[PHP class help]

It would be better to have  a separate class for the connection and queries.

You have created the $db object for the connection class correctly.

So for the second class which will contain all the functions for a query create an object like this:

  $query =new Db_mysql_statement();

 

[PHP class help]

The object

$query->

has not been created to access the functions in class Db_mysql_statement

[$_POST variables not $_POSTing...]

What exactly is this section of code supposed to do?

 //first, find out which set it's in
    if($i < count($setOne)){
       $set_num = '1';
        $temp_set_array = $setOne;
        $count = $i;
      }
   if(($i >= count($setOne)) && ($i < (count($setOne) + count($set_Two)))){
      $set_num = '2';
      $temp_set_array = $set_Two;
      $count = $i - count($setOne);
      }
   if($i >= (count($setOne) + count($set_Two))){
      $set_num = 'e';
      $temp_set_array = $encore;
      $count = $i - count($setOne) - count($set_Two);
      }
   

 

If you are trying to display the file paths , wont it be a better option to allow the user to select the file they want to upload?

you can provide an option to upload multiple files at a time.

[$_POST variables not $_POSTing...]

Did you check if the values are available in the second page? btw the band value is not being called in the else part

else{
   //If not, assign other variables
   $year = $_POST['year'];
   $month = $_POST['month'];
   $day = $_POST['day'];
   $date = create_date($year, $month, $day);
   $setOne = $_POST['setOne'];
   $set_Two = $_POST['setTwo'];
   $encore = $_POST['encore'];
   }

[$_POST variables not $_POSTing...]

where are the variable values assigned?. I dont see them in the code.

[$_POST variables not $_POSTing...]

Which form are you using to get the other variables if $_POST is not set?

[$_POST variables not $_POSTing...]

On second thought. nope. what ur doing is absolutely fine.

[$_POST variables not $_POSTing...]

check if its geting in the if loop (im sure u wud have done that)  and if it is just try changing the way the values are passed.eg:

 

<input type="hidden" name="name" value="<?php echo $value;?>">

 

echo them in the elements. I had a similar problem before.

 

[$_POST variables not $_POSTing...]

I guess its this part:

if($_POST && $_POST['submit'] == 'Submit'){

it should be if (($_POST)&&($_POST['submit']=='Submit')) {  }

[Has anyone worked with google contacts API? Please help]

Guys please, If you have not not worked on this please let me know.

[Has anyone worked with google contacts API? Please help]

Hi ,

I have a problem with the google contacts API. Iam using php functions to sent HTTP requests.

I keep getting the error 'Bad Request'

Would be grateful if someone could throw some light.

 

 

Thank You,

suzzane

[Has anybody worked with google contacts API? Need help]

Hi ,

I have a problem with the google contacts API. Iam using php functions to sent HTTP requests.

I keep getting the error 'Bad Request'

Would be grateful if someone could throw some light.

 

 

Thank You,

suzzane

[[SOLVED] Google data API -where do I post the question?]

Thank You! and happy holidays!

[[SOLVED] Google data API -where do I post the question?]

Hi,

I have a doubt on google data API. Would there be a section to post that. I went through all the sections but got confused.

Please help!

 

thank you

[Amazon Payments module]

could anyone tell me where to post a question about amazon payment.  Iam a php coder and use amazon payment service as my payment gateway.

                  I have posted the issue in the amazon site but dint get the response I wanted.

I just wanted to know if anyone here has worked on this before?

 

** plz dont mind that i posted this here. I just dint have a clue where to!!!

 

 

[Reading an image file with php file functions]

thank  you so much. It was this simple and I could not get a direct solution after so much googling.

 

Thanx again!

[Reading an image file with php file functions]

How do I read and output an image file using php.

I have a function that get the image path and then uses fread to read the image file.

 

 

This throws an error  " supplied argument is not a valid stream resource"  pointing to the fread

 

I found out we need the header("Content-type: image/jpg")

but not sure about the rest of the code.

 

Could anyone plz help?

 

thank you

[Wierd checkbox form submit problem (must click submit twice)]

Place the showinfo function on top. That will solve the problem.

[problem in registration form]

yeah you could use ajax.  If you would lik to stick with php for nw u can do that too witha little bit of javascript.

 

The form can be divided into two. have the login filed in a separate form and the rest of the fields in another form

eg:

 

 

        <?php
                 if(isset($_POST["check"]))
                      {
                           $user   = $_POST["user"];
                             }

?>
<form action="register.php" method="POST" name="form1">

       <tr> 
      <td class="textsr">Username*</td><td><input name="user" type="text" class="textbox" value="<?php echo $user;?>">
          <input name="check" type="submit" value="check availability" class="button">
                           </td>
  <tr>
     
   <td></td>
      
      <td class="textsr">
        <?php
if(isset($_POST["check"]) && ($_POST["user"]!="") && (trim($_POST["user"])!=''))
{$user=trim($_POST["user"]);

$res=mysql_query("select * from table where username='$user'");
$trows=mysql_num_rows($res);
if($trows==0)
{
  echo "<font color='green'>The username is available</font>";
  
  }
  else
  {
     echo "<font color='red'>The username is not available.<br>Please select another one</font>";
 }
}else
{
echo "";
} ?>
      </td>
  
</form>

 

This way you can have the message appear right under the username box.

 

 

[[SOLVED] New Hour, New Problem.]

Thers the foreach loop.

 

I guess you would have to go through a few basics to get a smooth flow in coding.

php.net is the best place for a start

[[SOLVED] New Hour, New Problem.]

Yes you can use a space. Use it just after u get the post value.

$HTML = $_POST['html'];

 

$url_array = explode(" ",$HTML);

[[SOLVED] New Hour, New Problem.]

Hi,

 

    Your code does not have the function to split more than 1 url. Everything is taken as  a single string. You could use the explode function.

Have a note for the users that they need to use a separator like comma for the urls

[Php GD library]

ok. Will try searching gogle. Hope to find a solution.

Thank you guys!!

[Php GD library]

Yes, Thats exactly what i want to do. And as I would  be reading the contents of a file and passing that as the string, there would a lot more tags than the bold. <html><head>....etcIs there a way to do that.

[Php GD library]

Hi friends,

 

    I had a doubt regarding the gd library functions in php. This is what iam trying to do:

 

I need to read contents from a html file and convert that to an image uding the gd functions. I can get the code to work using simple text(without html tags) but not without them.

  How do I convert the contents of the file which has html tags to an image.

This is the code i have been using:

 

<?php
header ("Content-type: image/png");
$handle = ImageCreate (400, 400) or die ("Cannot Create image");
$bg_color = ImageColorAllocate ($handle, 255, 0, 0);
$txt_color = ImageColorAllocate ($handle, 0, 0, 0);
ImageString ($handle, 5, 5, 18, "<b>hello</b>", $txt_color);
ImagePng ($handle);
?>

 

I would really appreciate it if someone could help.

Thank You