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Search option , to find info in my data base


dannybrazil

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Hello

im planing to build a site for restaurant information . ( style this one http://www.restaurants.co.il/)

 

i wanted to know how to make the search option( Name , Location , City , Category)  like in this site.

that the user can search his restaurant more efficiently.

 

i already know how to insert all the info to the DB but how to i give the order to search?

 

thanks alot for any help !!

 

Danny

 

 

 

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google 'mysql like'.  You want to use the like statement.

SELECT * FROM table WHERE table.field LIKE '%criteria%';

Just a basic example...

 

This will return all data from the table 'table' where the table.field contains the string 'criteria'.

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Hello

i found that code :

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

 

then 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 "<p>Please enter a search...</p>";
  exit;
  }

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

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

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

// Build SQL Query  
$query = "select * from the_table where 1st_field like \"%$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 "<h4>Results</h4>";
  echo "<p>Sorry, your search: "" . $trimmed . "" 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)) {
  $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 "<p>You searched for: "" . $var . ""</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["1st_field"];

  echo "$count.) $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
  $prevs=($s-$limit);
  print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< 
  Prev 10</a>&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 " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>";
  }

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

(from this url ; http://www.designplace.org/scripts.php?page=1&c_id=25)

My question is :
what would be in this file : action="search.php" ????

the only thing i can understand from it that it refers to the same php page , isnt it ?

Danny

 

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