[SOLVED] search box for website?

[artweb]

I need a search box for my website. So that a user can come and find things on my site just by typing in a word. When they click search it will give them a list of pages on my site that has to do with that word they searched for, with the links to the pages. I use mysql and php. I don't want to use google or any pre made search.

[ratcateme]

Hi if you have a database with all your pages in it you could make a search box like this

 

<table>
<?php
$con=mysql_connect('host','user','pass');
mysql_select_db('site_info',$con);
$query="SELECT * FROM `pages` WHERE `page` LIKE ('%".$_REQUEST['search']."%');";
$result=mysql_query($query,$con);
while($row=mysql_fetch_array($result)){
echo '
  <tr>
    <td><a href="'.$row['page_url'].'">'.$row['page_name'].'</a></td>
<td>'.$row['page_description'].'</td>
  </tr>';
}
?>
</table>
<form action="search.php" method="get">
<center><input type="text" name="search" /><input type="submit" value="Search"  /></center>
</form>

 

Scott.

[artweb]

Ok, 

I did what you said. I made a table in my database called site_info and put alot of my sites urls and page names into it. Then I changed the php to match my database. But I keep getting this error.

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /html/search.php on line 308

 

here's my code:

 

<?php

$con=mysql_connect('databasename','password','username');

mysql_select_db('site_info',$con);

$query="SELECT * FROM site_info WHERE page_url,page_names LIKE ('%".$_REQUEST['search']."%');";

$result=mysql_query($query,$con);

while($row=mysql_fetch_array($result)){  This line 308 for me

echo '

  <tr>

    <td><a href="'.$row['page_url'].'">'.$row['page_name'].'</a></td>

 

  </tr>';

}

?>

 

        </table>

<form action="search.php" method="get">

<center><input type="text" name="search" /><input type="submit" value="Search"  /></center>

</form>

 

Thank you for your help.

[Wolphie]

Try this

 

<?php
$con = mysql_connect('databasename', 'password', 'username');
mysql_select_db('site_info', $con);

$query = "SELECT * FROM `site_info` WHERE `page_url`, `page_names` LIKE ('%" . $_POST['search'] . "%')";
$result = mysql_query($query, $con) or die('Error: ' . mysql_error()); // Always use error handling
while($row = mysql_fetch_array($result)) {   This line 308 for me
   echo '
    <tr>
      <td><a href="' . $row['page_url'] . '">' . $row['page_name'] . '</a></td>
    </tr>
  </table>';
}

if(isset($_POST['submit'])) {
  echo '
   <form action="' . $_SERVER['PHP_SELF'] . '" method="post">
     <center><input type="text" name="search" /><input type="submit" value="Search"  name="submit" /></center>
   </form>';
}
?>

[Wolphie]

Oops that's wrong

 

<?php
$con = mysql_connect('databasename', 'password', 'username');
mysql_select_db('site_info', $con);

if(!isset($_POST['submit'])) {
  echo '
   <form action="' . $_SERVER['PHP_SELF'] . '" method="post">
     <center><input type="text" name="search" /><input type="submit" value="Search"  name="submit" /></center>
   </form>';
} else {
  $query = "SELECT * FROM `site_info` WHERE `page_url`, `page_names` LIKE ('%" . $_POST['search'] . "%')";
  $result = mysql_query($query, $con) or die('Error: ' . mysql_error()); // Always use error handling
  while($row = mysql_fetch_array($result)) {   This line 308 for me
     echo '
      <tr>
        <td><a href="' . $row['page_url'] . '">' . $row['page_name'] . '</a></td>
      </tr>
    </table>';
  }
}
?>

[artweb]

Ok,

I did that, and it helped but now I get a new error.

 

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' page_name, LIKE ('%blaster%')' at line 1

 

My database table is set up like this.

 

CREATE TABLE site_info

(

id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,

page_url varchar(100),

page_name varchar(100)

);

 

Thank you again for all your help.

