MySQL/PHP - Query Form

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As some of you already know, I'm trying to create a "query form" that will
allow me to display the results from MySQL to my screen in IE. I grabbed the
below code from the net hoping that I could tweak it for my needs. I'm using
MySQL, PHP and IIS and they all are running fine. As the code is, it will
display the form, but it won't display my result(s). I get "Parse error:
syntax error, unexpected ')' in c:\Inetpub\wwwroot\form.php on line 26" as
an error. My goal is to display my results in a table (similar to an Excel
spreadsheet). In a previous post, Michael Fesser has been a big help with
some of his suggestions and tweaks. Any other suggestions? I have to admit
that I know very little about PHP, but I do know how to make sense of code



<title> search script</title>
<meta name="author" content="Steve R, /">
<!-- / -->

<form name="form" action="form.php" method="get">
  <input type="text" name="q" />
  <input type="submit" name="Submit" value="Search" />

<!-- / -->



  // Get the search variable from URL

$trimmed = isset($_GET['q']) ? trim($_GET['q']) : '';

// rows to return

// check for an empty string and display a message.
if ($trimmed == "")
  echo "<p>Please enter a search...</p>";

// check for a search parameter
if (!isset($trimmed))
  echo "<p>We dont seem to have a search parameter!</p>";

//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("localhost","username","password"); //(host, username,

//specify database ** EDIT REQUIRED HERE **
mysql_select_db("database") or die("Unable to select database"); //select
which database we're using

// Build SQL Query
$query = "select * from sales_report where repfirstname = \'%$trimmed%\'";
//EDIT HERE and specify your table and field names for the SQL query


// If we have no results, offer a google search as an alternative

if ($numrows == 0)
  echo "<h4>Results</h4>";
  echo "<p>Sorry, your search: &quot;" . $trimmed . "&quot; returned zero

// google
 echo "<p><a href=\""">"
  . $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)) {

// 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: &quot;" . $trimmed . "&quot;</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["repfirstname"];

  echo "$count.)&nbsp;$title" ;
  $count++ ;

$currPage = (($s/$limit) + 1);

//break before paging
  echo "<br />";

  // next we need to do the links to other results
  if ($s>=1) { // bypass PREV link if s is 0
  print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$trimmed\">&lt;&lt;
  Prev 10</a>&nbsp&nbsp;";

// calculate number of pages needing links

// $pages now contains int of pages needed unless there is a remainder from

  if ($numrows%$limit) {
  // has remainder so add one page

// check to see if last page
  if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {

  // not last page so give NEXT link

  echo "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$trimmed\">Next 10

$a = $s + ($limit) ;
  if ($a > $numrows) { $a = $numrows ; }
  $b = $s + 1 ;
  echo "<p>Showing results $b to $a of $numrows</p>";


Re: MySQL/PHP - Query Form

Quoted text here. Click to load it

Look at those lines.
"//" begins a single line comment.
Text "password)" should be the part of the comment, so move it to
the end of previous line, or change commenting style to multiline
goes here


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