New problem ;). I got script running and showing projects, now i'm trying to
show category for every project from another table. I want to do this in new
page. Everything is fine except ;))... it shows the data in according with
how many projects i have. So if i have 3 project da data from category table
is shown three times, if two projects it's shown two times etc.
The code is following:
<?php
require('config .php');
$conn = mysql_connect(S QL_HOST, SQL_USER, SQL_PASS) or
die ('Could not connect to MySQL database.' .mysql_error()) ;
mysql_select_db (SQL_DB, $conn);
$display_block = "<h1>$project_n ame</h1>
<P>Select a Category to see details.</P>";
//validate category
$cat_id_info = isset($_GET['id']) ? intval($_GET['id']) : 0;
$get_info = "SELECT
category.id,
category.projec t_id,
category.cat_na me,
category.cat_de sc,
project.project _name,
project.id
FROM
category,
project
WHERE
category.projec t_id = $cat_id_info
ORDER BY category.cat_na me";
$result = mysql_query($ge t_info) or die (mysql_error()) ;
$num_category = mysql_num_rows( $result);
$display_block .= "<a href=\"proba_li nk.php\">Back</a></strong></p>";
if (mysql_num_rows ($result) < 1) {
$display_block = "<P><em>BLABLA. </em><br>
<a href=\"proba_li nk.php\">Back</a></strong></p>";
} else {
while ($cat = mysql_fetch_arr ay($result)) {
$cat_id = $cat[id];
$cat_name = strtoupper(stri pslashes($cat[cat_name]));
$cat_desc = stripslashes($c at[cat_desc]);
$display_block .= "<p><strong>$ca t_name</strong>
<br>$cat_desc </p>";
}
}
?>
<HTML>
<HEAD>
<TITLE>Navtec Marine</TITLE>
</HEAD>
<BODY>
<?
print $display_block;
?>
</BODY>
</HTML>
Tnx in advance, best regards
Igor Slivka
show category for every project from another table. I want to do this in new
page. Everything is fine except ;))... it shows the data in according with
how many projects i have. So if i have 3 project da data from category table
is shown three times, if two projects it's shown two times etc.
The code is following:
<?php
require('config .php');
$conn = mysql_connect(S QL_HOST, SQL_USER, SQL_PASS) or
die ('Could not connect to MySQL database.' .mysql_error()) ;
mysql_select_db (SQL_DB, $conn);
$display_block = "<h1>$project_n ame</h1>
<P>Select a Category to see details.</P>";
//validate category
$cat_id_info = isset($_GET['id']) ? intval($_GET['id']) : 0;
$get_info = "SELECT
category.id,
category.projec t_id,
category.cat_na me,
category.cat_de sc,
project.project _name,
project.id
FROM
category,
project
WHERE
category.projec t_id = $cat_id_info
ORDER BY category.cat_na me";
$result = mysql_query($ge t_info) or die (mysql_error()) ;
$num_category = mysql_num_rows( $result);
$display_block .= "<a href=\"proba_li nk.php\">Back</a></strong></p>";
if (mysql_num_rows ($result) < 1) {
$display_block = "<P><em>BLABLA. </em><br>
<a href=\"proba_li nk.php\">Back</a></strong></p>";
} else {
while ($cat = mysql_fetch_arr ay($result)) {
$cat_id = $cat[id];
$cat_name = strtoupper(stri pslashes($cat[cat_name]));
$cat_desc = stripslashes($c at[cat_desc]);
$display_block .= "<p><strong>$ca t_name</strong>
<br>$cat_desc </p>";
}
}
?>
<HTML>
<HEAD>
<TITLE>Navtec Marine</TITLE>
</HEAD>
<BODY>
<?
print $display_block;
?>
</BODY>
</HTML>
Tnx in advance, best regards
Igor Slivka
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