Select and Display MySQL Queries on same page

[roninmedia]

[php:1:b3cecffc86]<?

@ $db = mysql_pconnect(\"localhost\");

 

if (!$db)

{

echo \"Error: Could not connect to database. Please try again later.\";

exit;

}

 

mysql_select_db(\"mysql\");

$manga_sql = \"select * from manga

where title is not null

order by title desc\";

 

$manga_result = mysql_query($manga_sql);

if (mysql_num_rows($manga_result)) {

 

$manga = mysql_fetch_array($manga_result);

echo \"<a href=\"review.php?id=$manga[id]\">\";

echo $manga[title];

echo \"</a>\";

 

}

 

 

?>

 

This is the mysql query that is run on index.php. It displays all the reviews I have in a databse and visitors are then sent to review.php (the code for it shown below) which displays the review they queried.

 

<?php

 

/* just to be safe with register globals */

$id =& $HTTP_GET_VARS[\'id\'];

 

$sql = \'SELECT *

FROM manga

WHERE id=\'.$id;

 

/* then run the query and get the results etc... */

 

?>

[/php:1:b3cecffc86]

 

How would I integrate the two codes on to one single php file so the original mysql query and the query that is selected is on the same php page?

[DylanBlitz]

Give this a shot. Haven\'t tested it but it should work

 

[php:1:0fd7680877]<?php

@ $db = mysql_pconnect(\"localhost\");

 

if (!$db)

{

echo \"Error: Could not connect to database. Please try again later.\";

exit;

}

 

$id = $HTTP_GET_VARS[\'id\'];

 

if ((!$id) || ($id = \'\'))

{

mysql_select_db(\"mysql\");

$manga_sql = \"select * from manga

where title is not null

order by title desc\";

 

$manga_result = mysql_query($manga_sql);

if (mysql_num_rows($manga_result))

{

 

$manga = mysql_fetch_array($manga_result);

echo \"<a href=\"review.php?id=$manga[id]\">\";

echo $manga[title];

echo \"</a>\";

 

}

 

} else {

 

$sql = \'SELECT * FROM manga WHERE id=\'.$id;

 

/* then run the query and get the results etc... */

 

}

?>

[/php:1:0fd7680877]