  //we will redirect the user to another page where we will make sure they're logged in
  header( "Location: checkLogin.php" );

  }

  }
  else {

  //if nothing is returned by the query, unsuccessful login code goes here...

  echo 'Incorrect login name or password. Please try again.';
  }
  }
  ?>

     

    Ray

    [SOLVED] Please help - i'm stumped

    in PHP Coding Help

    Posted

    ok couple things

     

    you need to add this

    //$query="SELECT * FROM contacts, booking_form WHERE contacts_id = $record AND bf_id = $bf_record";
$query="SELECT * FROM contacts WHERE contacts_id = $record";
$result=mysql_query($query);
if(!$result) {
	print("Query Error: ".mysql_error());
}
  $row = mysql_fetch_assoc($result); // add this line

     

    Also need to change the variable for the issue loop

    while($irow=mysql_fetch_assoc($issues_result)) {  // change this to $irow and the variables below
$selected = in_array($irow['issue_number'], $bf_issues_booked) ? "checked=\"checked\"" : "";


$issues[] = '<input name="issue_number[]" type="checkbox" value="' .$irow['issue_number']. '" tabindex="11" ' . $selected . ' />Issue <strong>' .$irow['issue_number'] . '</strong> ' . $irow['issue_month'] . ' ' . $irow['issue_year'] ;



}

     

    That is what I found so far, let me know

     

    Ray

     

    [SOLVED] Please help, loop error with existing update script

    in PHP Coding Help

    Posted

    look like it will run youjust have so syntax errors

     

    //Run the count information
$count = 10;
for($i = 1; $i <= $count; $i++)
{
  $cost = number_format($_REQUEST["cost_".$i], 2, ".", "");
  $size = $_REQUEST['size_'."$i"];
  if ($cost > 0 && $size > 0){
   mysql_query("UPDATE `sizes` SET `size` = '$size', `cost` = '$cost' WHERE `refnum` = '$sizeref'");
   }
}
mysql_close($con);

     

    Also how do you know the loop needs to run 10 times?? The only thing is you will be updating the same record on every loop. Why would you do this??

     

    You may want to pass the count value from the form to the update script.

     

    Ray

    [SOLVED] Why does my code not display all my database info?

    in PHP Coding Help

    Posted

    you are closing your loop before you echo anything out, so when you do echo anything it will be the last row in the table

     

    	<?php
		 $links = @mysql_query("SELECT * FROM ".$mysql_pretext."_documents WHERE document_type_id = '2'");
if (!$links) {
echo("Error retrieving student life documents from the database!<br>Error: " . mysql_error());
exit();
}
$number = 0;
?>
<table>
  <tr>
    <td height="30" colspan="2" bgcolor="#A0C4EC" class="search"><span class="style1">Student Life Library</span></td>
  </tr>
<?php
while ($link = mysql_fetch_array($links)) {
$ids1 = $link["ID"];
$titles1 = $link["title"];
$type = $link["type"];
$date = $link["date"];
$filename = $link["name"];
$addedby = $link["added_by"];
$number++;
?>
  <tr>
    <td width="6%" height="30" valign="middle"><img src="images/pdf_icon.png" width="20" height="20" /></td>
    <td width="94%" height="30" valign="middle" class="text"><a href="getdocument.php?id=<?php echo($ids1); ?>"><?php echo($filename); ?></a> - <?php echo($titles1); ?></td>
  </tr>
<?php
}
?>
</table>

     

    Ray

    Problem in sending data from drop down list to mysql...

    in PHP Coding Help

    Posted

    Not sure what else it can be. I have this

    <?php

import_request_variables("pg", "form_");
$hostname = "localhost";
$db_user = "root";
$db_password = "*******";
$db_table = "customer";

//$db = mysql_connect($hostname, $db_user, $db_password);
//mysql_select_db("rdms",$db);
?>
<html>
<head>
<title>Adding New Contact in Record</title>
</head>
<link rel="stylesheet" type="text/css" media="screen" href="abc.css" />
<body>
<fieldset>
<legend>
New Customer Entry
</legend>
<img src="header.jpg"
align="top">
<h2>Adding New Contact in Record</h2><?php
if (isset($_REQUEST['Submit']))
{

