[SOLVED] Help needed related where condition in select statement

[huzefa]

i have a column vmkey in table wvideos, in that column i have master key of all videos like 'cW3z70WMCTg' for the youtube video of

 

Now i want to get the url from users favourite video & want to check in my database if that already exist in database or not

 

i m checking it by following condition

 

$sql = "select * from wvideos where '".$_POST['surl']."' LIKE vmkey";

 

but its not work, urgent help is needed

 

thank you

 

Huzefa

[dannyb785]

change

$sql = "select * from wvideos where '".$_POST['surl']."' LIKE vmkey";

 

to

 

$surl = addslashes($_POST['surl']);
$sql = "select * from wvideos where vmkey LIKE '%$surl%' ";

 

That should do it(provided the column name with the vmkeys's name is 'vmkey'. ).

[huzefa]

$surl = addslashes($_POST['surl']);

$sql = "select * from wvideos where vmkey LIKE '%$surl%' ";

 

Sorry but its not working, i have extra characters in $_POST['surl'] not in vmkey

[dannyb785]

$surl = addslashes($_POST['surl']);

$sql = "select * from wvideos where vmkey LIKE '%$surl%' ";

 

Sorry but its not working, i have extra characters in $_POST['surl'] not in vmkey

 

Are you putting $sql into a query?

 

[huzefa]

yah

 

$result = mysql_query($sql) or die('Error connecting to database, Please try later');

if (mysql_num_rows($result) > 0){

// further processing here

}

 

 

 

[dannyb785]

how about changing the die to "or die(mysql_error()); "

 

that should tell you your error

[huzefa]

i changed to "die(mysql_error());" but not show any error message

[huzefa]

I have in column vmkey : cW3z70WMCTg

& if user submit  :

 

than i want to show a message that video already added

 

experts  help is needed

 

Thank you

 

Huzefa

 

 

[huzefa]

is any body have a solution of my problem?

[dannyb785]

yah

 

$result = mysql_query($sql) or die('Error connecting to database, Please try later');

if (mysql_num_rows($result) > 0){

// further processing here

}

 

 

 

 

What's the "further processing" ?

 

And I didnt realize you needed to search for an exact entry. Don't to the LIKE. Change it to:

$surl = addslashes($_POST['surl']);
$sql = "select * from wvideos where vmkey='$surl' ";

 

If this doesn't work, then either your form is not putting the variable into $_POST['surl'] or there is some issue with your database and the values in it or the way it's setup

[huzefa]

$surl = addslashes($_POST['surl']);

$sql = "select * from wvideos where vmkey='$surl' ";

 

how its possible?

 

value in  $surl  is

& in vmkey column is cW3z70WMCTg

 

so how its locate as true

 

i also checked the variable & all others

 

 

 

 

 

[dannyb785]

^ omfg.

 

 

 

THIS is why I wish people would be more specific

 

Christ. I swear, I love helping people for free... but this is ridiculous.

 

 

Think about it... You're trying to find xxx value in http://www.youtube.com/xxx(or whatever) so would it make sense to search for 'xxx' in 'http://www.youtube.com/xxx' or search for 'http://www.youtube.com/xxx' in 'xxx' ? Because you're looking for 'http://www.youtube.com/xxx' in 'xxx' which obviously would never happen. Had you told us(or me) to begin with that $surl would have the entire url, it would've taken 2 seconds to say "just remove the part in front of the code you're looking for.

 

Once again. If you have a row in the database that says "bob" and then a variable b that says "bobby". Would it make sense to search for 'bobby' in the database to try and find 'bob' ?

[sasa]

try

<?php
$url = 'http://www.youtube.com/watch?v=cW3z70WMCTg';
$url = parse_url($url);
parse_str($url['query'],$out);
$url = addslashes($out['v']);
$sql = "select * from wvideos where vmkey='$url' ";
?>

[huzefa]

<?php

$url = '

$url = parse_url($url);

parse_str($url['query'],$out);

$url = addslashes($out['v']);

$sql = "select * from wvideos where vmkey='$url' ";

?>

 

Thanks sasa

 

Its working properly for youtube video, but what about google video, there is 'docid' variable & a lot of more video provider also on the net.

 

so is it not possible to check database value in the variable in where condition?

 

Thanks

 

Huzefa

[sasa]

try

<?php
$url = 'http://www.youtube.com/watch?v=cW3z70WMCTg';
$sql = "select * from wvideos where locate(vmkey,'$url') ";
?>

[huzefa]

Thank you sir,

 

now my problem has been solved with it.

 

Thanks once again

 

Huzefa