PHP code no longer works - diff btwn PHP4 and PHP5?

Beginners - Learning PHP
#1

I have a hosting site on a GoDaddy server - using the same php code and same tab-delimited data files. Godaddy no longer supports php4 - and my code no longer works. If some body could take a look and suggest what I might be doing wrong I’d be very appreciative!

The error I’m getting is “Warning: Invalid argument supplied for foreach() in /home/content/n/i/b/nibrs/html/cgi/property.php on line 43”

Code follows:

[code]

Crime Reporting Unit Property Report

Note: The Following table is a property stolen, with the data converted from NIBRS cases reported by this agency for the year in question.

<?php

$filename = “/home/content/n/i/b/nibrs/html/datafiles/prop.dat”; //here’s the filename

$id = fopen($filename, “r”); //open the file
while ($data = fgetcsv($id, filesize($filename),"\t")) //start a loop

echo “
$data[0]\n”;

if ($data[1] == “$ori” and $data[2] == “$year”)
$table[] = $data; //put each line into its own entry in the $table array
fclose($id); //close file

echo “

\n”;
echo “<td align='center’rowspan=‘2’>ORI<td align='center’rowspan=‘2’>Year<td align='center’rowspan=‘2’>Month<td align='center’colspan=‘7’>Robbery by Location<td align='center’colspan=‘2’>Burglary by Location<td align='center’colspan=‘3’>Larceny by Value<td align='center’colspan=‘9’>Nature of Larceny\n”;
echo "\n"; foreach($table as $row)

{
echo “

”;
foreach($row as $data)
echo “<td align='center’align=‘right’>$data”;
echo “\n”;
}

echo “

AgencyMonths
Hiway Comm. House Gas Station Conv. Store Residence Bank Misc Residence Non- Residence $200 and Over $50 - $200 Under $50 Pocket- picking Purse Snatching Shop lifting From Motor Veh MV Parts/Access Bicycle From Building Coin- op All Other Reporting
\n”;

?>

[/code]

Thanks!

Dan B

#2

Is the file path still correct and got all the right data in it?

#3

Yes it does! To test that out, I removed the ‘#’ from this line: # echo “
$data[0]\n”; and ran it - and got all 500o+ lines of data (well, the first item in the lines of data).

But thanks for the response - any other suggestions???

#4

Ahh, i reckon it’s because you haven’t initialised the $table variable.

[php]$id = fopen($filename, “r”); //open the file
$table = array();
while ($data = fgetcsv($id, filesize($filename),"\t")) //start a loop

echo “
$data[0]\n”;

if ($data[1] == “$ori” and $data[2] == “$year”)
$table[] = $data; //put each line into its own entry in the $table array
fclose($id); //close file[/php]

#5

Thanks for the hint - I think there is a more basic problem in the code.
Here’s the situation - the user is shown a web page with a form in it:

Agency:
Year: 2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
2010

But if I comment out the line "if ($data[1] == "$ori… the php will show all the lines of data - so I think the ‘ori’ and ‘year’ values are not being sent from the html into the php.

Is there a way I can check to see if the ‘select’ values in the form are being sent to the php???

#6

on the PHP file that you will process that value, echo to see if it contains the selected value
[php]
print_r($_REQUEST[‘year’]);
[/php]

Sponsor our Newsletter | Privacy Policy | Terms of Service