unexpected T_LNUMBER??

[mslauren]

Hi,

I have a basic php code that allows the user to search and display data from a database. I have searched and searched the entire code for errors, but I cannot find any.  When I try to run the code (do a search) it gives me this error:

Parse error: syntax error, unexpected T_LNUMBER, expecting ',' or ';' on line 99

 

This is my line 99:

echo ' <a href="'.$_SERVER['PHP_SELF']."?s=$news&q=$var\">Next 10 >></a>";

 

and this is my whole code:

<?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","xxxxxx","xxxxx");


mysql_select_db("xxxxxx") or die("Unable to select database"); 


// Build SQL Query  
$query = "SELECT lastname, firstname, state, zip, jobtype, otherjobtype, nightavail, weekendavail, ptft, objective, resume FROM data WHERE jobtype LIKE '%" . $trimmed . "%' ORDER BY lastname"; 

$numresults=mysql_query($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["1st_field"];

  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) {
  $prevs=($s-$limit);
  print ' <a href="'.$_SERVER['PHP_SELF'].'?s=$prevs&q=$var"><< Prev 10</a>&nbsp ';
  }

// 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>";

?>

 

I am completely new to php. If anyone could give me a hint I'd greatly appreciate it...

 

[teynon]

You have some crazy strings going here:

 

echo ' <a href="'.$_SERVER['PHP_SELF']."?s=$news&q=$var\">Next 10 >></a>";

 

 

Try

echo " <a href=\"{$_SERVER['PHP_SELF']}?s={$news}&q={$var}\">Next 10&gt&gt</a>";

[kickstart]

Hi

 

I have tried to run your script there are didn't get any initial errors like that (just runtime ones as I hadn't set the table up).

 

Are you sure that is the script that is giving you the error?

 

All the best

 

Keith