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webguync

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Posts posted by webguync

    not sure why I am getting an error with this SQL

    in MySQL Help

    Posted

    Hi, I am trying to create we table by copying and pasting the following into PHPMyAdmin. Not sure why, but I get a MySQL error.

     

    //Comments Table
CREATE TABLE comments
(
com_id INT PRIMARY KEY AUTO_INCREMENT,
com_name VARCHAR(100),
com_email VARCHAR(100),
com_dis TEXT,
post_id_fk INT,
FOREIGN KEY(post_id_fk) REFERENCES posts(post_id)
);

     

    anyone know why?

    display include only if value is posted

    in PHP Coding Help

    Posted

    so after reading these responses I believe this is what the form and if/else code should look like?

     

                	<form method="post" <?php echo filter_var($_SERVER['PHP_SELF'], FILTER_SANITIZE_STRING);?>
Enter Zip Code:<input type="text" size="12" maxlength="12" name="zip"><br /> 
<input id="submit" type="submit" name="Submit" value="Enter Zip"/>

</form>
<?php
$zip = $_POST['zip'];
echo $zip;
/* check to see if its set, if its numeric, if it has proper length  etc etc */
/* if the above check produce and error message ie $error_mess = some content */
/* then send back to form */
if (!isset($_POST['zip'])) {
    echo "please fill in a zip code value";
}
else{
include ('weather.php');
}
?>

     

     

    a better way to do date modified script?

    in PHP Coding Help

    Posted

    I have a small script on the bottom of all my pages on my site which list the date and time the page was modified. I want to make the script an include and just put that on the page, but as it is now scripted that won't work since you have to change the page name for every page. Is there a way to change this so it can be an include?

    $last_modified = filemtime("pagename.php");
print("Last Modified ");
print(date("m/j/y h:i", $last_modified));?>

    not sure why I am getting an undefined variable error

    in PHP Coding Help

    Posted

    Hi, I am testing out some simple code while trying out PHP OOP to connect to the MySQL DB. I am getting an undefined variable error, but not sure why.

     

    the test code is:

    <?PHP
error_reporting(E_ALL);
 include('includes/db.php');
 $db = new db();


 $db->query("SELECT title FROM blog_posts WHERE id = '7'");
 if($sql)
 {
	  while($r = mysql_fetch_array($sql))
	  {
		   echo $r['title'];
		   
	  }
 }
?>

     

    and the code to connect to the DB is:

    <?PHP
class db
{
 private $hostname;
 private $username;
 private $password;
 private $database;
 private $connect;
 private $select_db;

 public function db()
 {
	  $this->hostname = "";
	  $this->username = "";
	  $this->password = "";
	  $this->database = "";
 }

 public function open_connection()
 {
	  try
	  {
		   $this->connect = mysql_connect($this->hostname,$this->username,$this->password);
		   $this->select_db = mysql_select_db($this->database);
	  }
	  catch(exception $e)
	  {
		   return $e;
	  }
 }

 public function close_connection()
 {
	  try
	  {
		   mysql_close($this->connect);
	  }
	  catch(exception $e)
	  {
		   return $e;
	  }
 }

 public function query($sql)
 {
	  try
	  {
		   $this->open_connection();
		   $sql = mysql_query($sql);
	  }
	  catch(exception $e)
	  {
		   return $e;
	  }
	  $this->close_connection();
	  return $sql;
 }
}
?>

     

    the specific error is...

     

    Undefined variable: sql in E:\DBTest.php on line 9

    display include only if value is posted

    in PHP Coding Help

    Posted

    thanks, is this the way to do the variable check? I ask b/c it doesn't seem to be working. I don't get the included code.

    <form method="post" action="<?php echo $PHP_SELF;?>">
Enter Zip Code:<input type="text" size="12" maxlength="12" name="zip"><br /> 
<input id="submit" type="submit" name="Submit" value="Enter Zip"/>

</form>
<?php
$zip = $_POST['zip'];

/* check to see if its set, if its numeric, if it has proper length  etc etc */
/* if the above check produce and error message ie $error_mess = some content */
/* then send back to form */
if (!isset($zip)) {
    echo "please fill in a zip code value";
}
else{
include ('weather_report.php');
}
?>

     

    display include only if value is posted

    in PHP Coding Help

    Posted

    Hi, I only want to display some included PHP code only if a value is displayed from  form input. The value is a zip code which is used to display weather information, but the app produces an error when you first enter the page, because their is no value for the zip variable. I know I need an if/else statement, but not sure what to put in it.

     

    The form code is this

    <h3>Enter your zip code to receive weather info.</h3>
            	<form method="post" action="<?php echo $PHP_SELF;?>">
Enter Zip Code:<input type="text" size="12" maxlength="12" name="zip"><br /> 
<input id="submit" type="submit" name="Submit" value="Enter Zip"/>

</form>

<?php


// this is how we bring in the weather file

include 'weather_report.php' 

?>

     

    the zip variable is set-up in the weather_report.php file as follows

    $zip = $_POST["zip"];

     

    but like I said until that value is set, an error is thrown by the weather app code.

     

     

    help with changing variable via form

    in PHP Coding Help

    Posted

    Hi, I have a weather app, coded which creates weather based on a set zip code. I want to create a form which when the user fills out their local code and clicks the send button that will change the zip code variable. Currently it is set up like this

    $zip = '90210'; //change to your zipcode

     

    need help with getting started with this. Thanks!

    PHP sendmail

    in PHP Coding Help

    Posted

    shouldn't this work? or is sendmail() the same as mail()?

     

    function sendmail($to, $subject, $message, $from) {
$headers = "MIME-Version: 1.0" . "\r\n";
$headers .= "Content-type:text/html;charset=iso-8859-1" . "\r\n";
$headers .= "webguync@inspired-evolution.com" . "\r\n";


$result = sendmail($to,$subject,$message,$headers);

if ($result) return 1;
else return 0;
}

    PHP sendmail

    in PHP Coding Help

    Posted

    do you mean in the function?

    function mail($to, $subject, $message, $from) {
$headers = "MIME-Version: 1.0" . "\r\n";
$headers .= "Content-type:text/html;charset=iso-8859-1" . "\r\n";
$headers .= 'From: ' . $from . "\r\n";

$result = mail($to,$subject,$message,$headers);

if ($result) return 1;
else return 0;
}

     

    not really sure I am understanding what to change. I added the error reporting, and not getting any errors.

    PHP sendmail

    in PHP Coding Help

    Posted

    hmm, what is wrong with this code?

    <?php
$to = 'myemail@gmail.com'
$subject = 'this is a test'
$message = 'I rock at the phps'
mail($to, $subject, $message);
?>

     

    getting a 'Parse error: syntax error, unexpected T_VARIABLE in \test.php on line 3'

     

     

    PHP sendmail

    in PHP Coding Help

    Posted

    sorry, I just get the else echo, not a PHP error

    else echo 'Sorry, unexpected error. Please try again later';

     

    how would I set the simple script to go to a  specific email, because I think the email I have through the hosting company is not one that I typically use.

     

     

     

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