str_replace()

seabird

Hi everyone,

I have a PHP generated Jumpmenu on my site, using tablenames as group-headers.

I want my table about_me to show as about me.

I've been told to use str_replace() but have no clue how?

Can anyone help me? Here's my code

(btw not all tables have a underscore.... is that a problem?)

<?PHP
$link = mysql_connect();
$db=content;

if (! $link)
die( "couldn't connect");
mysql_select_db($db)
or die ("won't open test:".mysql_error() );

$db_res=mysql_list_tables("$db",$link);
$num=mysql_num_rows($db_res);
for($x=0; $x<$num; $x++)
{
$table=mysql_tablename($db_res,$x);
/////////Don't want underscore here////////////////
print "<option>".$table."</option><br>
////////That way my link shows the space/////////
{
$query = "SELECT DISTINCT * FROM $table";
$result = mysql_query($query) or die("Error in query");

while($row = mysql_fetch_assoc($result))
{
print "<option value=".$row['Field2']."> -".$row['Field1']."</option><br>\n";
}
}
}
?>

Jacco

ultraslacker

Check out the man page for str_replace() - that should get you started.

http://www.php.net/manual/en/function.str-replace.php

seabird

Sorry for being a zero, but I'm still lost after reading that page. All I know know is that I need some arguments, but don't know which ones.

str_replace( "%%" , "% %" , "%%")

was my best attempt replacing About_me to About me, but all I got was a error.

Can someone pls just give me the answer to this one, after puzzeling for 6 hours on my jumpmenu I don't feel like puzzeling this one now.

Thanx in advanse,
Jacco

seabird

I had some help elsewhere I found my solution:

$table=mysql_tablename($db_res,$x);
$table = str_replace("", " ", $table); // replaces all "" with " " (space)
print "<option>".$table."</option><br>";
$table = str_replace(" ", "", $table); // replaces all " " with ""

Thanx anyways