Displays nothing?

[3raser]

<?php

$code = $_POST['code'];

$dbhost = "****";
$db = "****";
$dbuser = "****";
$dbpassword = "****";

//connecting to the database
$connect = mysql_connect("$dbhost","$dbuser","$dbpassword") or die("Connection failed!");
mysql_select_db("$db") or die("Database fail!");

//extract
$extract = mysql_query("SELECT * FROM info WHERE vipcode='$code'");
$numrows = mysql_num_rows($extract);

while ($row = mysql_fetch_assoc($extract))

{

if (!$row[levelofproject] || !$row[name] || !$row[id]){
   echo "Invalid code! <a href='http://www.srl.comoj.com'>Home</a>";
} else {

if ($row[progress] > 70) {
   $progress = "green";
}
       
echo "Progress: <span style='color:". $progress ."'>". $row[progress] ."%</span>";
}
}


?>

 

And my HTML code:

 

<link rel="stylesheet" type="text/css" href="style.css" />
<br><br><form action='info.php' method='POST'><input type='text' name='code' size='16' align='center'><a href='info.php' type='submit'><img src='Go.png' align='center' border='0'></a></img></form>

<Br><br><br><br><br><b><font family='arial'>Warning: Best viewed in Google Chrome</font></b>

 

When I type in a number, and then click the button it sends me to a blank page.......? It DOES send me to info.php, but it's a blank page....

[/code]

[Fergal Andrews]

Hi Justin,

Using a hyperlink to a php page will not submit form data, so when your link goes to info.php there is no form data. The SQL statement therefore will be

 

SELECT * FROM info WHERE vipcode=''

 

which will return no results unless you have a record with a null code value. The conditional statement

 

if (!$row[levelofproject] || !$row[name] || !$row[id]){

 

therefore fails and nothing is output to the screen.

 

All the best,

Fergal

You need to use a proper method of submitting the form, either a submit input or javascript and put a form action with info.php as the value in the form tag.

Fergal

[teamatomic]

Try using a submit button instead of a link with graphic, if you want an image try:

<input type="image" name="submit" src="button.png" />

 

 

HTH

Teamtomic

[3raser]

I don't think that'll work....

[3raser]

Wtf, I can't edit my post?....

 

Well here is my new HTML code:

 

<link rel="stylesheet" type="text/css" href="style.css" />
<br><br><form action='info.php' method='POST'><input type='text' name='code' size='16'><input type='submit' value='Go'></form>

[3raser]

Bump

[shlumph]

We don't know what's on info.php.... is it the first block of code you posted? If so, there may be no records in your database, or an error on the page.

 

Try this:

ini_set('display_errors', E_ALL);

display_errors(1);

 

And also print out some debugging info on that page...

[3raser]

Ok, I put that code in. It shows this error:

 

Fatal error: Call to undefined function display_errors() in /home/a2662105/public_html/info.php on line 3

[ronvanb]

The code can't find the function display_errors().

Maybe you forgot to include the function in info.php

Or made a type mistake in the function name.

It just cant find the function.

If you call a function name that does not exist. You will get this error.

I think that is the solution.

Hope you will get it to work

 

[3raser]

The code can't find the function display_errors().

Maybe you forgot to include the function in info.php

Or made a type mistake in the function name.

It just cant find the function.

If you call a function name that does not exist. You will get this error.

I think that is the solution.

Hope you will get it to work

 

I'm not sure of what might be wrong. I'll look though.

[jl5501]

The code to report all errors is this

 

ini_set('display_errors',1);
error_reporting(E_ALL);

 

Which is slightly different to what you have been given, so change it to that

[3raser]

I guess it has something to do with $code

 

Notice: Undefined index: code in /home/a2662105/public_html/info.php on line 5

[jl5501]

It is saying there is no $_POST['code'] in $_POST

 

So you need to see what is in there

 

print_r($_POST) will show you

[3raser]

Alright, I added that. Now:

 

Parse error: syntax error, unexpected T_VARIABLE in /home/a2662105/public_html/info.php on line 8

[Andy-H]

 

<?php

if (isset($_POST['submit']))
{

  $code = isset($_POST['code']) ? intval($_POST['code']) : 0;

  $dbhost = "****";
  $db = "****";
  $dbuser = "****";
  $dbpassword = "****";

  //connecting to the database
  $connect = mysql_connect($dbhost, $dbuser, $dbpassword) or die("Connection failed!");
  mysql_select_db($db, $connect) or die("Database fail!");

  //extract
  $extract = mysql_query("SELECT * FROM info WHERE vipcode='$code'")or die("Error with query: " . mysql_error());
  $numrows = mysql_num_rows($extract);

  if ($numrows != 0)
  {
    echo "Invalid code! <a href='[url=http://www.srl.comoj.com]http://www.srl.comoj.com[/url]'>Home</a>";
  } else {

    while ($row = mysql_fetch_assoc($extract))
    {

      if ($row['progress'] > 70) {
         $progress = "green";
      }
       
    echo "Progress: <span style='color:". $progress ."'>". $row['progress'] ."%</span>";
    }
  }
}
?>

 

<link rel="stylesheet" type="text/css" href="style.css" >
<br >
<br >
<form action='info.php' method='POST'>
  <input type='text' name='code' size='16'>
  <input type='submit' name='submit' value='Go'>
</form>

[sasa]

try

<link rel="stylesheet" type="text/css" href="style.css" />
<br><br>
<form action='info.php' method='POST' name='myform'>
<input type='text' name='code' size='16' align='center'>
<a href='javascript: document.myform.submit()' type='submit'>
	<img src='Go.png' align='center' border='0' />
</a>
</form>
<br><br><br><br><br>
<b><font family='arial'>Warning: Best viewed in Google Chrome</font></b>