Another (quite) newbie question

[Joop]

Hi,

I need to make a WHERE name LIKE %textvalue% query in my database. First i tried this:

if ($Search == Search) 
{ $eps = mysql_query("SELECT naam, id, date FROM apple WHERE naam LIKE \"%$sid\""); } 

wich resulted in a my_sql error, i thought this was either a quotation problem or the $sid wasn't in the mysql string, that last one was easy to find out, i removed the mysql_query function and echo("$eps"); wich stated the query correctly.

Then I reconstructed with echo's and stuff to make sure the quotes wasn't a problem wich alse came out negative..

Can someone help me out here?, it's probebly a minor mistake.


(Just in case, Search == Search chks if the search button is pressed and $sid is a value you enter in a form)

[petesmc]

Hi,

Try this:

if ($Search == "Search") 
{ $eps = mysql_query("SELECT naam, id, date FROM apple WHERE naam LIKE '%$sid'"); } 

[Joop]

Ah yes, I forgot to some quotes here and then,. but the code U stated didn't work either

I think I'm making a major error here, let me first paste some code over here

if ($Search == "Search") {
$sql = "SELECT naam FROM appel";
$eps = mysql_query("$sql"); 

This wil put the data of the column naam from appel into $eps, now I can get put this info into paragraphs (or tables)

while ($row = mysql_fetch_array($eps) ) {
$naam=$row["naam"]; echo("<p>". $naam . "</p>"); 

And close the functions

} } 

Now i wish to search the database for specific data, and I've searched the mysql reference to make that happen

All i need to do is add a WHERE naam LIKE %$sid% line in the mysql console wich works perfectly, now adding it into my php code.. i chanched the $sql line into:

$sql = "SELECT naam FROM appel" . 
"WHERE naam LIKE %$sid%"; 

wich gives me an undefined call error,.. what's wrong with my aproach?, are there alternatives?

[petesmc]

Hi,

You must put single quotes around %$sid% so '%$sid%'.
Also, make sure that sid actually exists.

[oodie]

Originally posted by Joop

$sql = "SELECT naam FROM appel" . 
"WHERE naam LIKE %$sid%"; 

You forget to put a space here between appel and WHERE. Also like petesmc said, put space around %$sid%

$sql = "SELECT naam FROM appel" . 
" WHERE naam LIKE '%$sid%'"; 

If the above fails, try to put $sid outside the quote like this.

$sql = "SELECT naam FROM appel" . 
" WHERE naam LIKE '%". $sid. "%'";