$form_cust_country = $_POST['customer_country'];
$sql = "INSERT INTO $db_table(customer_id, customer_name, customer_add, customer_add1, customer_phone, customer_mobile,
customer_fax, customer_city, customer_county, customer_postcode, customer_country, customer_email,
customer_position, customer_website, customer_comment)
values ('$form_cust_id', '$form_cust_name','$form_cust_add','$form_cust_add 1','$form_cust_phone',
'$form_cust_mobile','$form_cust_fax','$form_cust_city','$form_cust_county','$form_cust_postcode',
'$form_cust_country','$form_cust_email','$form_cust_position','$form_cust_website','$form_cust_comment')";
echo $sql."<br />";
if($result = mysql_query($sql ,$db))
{
echo "Thank you, Your information has been entered into our database";
}
else
{
echo "ERROR: ".mysql_error();
}

}
else
{
?>

<form method="post" action="">

<table width="470">
<tbody>

<tr>
<td>ID:</td>
<td><input type="text" name="cust_id" class="abc"></td>
</tr>
<tr>
<td> Name:</td>
<td><input type="text" name="cust_name" class="abc"></td>
</tr>
<tr>
<td>Address:</td>
<td><input type="text" name="cust_add" class="abc"></td>
</tr>
<tr>
<td>Line 2:</td>
<td><input type="text" name="cust_add1" class="abc"></td>
</tr>
<tr>
<td>Phone:</td>
<td><input type="text" name="cust_phone" class="abc"></td>
</tr>
<tr>
<td>Mobile:</td>
<td><input type="text" name="cust_mobile" class="abc"></td>
</tr>
<tr>
<td>Fax: </td>
<td><input type="text" name="cust_fax" class="abc"></td>
</tr>
<tr>
<td>City:</td>
<td><input type="text" name="cust_city" class="abc"></td>

</tr>

<tr>
<td>County:</td>
<td><input type="text" name="cust_county" class="abc"></td>
</tr>
<tr>
<td>Post code:</td>
<td><input type="text" name="cust_postcode" class="abc"></td>
</tr>
</tbody></table>
<table width="450">
<tbody>
<tr>
<td>Country:</td>

<select name="customer_country" class="abc">

<option value="" selected="selected">Select a Country</option>

<option value="United States">United States</option>

<option value="United Kingdom">United Kingdom</option>
</select>
</tr>

<tr>
<td>Email:</td>
<td><input type="text" name="cust_email" class="abc"></td>
</tr>
<tr>
<td>Position: </td>
<td><input type="text" name="cust_position" class="abc"></td>
</tr>
<tr>
<td>Website:</td>
<td><input type="text" name="cust_website" class="abc"></td>
</tr>
<tr>
<td>Comments:</td>
<td><input type="text" name="cust_comment" class="abc"></td>
</tr>

</tbody></table>
</fieldset>




<input type="submit" name="Submit" value="Submit">
</form>
<?php
}
?>
</body>
</html>

     

    and when I echo the sql I get this

    INSERT INTO customer(customer_id, customer_name, customer_add, customer_add1, customer_phone, customer_mobile, customer_fax, customer_city, customer_county, customer_postcode, customer_country, customer_email, customer_position, customer_website, customer_comment) values ('1', 'Hello','222 nnjjh ave','222 nnjjh ave 1','4015554444', '5555555555','1544455654','somecity','some county','025558', 'United States','sjklkgh@jklghlkjh.com','position','www.domain.com','jklglkg')

     

    Ray

    Problem in sending data from drop down list to mysql...

    in PHP Coding Help

    Posted

    try and echo out the query and see what you get

     

    $sql = "INSERT INTO $db_table(customer_id, customer_name, customer_add, customer_add1, customer_phone, customer_mobile,
customer_fax, customer_city, customer_county, customer_postcode, customer_country, customer_email,
customer_position, customer_website, customer_comment)
values ('$form_cust_id', '$form_cust_name','$form_cust_add','$form_cust_add 1','$form_cust_phone',
'$form_cust_mobile','$form_cust_fax','$form_cust_city','$form_cust_county','$form_cust_postcode',
'$form_cust_country','$form_cust_email','$form_cust_position','$form_cust_website', '$form_cust_comment')";
echo $sql."<br>";
$result = mysql_query($sql ,$db);
if($result)
{
echo "Thank you, Your information has been entered into our database";
}
else
{
echo "ERROR: ".mysql_error();
}

     

    Ray

     

    [SOLVED] Please help - i'm stumped

    in PHP Coding Help

    Posted

    What you need to do is grab the comma seperated values then create an array from them I use a function for this to get the data. Loop through all your issues then check to see if the id is in the array that you created. use the in_array() command to see if is is in the array.

