Help with querying a MySQL database

Pleb

hi, im new to php, and i'm very 😕 . For my site i decided to add a ringtone section, so i've made the actual database, but i am having trouble retrieving the info

heres the code ive got at the mo:

<html>
<head>
<title>CyberPlebs</title>
<STYLE type="text/css">
	#P { font: 9pt Verdana,Helvetica,Arial }
	#TD { font: 9pt Verdana,Helvetica,Arial }
    .noline {a:Verdana,Helvetica,Arial,Sans Serif;font-size:10pt;color:#000000;text-decoration:none;}
    A:hover { color:#3399FF; }
	</STYLE>

<style>
<!--
a{text-decoration:none}
//-->
</style>
</head>
<body bgcolor="#000000" text="#FFFFFF" link="#FFFFFF" vlink="#FFFFFF" alink="#FFFFFF">

<?php



$db = mysql_connect("localhost", "username", "password");

mysql_select_db("username_db",$db);

// display individual record

if ($id) {

   $result = mysql_query("SELECT * FROM ringtone WHERE id=$id",$db);

   $myrow = mysql_fetch_array($result);

   printf("Title: %s\n<br>", $myrow["title"]);

   printf("Band: %s\n<br>", $myrow["band"]);

   printf("RingTone: %s\n<br>", $myrow["ringtone"]);

   printf("Tempo: %s\n<br>", $myrow["tempo"]);

   printf("Category: %s\n<br>", $myrow["Category"]);

}

else {

// show employee list

   $result = mysql_query("SELECT * FROM ringtone",$db);

if ($myrow = mysql_fetch_array($result)) {

  // display list if there are records to display

  do {

    printf("<a href=\"%s?id=%s\">%s %s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["title"], $myrow[""]);

  } while ($myrow = mysql_fetch_array($result));

} else {

  // no records to display

  echo "Sorry, no records were found!";	

}
}
?>



</body>



</html>

this code kinda works, i the ID bit at the top i want to keep, but the bit below that i want changed

I want it changed so that if the query is ?Category=<a category name here> it brings up all the names of the different bands that have made a tune in that Category, and if there is no band, it shows the titles of the induvidual ringtone in that category

also,

I want it changed so that if the query is ?Band=<a band name here> it brings up all the titles of the ringtones made by that Band

also,

I want it changed so that if the query is nothing it brings up all the different Categories

... I know it's a lot to ask, but i would be very gratefull if anyone can make the code for me that does that.....Thanks in advance 🙂

malbera

it's not so dificult

I want it changed so that if the query is ?Band=<a band name here> it brings up all the titles of the ringtones made by that Band

$result = mysql_query("SELECT * FROM ringtone WHERE band='$_GET[band]' ",$db);

I want it changed so that if the query is nothing it brings up all the different Categories

this depends on what query returns nothing
anyway, something like this might help you

if (mysql_num_rows($result)==0) {
$result=mysql_query("select * from ringtone where category='others'");
}

hope that helps

Pleb

k, so i got this now,

<html>
<head>
<title>CyberPlebs</title>
<STYLE type="text/css">
	#P { font: 9pt Verdana,Helvetica,Arial }
	#TD { font: 9pt Verdana,Helvetica,Arial }
    .noline {a:Verdana,Helvetica,Arial,Sans Serif;font-size:10pt;color:#000000;text-decoration:none;}
    A:hover { color:#3399FF; }
	</STYLE>

<style>
<!--
a{text-decoration:none}
//-->
</style>
</head>
<body bgcolor="#000000" text="#FFFFFF" link="#FFFFFF" vlink="#FFFFFF" alink="#FFFFFF">

<?php



$db = mysql_connect("localhost", "username", "password");

mysql_select_db("username_db",$db);

// display individual record

if ($id) {

   $result = mysql_query("SELECT * FROM ringtone WHERE id=$id",$db);

   $myrow = mysql_fetch_array($result);

   printf("Title: %s\n<br>", $myrow["title"]);

   printf("Band: %s\n<br>", $myrow["band"]);

   printf("RingTone: %s\n<br>", $myrow["ringtone"]);

   printf("Tempo: %s\n<br>", $myrow["tempo"]);

   printf("Category: %s\n<br>", $myrow["Category"]);

}

else if {

// show employee list

$result = mysql_query("SELECT * FROM ringtone WHERE band='$_GET[band]'",$db);

if ($myrow = mysql_fetch_array($result)) {

  // display list if there are records to display

  do {

    printf("<a href=\"%s?id=%s\">%s %s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["title"], $myrow[""]);

  } while ($myrow = mysql_fetch_array($result));

