PHP Error, but where?

Rivulent

OK, should be my last question for a long while (And the lie detector tests go CRAZY the world over), but here's the problem. I am setting up a query for my colleagues to view information pulled from the table "form". Unfortunately, using this code, the page comes up blank, as if to signify a problem/typo in the php encoding. I spent the last hour or so looking it up and down and can't find anything obvious that I made for a mistake, so I'm starting to believe that the problem lies in the way I coded the query. It seems right, I used a while loops, etc. etc., but unfortunately, it's just not working. The query seems to work, becuase the page doesn't come up with a mysql_error(), rather the page is just blank.

//php before the <html> tag
<?php
$form_info = "SELECT * FROM form";



require_once('../Connections/MX6_Performance.php');
mysql_select_db("mxperf");
$query = mysql_query($query_info, $MX6_Performance);			

$form_results = mysql_query($form_info, $MX6_Performance) or die(mysql_error());


while ($form = mysql_fetch_assoc($form_results)) {
$fname = form['fname'];
$lname = form['lname'];
$address = form['address'];
$city = form['city'];
$zip = form['zip'];
$state = form['state'];
$phone = form['zip'];
$email = form['email'];
$year = form['year'];
$paid = form['paid'];

$form_block= "<table width=\"850\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">
  <tr> 
    <td>$fname $lname</td>
    <td>$address</td>
    <td>$city</td>
    <td>$state</td>
    <td>$zip</td>
    <td>$phone</td>
    <td>$email</td>
    <td>$year</td>
    <td>$paid</td>
  </tr>
</table>";
}
else $form_block="";


}

?>

//inside the body, inserts the form_block

<?php
 echo $form_block;
 ?>

//after the </html> tag, to close the query

<?php
mysql_close($form);

?>

Thank you all for your great help!

paulnaj

This line ...

$fname = form['fname']

... should be ...

$fname = $form['fname']

... ditto for the other calls to the $form hash.

Rivulent

Thanks for that. Unfortunately, I still face the same problem: the output html just comes up:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=Content-Type content="text/html; charset=windows-1252"></HEAD>
<BODY></BODY></HTML>

I do in fact have some html OUT of the php brackets, so even if the query wasn't operating, the html outside the php should still show up. The fact that nothing shows up tells me that theres is an error in the coding.

paulnaj

Show us the whole page. If it's a big file, save it as a .txt and upload it.

helz

looks to me like its executing the else , so its showing up nothing because:
else $form_block="";

btw why is that there?

also if it is executing the else part then it replaces whatever $form_block was populated with.

and another thing, because you are only outputting the data outside of the while loop, the only thing that will get outputted will be the last record in the database as each time it loops it will replace the the record previously...
a simple fix. move the while loop to the part where you want to output and then put the echo inside the while loop.
Actually you dont even need to assign it to a variable just output it straight away:

while ($form = mysql_fetch_assoc($form_results)) { 
$fname = form['fname']; 
$lname = form['lname']; 
$address = form['address']; 
$city = form['city']; 
$zip = form['zip']; 
$state = form['state']; 
$phone = form['zip']; 
$email = form['email']; 
$year = form['year']; 
$paid = form['paid']; 

echo "<table width=\"850\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\"> 
  <tr> 
    <td>$fname $lname</td> 
    <td>$address</td> 
    <td>$city</td> 
    <td>$state</td> 
    <td>$zip</td> 
    <td>$phone</td> 
    <td>$email</td> 
    <td>$year</td> 
    <td>$paid</td> 
  </tr> 
</table>"; 
}

<-- New Tool Alert -->
if you are pulling alot of fields from the database instead of doing:
$var1 = $array['var1'];
$var2 = $array['var2'];
$var3 = $array['var3'];
$var4 = $array['var4'];

you can simply use:
extract($array);
it will convert $array['var1'] into $var1

Rivulent

All great stuff, and I will surely use it in the future. I have updated the code to the way you have it.

<?php
$form_info = "SELECT * FROM form";



require_once('../Connections/MX6_Performance.php');
mysql_select_db("mxperf");

$form_results = mysql_query($form_info, $MX6_Performance) or die(mysql_error());

while ($form = mysql_fetch_assoc($form_results)) {  

$fname = $form['fname'];  

$lname = $form['lname'];  

$address = $form['address'];  

$city = $form['city'];  

$zip = $form['zip'];  

$state = $form['state'];  

$phone = $form['zip'];  

$email = $form['email'];  

$year = $form['year'];  

$paid = $form['paid'];  

echo "<table width=\"850\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\"> 
  <tr> 
    <td>$fname $lname</td> 
    <td>$address</td> 
    <td>$city</td> 
    <td>$state</td> 
    <td>$zip</td> 
    <td>$phone</td> 
    <td>$email</td> 
    <td>$year</td> 
    <td>$paid</td> 
  </tr> 
</table>";  

}
 ?>

Now, it's outputting the the initial table with the descriptions of what each field is (I made a table above the query to headline what each column is). But no information is coming out. I'm going to run some tests tomorrow and fix the problem, shouldn't be something major. I appreciate the help all.

BTW the reason I used the form_block way is so that I could have it set to a variable in which I can repeat later on if needed, rather than copy pasting the same coding.

UPDATE: Resolved! Info wasn't coming back becuase I originally made the document in one folder, and then moved it a folder deeper, and the ../ wasn't added on like it should have automatically. Either way, it's a joyous day indeed! 🙂