Search and Display

I have got a sql database that I would like to search then display the results.

The database is called: lever
The table is called: faults

Then i have the following fields in the table:

I would like a user to be able to search by incident and display the results for that row.

Is anyone willing to help me out.

1 Like

[php]mysql_query(‘SELECT * FROM faults WHERE incident LIKE "%’.mysql_escape_string($_GET[‘search’]).’%"’);[/php]

Thanks for that. How do I then display the results?

Whats wrong with this:

<?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 "

Please enter a search...

"; exit; } // check for a search parameter if (!isset($var)) { echo "

We dont seem to have a search parameter!

"; exit; } //connect to your database ** EDIT REQUIRED HERE ** mysql_connect("localhost","******","***"); //(host, username, password) //specify database ** EDIT REQUIRED HERE ** mysql_select_db("darcynet_lever") or die("Unable to select database"); //select which database we're using // Build SQL Query $query = "select * from faults where incident like "%$trimmed%" order by 1st_field"; // 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 "


"; echo "

Sorry, your search: "" . $trimmed . "" returned zero results

"; // google echo "

Click here to try the search on google

"; } // 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 "

You searched for: "" . $var . ""

"; // 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/$limit) + 1); //break before paging echo "
"; // next we need to do the links to other results if ($s>=1) { // bypass PREV link if s is 0 $prevs=($s-$limit); print " << Prev 10&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 " Next 10 >>"; } $a = $s + ($limit) ; if ($a > $numrows) { $a = $numrows ; } $b = $s + 1 ; echo "

Showing results $b to $a of $numrows

"; ?>

what is it doing/not doint?
put error_reporting(E_ALL) in the first line and tell us the errors u may get!
uncomment //$numrows=mysql_num_rows($numresults);

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/darcynet/public_html/search.php on line 35

Sorry, your search: “S11835” returned zero results

Click here to try the search on google

Couldn’t execute query

to get a mysql-error use:
$numresults=mysql_query($query) or die(mysql_error());

ah great, thanks for that. Found the problem and it now seems to work and displays:

[i]You searched for: “S11835”

Results1.) S11835

Showing results 1 to 1 of 1[/i]

if i want it to show other fields from the table, where do i put them. ie start date

once again thanks for your help

just use $row[“Name”] / $row[“Phone”] or however the field u wanna display is called

Hopefully this will be the last bit of help i need. I know im a bit stupid but what do I add to this:

[i]// now you can display the results returned
while ($row= mysql_fetch_array($result)) {
$title = $row[“incident”];

echo “$count.) $title” ;
$count++ ;

Dont worry i have managed to sort it. Thanks for all your help

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