Undefined Variable: PHP_SELF, pls help
Posted on
16th Feb 2014 07:03 pm by
admin
Hi,
Im a newbie on PHP / MySQL programming and Im running a script to search one field on my DB table. The results will show after I key in my search word but there's an error below that says: Undefined variable: PHP_SELF in C:wampwwwITInventoryApplicationITInventoryfilesSearchSerial.php on line 105. Also the link to the next page is not clickable.
When I change to $_SERVER['PHP_SELF'] I get:
Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING'.
Pls help me... thanks.
_________________________________________________ __________
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<?php
// Get the search variable from URL
$var = @$_GET['q'] ;
$trimmed = trim($var); //trim whitespace from the stored variable
// rows to return
$limit=10;
// check for an empty string and display a message.
if ($trimmed == "")
{
echo "<p>Please enter a search...</p>";
exit;
}
// check for a search parameter
if (!isset($var))
{
echo "<p>We dont seem to have a search parameter!</p>";
exit;
}
//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("localhost","root",""); //(host, username, password)
//specify database ** EDIT REQUIRED HERE **
mysql_select_db("itinventory") or die("Unable to select database"); //select which database we're using
// Build SQL Query
$query = "select * from assetsdb where SerialNo like "%$trimmed%"
order by SerialNo"; // EDIT HERE and specify your table and field names for the SQL query
$numresults=mysql_query($query);
$numrows=mysql_num_rows($numresults);
// If we have no results, offer a google search as an alternative
if ($numrows == 0)
{
echo "<h4>Results</h4>";
echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>";
// google
echo "<p><a href="http://www.google.com/search?q="
. $trimmed . "" target="_blank" title="Look up
" . $trimmed . " on Google">Click here</a> to try the
search on google</p>";
}
// next determine if s has been passed to script, if not use 0
if (empty($s)) {
$s=0;
}
// get results
$query .= " limit $s,$limit";
$result = mysql_query($query) or die("Couldn't execute query");
// display what the person searched for
echo "<p>You searched for: "" . $var . ""</p>";
// begin to show results set
echo "Results";
$count = 1 + $s ;
// now you can display the results returned
while ($row= mysql_fetch_array($result)) {
$title = $row["SerialNo"];
echo "$count.) $title" ;
$count++ ;
}
$currPage = (($s/$limit) + 1);
//break before paging
echo "<br />";
// next we need to do the links to other results
if ($s>=1) { // bypass PREV link if s is 0
$prevs=($s-$limit);
print " <a href="$PHP_SELF?s=$prevs&q=$var"><<
Prev 10</a>  ";
}
// calculate number of pages needing links
$pages=intval($numrows/$limit);
// $pages now contains int of pages needed unless there is a remainder from division
if ($numrows%$limit) {
// has remainder so add one page
$pages++;
}
// check to see if last page
if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {
// not last page so give NEXT link
$news=$s+$limit;
echo " <a href="$_SERVER['PHP_SELF']?s=$news&q=$var">Next 10 >></a>";
}
$a = $s + ($limit) ;
if ($a > $numrows) { $a = $numrows ; }
$b = $s + 1 ;
echo "<p>Showing results $b to $a of $numrows</p>";
?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
</body>
</html>
No comments posted yet
Your Answer:
Login to answer
252
5
Other forums
LinkedList help
Ok so I just learned quickly about lists, so I have a not too hard project I think, but am having a
Dynamic Data + Sql Server 2005 Enterprise?
Hi! I have just started to learn ASP.NET, and it looks like it is quite a lot to learn. Im not reall
storing results of a function - previous result overwritten with new result
I have created a function to validate input.
function validate_dimension($value,$name) {
<
DELETE rows based on content
I have a link in my rows
$bit="http://bit.ly/abcd";
$query = mysql_query("D
Problem with PHP code- simple contact form
I'm relativily new to PHP; I know HTML and CSS stuff but I have a problem- I have a contact form wit
Problem assigning value to variable in "IF" function
Does this script makes sense? I am trying to take the value that is set to "authenticat" a
Call db table from any PHP file
Hi,
I want to be able to call a database table that will be setup in another file called init
search function
HI guys,
if anyone could point us in the right direction of how to do this, or provide some t
fwrite error
Hi All,
Does anyone know what is causing the error in this code?
Code: <?
$error
Unable to retreve the values from Mysql Query
Hi,
Here is the php code that I have, Query is running properly in phpmyadmin and is resu