search query

[Shadeless]

A while i posted a search quero and some1 replied with a solution, but i can't get this solution get to work, is there any1 who could help me with this, or provide me an solution.

I'm trying here to make it possible to search within serveral fields.

The query gives a parse error at line 44,

thx in advance

$query = "SELECT * FROM $tablename WHERE 1=1 "


    if   ( $Date_Origin != "$Date_origin") {               //line 44
           $query .= "or Date_Origin LIKE '%".$Date_Origin."%'"
      }
    if   ( $Category != "" {
           $query .= "or Category LIKE '%".$Category."%'"
      }
    if   ( $description != "" {
           $query .= "or description LIKE '%".$description."%'"
      }
    if   ( $DY_nr != "" {
           $query .= "or DY_nr LIKE '%".$DY_nr."%'"
      }
    if   ( $Software != "" {
           $query .= "or Software LIKE '%".$Software."%'"
      }
    if   ( $Software_version != "" {
           $query .= "or Software_version LIKE '%".$Software_version."%'"
      }
    if   ( $Firmware != "" {
           $query .= "or Firmware LIKE '%".$Firmware."%'"
      }
    if   ( $Firmware_version != "" {
           $query .= "or Firmware_version LIKE '%".$Firmware_version."%'"
      }
    if   ( $Originator != "" {
           $query .= "or Originator LIKE '%".$Originator."%'"
      }
    if   ( $Keywords != "" {
           $query .= "or Keywords LIKE '%".$keywords."%'"
      }
    if   ( $Answer_keywords != "" {
           $query .= "or Answer_keywords LIKE '%".$Answer_keywords."%'"
      }

[J_ystick_FI]

porbably wrong column name.
query = "SELECT * FROM $tablename WHERE 1=1 "
1 is the name of the column or the value ??

hope this helps
😃 J@ystick_FI😃

[Shadeless]

i start to think this query is messed up ...

And yes 1=1 is indeed wrong 🙁,

what i want to try is some1 fills in a form.. with 10 search criteria

then the query needs to look in the table where the rows match tre criteria.

I have for example the fields, Date_Origin,m Category, Instument_Type_nr, Originator

If some1 fills in Date_Origin and Category, the results have to match these 2 fields

But i'm kinda lost of how to do this, it can't be to hard.. but well it is for me..

I know how to search for 1 keyword in serveral fields, but not how i can search different keywords in different fields ( 1 keyword match a "static" field

$keyword1=Date_origin, $keyword2=Category, $keyword=Orginator, etc

i had a query but that 1 always returned to allot results because if 1 keyword matched a within a field it would return a result.

I'm very sorry for explaining it in such bad way, i hope it makes any sense..

[lurius]

You could do it as follows:

if ($some_condition1 != '')
$queries[] = "some_col LIKE '%$some_condition1%'";
if ($some_condition2 != '')
$queries[] = "some_col LIKE '%$some_condition2%'";
...

then:
$query = "select something from somewhere";
if (is_array ($queries))
$query .= " where " . implode (' OR ', $queries);

-Lurius

[Shadeless]

i actualy understand it 🙂 🙂 🙂

many thanks from happy me🙂!!

[Shadeless]

hehe here i am again 🙂

ok the search query seems to work, alltough i now stumbled into a new problem :rolleyes:

I get a error when i fill somethin in in my form, after i press the submit button, or when i don't fill something in, the result page stays empty(as it should be).

So i have a problem with the mysql_numrows

the code so far is

<?php 

//trims received data from form on search_engine.html

//database connection variables

$host = "localhost";
$user = "root";
$password = "";
$pass = "";
$dbname = "knowledgebase";
$tablename = "information";
$link = mysql_connect($host, $user, $password);

 ///////// search query

    if ($Date_Origin != '')
    $queries[] = "Date_Origin LIKE '%$Date_Origin%'";
    if ($Category != '')
    $queries[] = "Category '%$Category%'";
    if ($Machine_Type_nr != '')
    $queries[] = "Machine_Type_nr LIKE '%$Machine_Type_nr%'";
    if ($description != '')
    $queries[] = "description '%$description%'";
    if ($DY_nr != '')
    $queries[] = "DY_nr '%$DY_nr%'";
    if ($Software != '')
    $queries[] = "Software '%$Software%'";
    if ($SOftware_version != '')
    $queries[] = "Software_version '%$Software_version%'";
    if ($Firmware != '')
    $queries[] = "Firmware '%$Firmware%'";
    if ($Firmware_version != '')
    $queries[] = "Firmware_version '%$Firmware_version%'";
    if ($Originator != '')
    $queries[] = "Originator '%$Originator%'";
    if ($Keywords != '')
    $queries[] = "Keywords '%$Keywords%'";
    if ($Answer_keywords != '')
    $queries[] = "Answer_keywords '%$Anser_keywords%'";

    $query = "select * from $tablename";
    if (is_array ($queries)) 
    $query .= " where " . implode (' OR ', $queries);

    $result = mysql_connect($host, $user, $pass);
  //////////end search query
$result = mysql_db_query ($dbname, $query, $link);

/// line 82 where i get the error
///Warning: Supplied argument is not a valid MySQL 
/// end few more lines, but can't get layout good in here ;P

$num=mysql_numrows($result);

mysql_close($link);

?>

below are some table's and stuff to display the rows on the page, don't think the problem is there

[lurius]

It's useful to echo query that does not work. By doing that, you could (hopefully) spot errors and such...

Your queries (not all of them) miss LIKE-word...

-Lurius

[Shadeless]

em.. 🙂 2 don't the rest do :p , wil try it thx 🙂

[Shadeless]

You are a genius 🙂 and i'm very stupid not to notice the missing of LIKE words 😕 😕

thx alottie man(or woman) 🙂

[Shadeless]

Hi there again,

1 more question,

how can i count the results,

so that i can display the number of results found. Can this be done within this query?

i've seen query's like SELECT COUNT .. etc

can this be done within the search query? how can i do it exactly, can i do it in a way so it effects all if lines? Hope it makes sence what i write here 🙂

thx in advance,

with kind regards,

Shadeless