Help sql select...

[ivanella]

I guys, i'm sorry for my english but i'm italian girl.

 

I need help for this script:

 

$sql = "SELECT link_title FROM film_links WHERE '{$attori}' LIKE CONCAT( '%', link_field6, '%' )";

 

it dosnt work, do you know why???

 

thanks for help....

[dptr1988]

I'm not sure if this is the problem but it looks like your values are reversed in the 'LIKE' part

 

Try this:

<?php
$sql = "SELECT link_title FROM film_links WHERE  link_field6 LIKE '%{$attori}%' ";
?>

[Rohan Shenoy]

Try if this works:

$sql = "SELECT link_title FROM film_links WHERE $attori LIKE  '%$link_field6%')";

[dptr1988]

Rohan Shenoy: 'link_field6' appears to be a SQL field name, rather then a PHP variable

 

[Rohan Shenoy]

^ya true, looks like to me too

but u never know what the db table structure is like.

[ivanella]

$link_field6 is a field of database

 

that conteins values like:

 

"Marco" or "Luca" or "Anna"

 

while

 

$attori

 

is a variabile with values like:

 

"Marco Luca Anna"

 

....thanks for help....

 

 

 

[dptr1988]

ivanella: Did you try my example? Did it work?

 

Rohan Shenoy:

 

Yes, and it's always good to try everything possible and get everyones ideas. More brains are better then one.

 

Note: If you want to get really, really, really, really, technical, there are 5 indications that the name 'attori' is a PHP variable and none of those 5 indicators appear on the 'link_field6' fieldname. Those find indicators are left and right single quotes, left and right curly brackets and the dollar sign. A SQL field name would not be quoted in a situation like this, would never have the curly brackets or the dollar sign. So if the person writing this code put all of those characters in on one PHP variable that indicated it was a PHP variable, there is a good chance that they would also at least include 1 of those indicators on the 'link_field6' name.  Also the word 'field' appears in 'link_field6' which also suggests that could be a SQL field name.

 

 

 

 

 

[dptr1988]

If your $attori variable contains multiple names like you mentioned, you can replace all of the spaces between the names with '%' before using it in the SQL query, so that it would match each word in it rather then the whole string.

 

<?php
$attori = str_replace(' ', '%', $attori);
?>

[Rohan Shenoy]

@dptr 1988

Thanks dude!

[soycharliente]

I think she said it would contain a name LIKE those. Not a list of names. I'm fairly confident that dptr's solution is what she needs.

 

<?php
$sql = "SELECT `link_title` FROM `film_links` WHERE `link_field6` LIKE '%{$attori}%'";
?>