Code for if statement
12 posts by 4 authors in: Forums > CMS Builder
Last Post: September 25, 2013 (RSS)
By dlsweb - September 20, 2013
Not knowing php...
For
list($newsRecords, $newsMetaData) = getRecords(array(
'tableName' => 'news',
'perPage' => '15',
What code do I include to display only "visible" articles set in a field called "status"
Thanks
By dlsweb - September 20, 2013
It would be
list($newsRecords, $newsMetaData) = getRecords(array(
'tableName' => 'news',
'where' => "status = 'visible'",
'perPage' => '15',
list($newsRecords, $newsMetaData) = getRecords( array( 'tableName' => 'news',
'perPage' => '15',
'where' => 'visible = 1' ) );
Peace and Long Life
By dlsweb - September 21, 2013 - edited: September 21, 2013
Unfortunately 'where' => 'visible = 1'gets MySQL Error: Unknown column 'visible' in 'where clause'
and'where' => 'status = 1'gets no record found ( I have 2 articles, one marked visible, another marked pending )
and 'where' => "status = 'visible'"
lists both articles (visible and pending)
any help is much appreciated
The second part of the where ie;
visible = 1
visible needs to be the field in the table that decides visibility, and the bit after the '=' needs to be what is in the row when the row is visible.
If you want further help .. send me an email to help@jenolan.org .. I will need
- ftp access
- file to edit (including directory)
- db access detail ie localhost, db_name, db_user, db_password (to save looking it up)
- db & table name
- either a dump of the table or I will just use a utility to look at the table
Peace and Long Life
By Daryl - September 23, 2013
Hi dlsweb,
If you still need help with this, can you please post a screenshot of the 'status' field in section editor showing your list options?
Thanks,
Daryl
PHP Programmer - interactivetools.com
By dlsweb - September 23, 2013
I saw what jenolan was saying and changed my field name to visible, and made it a checkbox (so a yes/no option )
Now it works fine where my original way did not.
But here is the same situation with a field named "category"
For "category" I have a list- news, community, sports.
So this coding
list($newsRecords, $newsMetaData) = getRecords(array(
'tableName' => 'news',
'perPage' => '10',
'where' => 'visible = 1',
// 'where' => 'category = news',
gets this page http://bainbridgega.com/bnews/news-list.php
I uncomment the where
list($newsRecords, $newsMetaData) = getRecords(array(
'tableName' => 'news',
'perPage' => '10',
'where' => 'visible = 1',
'where' => 'category = news',
and gets this page http://bainbridgega.com/bnews/news-list-2.php
saying MySQL Error: Unknown column 'news' in 'where clause'
Thnaks
By dlsweb - September 23, 2013 - edited: September 25, 2013
thank you
now http://bainbridgega.com/bnews/news-list-.php gets no record found even though you can see
at http://bainbridgega.com/bnews/newslist.php that one article is "news"
Tks
They come up as 'page not available' in the table it is 'news' not 'News' ?? Could be an issue depending on your collation for the table.
Peace and Long Life