search script problems

Shadeless

Below i have included a search form and the query going along with it.

I didn't included the show results of this script because i don't think it's nessercary.

The problem i have is that i cannot search in all the fields for the keyword. How can i make a selection in the search form where i can search in all the fields at once. i tried some things but these didn't worked. I cannot imagine it is something difficult, but heck i'm a newbie 🙁

Anywayz the search form here

<h5>Quick Search </h5>
<form action="\knowledgebase\search\results.php" method="post" name="search">
<strong>Sellect Search Category:</strong><br />
<select name="type" class="form">
<option value="id">id</option>
<option value="Date_Origin">Date_Origin</option>
<option value="Information_Date">Information_Date</option>
<option value="category">Category</option>
<option value="Machine_Type_nr">Instrument_Type_nr</option>
<option value="description">Description</option>
<option value="DY_nr">DY_nr</option>
<option value="Software">Software</option>
<option value="Software_version">Software_version</option>
<option value="Firmware">Firmware</option>
<option value="Firmware_version">Firmware_version</option>
<option value="Originatior">Originator</option>
<option value="Keywords">Keywords</option>
<option value="Long_Description">Long_Description</option>
<option value="Answer_keywords">Answer_keywords</option>
<option value="Full_answer">Full_answer</option>
</select>
<br /><br />
<strong>Search Keyword:</strong><br />
<input name="keyword" type="text" size="25" class="form" />
<br /><br />
<input type="reset" name="reset" value="Reset" title="click here to reset form" />
<input type="submit" name="action" value="submit" title="click here to submit form" />
</form>

And the search query


$host = "localhost";
$user = "root";
$password = "";
$dbname = "knowledgebase";
$tablename = "information";
$link = mysql_connect($host, $user, $password);
$query = "SELECT * FROM $tablename WHERE ".$type." LIKE '%".$keyword."%'";

$result = mysql_db_query ($dbname, $query, $link);

$num=mysql_numrows($result);

$keyword = stripslashes($keyword);

print("Your search for <strong>\"$keyword\"</strong> in the <strong>\"$type\"</strong> category returned: <strong>".$num_results."</strong> results<br /><br />\n");

mysql_close($link);

If any1 could help me i would be a very happy newbie 🙂

thx in advance

with kind regards

archangel_617b

Easy-peasy... well, kinda. Searching is never easy but what you can do is just add a type like 'All Fields'. Then, on the PHP page build two different queries. One which is based on a signle field and another based on all fields. Something like:

if ($type != 'All Fields')
{
    $query = "SELECT * FROM $tablename WHERE ".$type." LIKE '%".$keyword."%'";
}
else
{
    $query = 'SELECT * FROM $tablename WHERE ';
    $query .= 'Date_Origin LIKE "%'.$keyword.'%" OR ';
    $query .= 'Information_Date LIKE "%'.$keyword.'%" OR ';
    // and add criteria for each field
}

Hope that clears things up for you.

Shadeless

thx for ur help so far, but i haven't managed to get it to work yet.

i tried it but i now get the folowing error when trying the all fields selection

I replaced the first query with the query u gave, and i extended the criteria fields. I don't get any error with the "old" selections, only with the new, all fields 1 ( i just named it all, but tried with all fields befor that)

the error:

Warning: Supplied argument is not a valid MySQL result resource in C:\apache\htdocs\knowledgebase\search\results.php on line 73
Your search for "1" in the "All" category returned: results

$result = mysql_db_query ($dbname, $query, $link);

$num=mysql_numrows($result);   //(line 73)

$keyword = stripslashes($keyword);

print("Your search for <strong>\"$keyword\"</strong> in the <strong>\"$type\"</strong> category returned: <strong>".$num_results."</strong> results<br /><br />\n");

mysql_close($link);

ow yes i modified the 2nd select query a bit because else it won't reconize $tablename but i don't think that is the problem.

archangel_617b

Hmm... Now we get to learn about debugging then. Take a peek at [man]mysql_error/man. When you do any mysql operation you should check for valid result and display the mysql error if it has failed. Eg:

$result = mysql_connect($host, $user, $pass);
if (!$result)
    die('There was an error in mysql_connect:<br />'.mysql_error());

And you can check any value returned by any of the mysql functions like this. Importantly, you should check the result returned by mysql_query since it will print any SQL syntax errors that are in there.