Export categories with tree structure - php

Question

It's common to use ID and parent to store the categories tree. In my case, a knowledge base, in article table every record(article) belong to a category(with a field catid), in categories table every record has a ID and parent to save the tree structure.
article table
id | catid | date | subject | content
categories table
id | name | parent
The problem is when export article to csv or excel,
id | category | date | subject | content
how to keep the whole categories tree, just like when a article belong to son category, I want the category like
grandpa/father/son/, or grandpa : father : son
then can open the csv in excel, menu -> data -> text to column, seperate by ":", you will get a whole tree category not only current category.
work code sample:
$query = 'SELECT `a`.`id` AS `ID`, concat(CASE WHEN ISNULL(`f`.`name`) THEN "" ELSE concat(`f`.`name`,":") END,CASE WHEN ISNULL(`e`.`name`) THEN "" ELSE concat(`e`.`name`,":") END,CASE WHEN ISNULL(`d`.`name`) THEN "" ELSE concat(`d`.`name`,":") END,CASE WHEN ISNULL(`c`.`name`) THEN "" ELSE concat(`c`.`name`,":") END,CASE WHEN ISNULL(`b`.`name`) THEN "" ELSE `b`.`name` END) AS `Category`, `a`.`subject` AS `Subject`,`a`.`content` AS `Content`,`a`.`dt` AS `Date Created`,
FROM articles as a
LEFT JOIN categories as b ON a.catid = b.id
LEFT JOIN categories as c ON b.parent = c.id
LEFT JOIN categories as d ON c.parent = d.id
LEFT JOIN categories as e ON d.parent = e.id
LEFT JOIN categories as f ON e.parent = f.id
WHERE (DATE(`dt`) BETWEEN \'' . $date_from . '\' AND \'' . $date_to . '\') order by id asc';
hope it help for others which want to export tree structure fields.

Answer 1

You have to parse the result after querying the database:
$query = 'query code here';
$result = mysql_query($query);
while( $row = mysql_fetch_array($result) ) {
// do something with that row
}
Besides, mysql_x functions are deprecated. Use PDO or mysqli instead.

Answer 2

<?php
$conn = mysql_connect("localhost","YOUR DB USER NAME","YOUR DB PASSWORD");
if(!$conn) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("YOUR DB NAME", $conn);
$result = mysql_query("YOUR QUERY HERE");
while($row = mysql_fetch_array($result)) {
echo $row['ID'];
-----
-----
}
mysql_close($conn);

Answer 3

I assume, you would like to execute the SQL code in PHP, that you designed in phpMyAdmin - which works well. The most common problems are with the quotes. Learn about String quotes here
As I see you use ' to frame the query code, but you use it inside the query as well. That must have produced an awful series of syntax errors. You may use " to frame the query code while setting it to a variable.
If you get nothing, than maybe the SQL command could have executed, but the result was empty. You can test this with the var_dump() function. Also check that you have all the error reporting switched on. Learn about error reporting here
Also, I'd suggest you to examine and use the PDO class instead of the mysql functions, which would throw you handsome exceptions you could catch and deal with the errors elegantly. The PDO
will also secure your code in some way.