} else {

  // no records to display

  echo "Sorry, no records were found!";	

}
else {

// show employee list

   $result = mysql_query("SELECT * FROM ringtone",$db);

if ($myrow = mysql_fetch_array($result)) {

  // display list if there are records to display

  do {

    printf("<a href=\"%s?id=%s\">%s %s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["title"], $myrow[""]);

  } while ($myrow = mysql_fetch_array($result));

} else {

  // no records to display

  echo "Sorry, no records were found!";	

}
}
?>



</body>



</html>

but im getting this error:

Parse error: parse error, expecting `'('' in ringtone.php on line 47

and i don't know what i need to put and where i need to put it

malbera

which line is #47??

Pleb

dunno...

malbera

try to figure it out (get some good editor) then post pls again

Pleb

Its saying that the line in bold is the line with the problem:

printf("RingTone: %s\n<br>", $myrow["ringtone"]);
printf("Tempo: %s\n<br>", $myrow["tempo"]);
printf("Category: %s\n<br>", $myrow["Category"]);
}
else if {
// show employee list
$result = mysql_query("SELECT * FROM ringtone WHERE band='$_GET[band]'",$db);
if ($myrow = mysql_fetch_array($result)) {

abda53

the editor is use is cedit (crimson editor) its free..
http://www.crimsoneditor.com

you can go to a specific site, and also search for a word/string in multiple documents all at once..

TouchTheClouds

is it possible to compare data from two different rows of the same mysql table without getting them to php and then sending back like this:

$first = mysql_query("blabla"); //get data from one row
$second = mysql_query("blabla"); //from another
if ($first > $second) blabla;

(it is what i have now)

I need something like this:

$check = mysql_query("SELECT * FROM table Where row1 > row2");

?
thanks

Pleb

k, ive got this script:

<html>
<head>
<title>CyberPlebs</title>
<STYLE type="text/css">
	#P { font: 9pt Verdana,Helvetica,Arial }
	#TD { font: 9pt Verdana,Helvetica,Arial }
    .noline {a:Verdana,Helvetica,Arial,Sans Serif;font-size:10pt;color:#000000;text-decoration:none;}
    A:hover { color:#3399FF; }
	</STYLE>
<style>
<!--
a{text-decoration:none}
//-->
</style>
</head>
<body  style="font-family: Comic Sans MS" bgcolor="#000000" text="#FFFFFF" link="#FFFFFF" vlink="#FFFFFF" alink="#FFFFFF">
<?php
$db = mysql_connect("localhost", "username", "password");
mysql_select_db("username_db",$db);
// display individual record
if ($id) {
   $result = mysql_query("SELECT * FROM ringtone WHERE id=$id",$db);
   $myrow = mysql_fetch_array($result);
   printf("Title: %s\n<br>", $myrow["title"]);
   printf("Artist/Band: %s\n<br>", $myrow["artist/band"]);
   printf("RingTone: %s\n<br>", $myrow["ringtone"]);
   printf("Tempo: %s\n<br>", $myrow["tempo"]);
   printf("Category: %s\n<br>", $myrow["Category"]);
}
else if ($Category) {
$result = mysql_query("SELECT * FROM ringtone WHERE Category='$_GET[Category]'",$db);
    if ($myrow = mysql_fetch_array($result)) {
      // display list if there are records to display
      do {
        printf("<a href=\"%s?id=%s\"><b>%s</b>: %s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["artist/band"], $myrow["title"]);
      } while ($myrow = mysql_fetch_array($result));
    } else {	
      echo "Sorry, no records were found!";
} 
}
else {
        printf("&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<font size="2" color="#3399FF">Category List</font><br>\n", $PHP_SELF); 
        printf("<a href=\"%s?Category=TV\">TV Tunes</a><br>\n", $PHP_SELF);
        printf("<a href=\"%s?Category=Films\">Film Tunes</a><br>\n", $PHP_SELF);
        printf("<a href=\"%s?Category=Rock\">Rock Tunes</a><br>\n", $PHP_SELF);
}
?>


</body>



</html>

but when i run it i get an error message,

parse error: parse error in ringtones.php on line 43

but the error message doesn't appear if i delete the last else statement, why do i get the error message?? Please help

Pleb

somebody...please reply....PLEASE

abda53

 <font size="2" color="#3399FF">Category

try taking out the quotes...

abda53

either take out the " " or make it a \" \"
you probably just overlooked it

Pleb

Thank you 2 sooooo much, i couldn't see that for the life of me...

abda53

no prob.. you can send me $5 through paypal at abda53@itookmyprozac.com
🆒