[Wolphie]

Sorry typo, i spelled it "names" rather than "name"

 

Use this:

 

<?php
$con = mysql_connect('databasename', 'password', 'username');
mysql_select_db('site_info', $con);

if(!isset($_POST['submit'])) {
  echo '
   <form action="' . $_SERVER['PHP_SELF'] . '" method="post">
     <center><input type="text" name="search" /><input type="submit" value="Search"  name="submit" /></center>
   </form>';
} else {
  $query = "SELECT * FROM `site_info` WHERE `page_url`, `page_name` LIKE ('%" . $_POST['search'] . "%')";
  $result = mysql_query($query, $con) or die('Error: ' . mysql_error()); // Always use error handling
  while($row = mysql_fetch_array($result)) {   This line 308 for me
     echo '
      <tr>
        <td><a href="' . $row['page_url'] . '">' . $row['page_name'] . '</a></td>
      </tr>
    </table>';
  }
}
?>

[artweb]

I seen that too, but my error still is the same.  ???

[Wolphie]

<?php
$con = mysql_connect('databasename', 'password', 'username');
mysql_select_db('site_info', $con);

if(!isset($_POST['submit'])) {
  echo '
   <form action="' . $_SERVER['PHP_SELF'] . '" method="post">
     <center><input type="text" name="search" /><input type="submit" value="Search"  name="submit" /></center>
   </form>';
} else {
  $query = "SELECT * FROM `site_info` WHERE `page_url`, `page_name` LIKE '%" . $_POST['search'] . "%'";
  $result = mysql_query($query, $con) or die('Error: ' . mysql_error()); // Always use error handling
  while($row = mysql_fetch_array($result)) {   This line 308 for me
     echo '
      <tr>
        <td><a href="' . $row['page_url'] . '">' . $row['page_name'] . '</a></td>
      </tr>
    </table>';
  }
}
?>

I think it may be the brackets. Try that

 

[artweb]

Taking the Brackets didn't change my nice little error. :-\

Thank you for your help, if you know anything else to do, please let me know.

[Wolphie]

<?php
$con = mysql_connect('databasename', 'password', 'username');
mysql_select_db('site_info', $con);

if(!isset($_POST['submit'])) {
  echo '
   <form action="' . $_SERVER['PHP_SELF'] . '" method="post">
     <center><input type="text" name="search" /><input type="submit" value="Search"  name="submit" /></center>
   </form>';
} else {
  $query = "SELECT * FROM `site_info` WHERE `page_url`, `page_name` LIKE '%" . $_POST['search'] . "%'";
  $result = mysql_query($query, $con) or die('Error: ' . mysql_error()); // Always use error handling
  while($row = mysql_fetch_array($result)) {
     echo '
      <tr>
        <td><a href="' . $row['page_url'] . '">' . $row['page_name'] . '</a></td>
      </tr>
    </table>';
  }
}
?>

 

Try that WITHOUT changing a thing, then copy and paste the error here.

[artweb]

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' `page_name` LIKE '%Airless Blaster%'' at line 1

[Wolphie]

Try removing page_url from the query and try just a single field.

[craygo]

you can't put 2 fields to look in 1 like call

 

$query = "SELECT * FROM `site_info` WHERE `page_url` LIKE '%" . $_POST['search'] . "%' OR `page_name` LIKE '%" . $_POST['search'] . "%'";

 

Also WHERE statements are always seperated by AND, OR not a comma

 

Ray

[Wolphie]

Yeah, that's what i thought however i wasn't sure. Thanks for the clarification.

[artweb]

You guys are the best! Excellent, It works perfect. Thanks for your help.

[Wolphie]

Our pleasure, please remember to click "Topic Solved"

[phpSensei]

I don't think he has clicked it, if you don't know what the Topic Solved button is, refer to Wolphie's signature.

[artweb]

You guys are the best! Excellent! It works perfect. Thank you!