Question

table: taxonomy_index

nid          tid   
2              1
3              1
3              4
3              5
4              6
4              1
4              3
4              7

table: taxonomy_term_data

tid           vid          name
1              2           java
2              2            php
3              2            c
4              1           tag1
5              1            tag2
6              1            tag3
7              1            tag4
8              1            tag5

now i want to according to nid=$nid get the name where vid=2.? how do i do?

the following is my query code. but it’s wrong.

$result = mysql_query('select tid,name form taxonomy_index as ti left join taxonomy_term_data as ttd on ti.tid=ttd.tid where vid=2 and nid=$nid')

Answer 1

My guess is that you need to replace your single quotes ' with double quotes ". This is because PHP does not expand variables surrounded by single quotes.

$result = mysql_query("select tid,name from taxonomy_index as ti left join taxonomy_term_data as ttd on ti.tid=ttd.tid where vid=2 and nid=$nid")

EDIT:

You also have from misspelled as form