'quotation problem'

[Joop]

Hi again,

I'm having a little quotation problem, i've read a nice tutorial about variables and how to use them and now knowing this I thought I could make my (not complex) database a searchable one.

Now i've made a <form> wich put the search text in a $sid variable and gets it.. (buttonname search value search)
then i've used "

if ($search == search) { $cruel=SELECT naam, links, rechts, date FROM appel . WHERE naam LIKE $sid;

$db2 = mysql_query($cruel); }

This isn't gonna work, since a mysql_query should be(" enter a mysql console command here ")

Is there something i'm missing?

[Joop]

Uhm, i could also ask, how to put a variable in a mysql command

oh because I was just bashing out code, I forgot the % around $id

[jazztie]

I don't understand what the problem is...

but shouldn't the code look like this:

if ($search == search) { 
  $cruel="SELECT naam, links, rechts, date FROM appel . WHERE naam LIKE $sid"; 

  $db2 = mysql_query($cruel); 
} 

You can read the values in $db2 like this:

$naam     = mysql_result($db2,0,'naam');
$links     = mysql_result($db2,0,'links');
$rechts     = mysql_result($db2,0,'rechts');
$date     = mysql_result($db2,0,'date');

print "$naam, $links, $rechts, $date"; 

You can now print the variables

[Joop]

For sql, u need to quote your LIKE input.. like this:

-->SELECT Text FROM table1
-->WHERE Text LIKE "%this%";

This will output every row of text containing the word this.

Now, i want to put "%$Search%" in a variable. This way you can interactivly search the database..

I happen to know another way around, but that's copying an pasting code.. that will help untill my nextproblem.. so I'm not learning here..

[jazztie]

Put \ in front of "
like: WHERE Text LIKE \"%this%\" (see beneath how it looks in PHP)

WHERE Text LIKE \"%this%\" 

The parser will disregard the quotes...

[Joop]

Cool thanx!!

[jazztie]

No probs...

It goes for any special character that is also used in PHP coding...

Good luck with your search engine