     

    // Define the array variable

$bf_issues_booked = explode(",",$row['bf_issue_number']);

// just to see if there is any result

echo $bf_issues_booked;

// A loop used to populate my form with the checkboxes and issue name, number etc. This works fine

while($row=mysql_fetch_assoc($issues_result)) {
$selected = in_array($row['issue_number'], $bf_issues_booked) ? "selected" : "";

$issues[] = '<input name="issue_number[]" type="checkbox" value="' .$row['issue_number']. '" tabindex="11" $selected />Issue <strong>' .$row['issue_number'] . '</strong> ' . $row['issue_month'] . ' ' . $row['issue_year'] ;

}

     

    Ray

    Problem in sending data from drop down list to mysql...

    in PHP Coding Help

    Posted

    your values are messed up. there are spaced in your values and you forgot a comma.

     

    $sql = "INSERT INTO $db_table(customer_id, customer_name, customer_add, customer_add1, customer_phone, customer_mobile,

    customer_fax, customer_city, customer_county, customer_postcode, customer_country, customer_email,

    customer_position, customer_website, customer_comment)

    values ('$form_cust_id', '$form_cust_name','$form_cust_add','$form_cust_add 1','$form_cust_phone',

    '$form_cust_mobile','$form_cust_fax','$form_cust_c ity','$form_cust_county','$form_cust_postcode',

    '$form_cust_country','$form_cust_email','$form_cus t_position','$form_cust_website,''$form_cust_comme nt')";

     

    Ray

    [SOLVED] PHP mySql session Login help please

    in PHP Coding Help

    Posted

    session_register is may not work, it requires register_globals to be on and they are off by default

     

    http://us.php.net/manual/en/function.session-register.php

     

    what you want to so is use $_SESSION instead

    <?PHP 
//check that the user is calling the page from the login form and not accessing it directly 
//and redirect back to the login form if necessary 
if (!isset($username) || !isset($password)) { 
header( "Location: http://localhost/test2/login.htm" ); 
} 
//check that the form fields are not empty, and redirect back to the login page if they are 
elseif (empty($username) || empty($password)) { 
header( "Location: http://localhost/test2/login.htm" ); 
} 
else{ 

//convert the field values to simple variables 

//add slashes to the username and md5() the password 
$user = addslashes($_POST['username']); 
$pass = md5($_POST['password']); 


//set the database connection variables 

$dbHost = "localhost"; 
$dbUser = "root"; 
$dbPass = ""; 
$dbDatabase = "randv"; 

//connet to the database 

$db = mysql_connect("$dbHost", "$dbUser", "$dbPass") or die ("Error connecting to database."); 

mysql_select_db("$dbDatabase", $db) or die ("Couldn't select the database."); 

$result=mysql_query("select * from users where username='$user' AND password='$pass'", $db); 

//check that at least one row was returned 

$rowCheck = mysql_num_rows($result); 
if($rowCheck > 0){ 
while($row = mysql_fetch_array($result)){ 

  //start the session and register a variable 

  session_start(); 
  $_SESSION['username'] = $row['username']; 
  $_SESSION['logged_in'] = 1;

  //successful login code will go here... 
  echo 'Success!'; 

  //we will redirect the user to another page where we will make sure they're logged in 
  header( "Location: checkLogin.php" ); 

  } 

  } 
  else { 

  //if nothing is returned by the query, unsuccessful login code goes here... 

  echo 'Incorrect login name or password. Please try again.'; 
  } 
  } 
  ?>

     

    and

    <?php 

//start the session 
session_start(); 

//check to make sure the session variable is registered 
if(isset($_SESSION['username']) && $_SESSION['logged_in'] == 1){ 

//the session variable is registered, the user is allowed to see anything that follows 

echo 'Welcome, you are still logged in.'; 

} 
else{ 

//the session variable isn't registered, send them back to the login page 
header( "Location: http://localhost/test2/login.htm" ); 
} 

?>

     

    Also look here for how to properly destroy a session

     

    http://us.php.net/manual/en/function.session-destroy.php

     

    Ray

    Problem in sending data from drop down list to mysql...

    in PHP Coding Help

    Posted

    First fix your query and get back to us, cause once I fixed it everything is fine

    $sql = "INSERT INTO $db_table(customer_id, customer_name, customer_add, customer_add1, customer_phone, customer_mobile,
customer_fax, customer_city, customer_county, customer_postcode, customer_country, customer_email,
customer_position, customer_website, customer_comment)
values ('$form_cust_id', '$form_cust_name','$form_cust_add','$form_cust_add 1','$form_cust_phone',
'$form_cust_mobile','$form_cust_fax','$form_cust_city','$form_cust_county','$form_cust_postcode',
'$form_cust_country','$form_cust_email','$form_cust_position','$form_cust_website', '$form_cust_comment')";

     

    Ray

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