TalkPHP - [SOS]a search for a database ,double click here

TalkPHP (http://www.talkphp.com/forums.php)

- Absolute Beginners (http://www.talkphp.com/absolute-beginners/)

- - [SOS]a search for a database ,double click here (http://www.talkphp.com/absolute-beginners/1881-sos-search-database-double-click-here.html)

webtuto

01-06-2008 09:37 PM

[SOS]a search for a database ,double click here

hi ,
well im trying to do a search form for my database here is the code =>

PHP Code:


		
			
<?php



if(!empty($_POST['search']) ) {

$con = mysql_connect("localhost","root","") ;

$db = mysql_select_db("test") ;

$sql = "select * from webt where name='%$_POST[search]%' ";

$res = mysql_query($sql) ;

if (mysql_num_rows($res) == 0 ) {

          echo "No result";

 }else{ while ($row = mysql_fetch_array($res) ) {

       

      echo $row[name]."<br><br>" ; 

       }

       

}

}else{ echo"full the form"; } 



?>

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

<b>Search</b><input type="text" name="search" >

<input type="submit" value="Search" >

</form>

and in the database there is name='aa'
name is the name of the column
anyway when i delete "%" the code works great but when i add them it douesnt work *!*

bluesaga

01-06-2008 09:46 PM

% is a wildcard, that should only be used for LIKE mysql searches...

PHP Code:


		
			
$db = mysql_select_db("test") ;
$sql = "select * from webt where name='%$_POST[search]%' ";
$res = mysql_query($sql) ;

Should probably be:

PHP Code:


		
			
$db = mysql_select_db("test") ;
$sql = "select * from webt where name LIKE '%$_POST[search]%' ";
$res = mysql_query($sql) ;

please note, your script is FAR from secure and you should look at some topics here regarding:
http://www.talkphp.com/tips-tricks/1...rotection.html

Wildhoney

01-06-2008 09:47 PM

The actual syntax for that is like so:

sql Code:

SELECT
    myColumn
FROM
    myTable
WHERE
    myColumn
LIKE
    '%findThis%'

Edit: Meh young William, meh. Nothing else to add.

webtuto

01-06-2008 10:07 PM

thanks it works but for exemple you search "aaa" , i want it to give me the result from database as "aa"
but it douesnt :o
it give "no result"

EyeDentify

01-12-2008 01:15 PM

@webtuto

the % procent means that it replaces "anything" so for example you want to find a name ending with "Harrison" and you would search for "%Harrison". See the % in the beginning ?

Switch place of the % if you want to search for the name beginning with "Harrsion" as so "Harrison%". Catching on now ?

And if you want to search for a name containing "Harrison" then it would be "%Harrison%".

I belive you can use a ? questionmark for replacing a single char.
Correct me if im out on a limb here ?

And see Wildhoney and bluesaga´s post above.

I had trouble with the LIKE thing in my early days of MySQL queries to.

Good Luck.

/EyeDentify

All times are GMT. The time now is 09:46 AM.