View Full Version : Problem with searching database with php code

07-22-2007, 05:48 PM
Hi, I'm having a problem, the error I get is 'couldn't execute query'. Hope someone can help me. (database connection seems to be fine). It's to search 14 countries for various articles. Any help would really be appreciated. Thanks:)

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

// Get the search variable from URL
$var = @$_GET['q'] ;
$trimmed = trim($var); //trim whitespace from the stored variable

// 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($var))
echo "<p>We dont seem to have a search parameter!</p>";

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

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

// Build SQL Query
$query = "select * from files where path like \"&#37;$trimmed%\"
order by id";


// 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 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)) {

// 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;" . $var . "&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["id"];

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=$prevs&q=$var\">&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 division

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=$var\">Next 10 &gt;&gt;</a>";

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


07-24-2007, 01:54 PM
Is there really no one out there that can help me with this?:confused:

07-25-2007, 11:43 PM
The issue is in this line: $query .= " limit $s,$limit";
The other query isn't generating an error, so it must be that.
I'm not sure what the problem is, exactly.
You might want to set limit as a string, not a number, like:
$limit = '10';, since that may be treated differently by PHP.

Is there any more debugging info you can give us?

Try typing in the query manually as it should be sent, and see if that works. If not, then your query idea is wrong; if it does work, then it is being generated incorrectly.

Now, one other option is that $limit may be more than the number of rows, and that might give an error. I'm not sure about this.

if ($numrows < 10) { $limit = $numrows; }

Actually, I think the issue has to do with your if statement ending--
if (no results) {
...display error....
...now search...

The search query is outside the if, so it will still execute if there are no results... error.

You may just be able to move the } and fix it.