Markie
Posts: 251
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| Posted: 08/06/2008, 6:30 AM |
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I thought this was simple, but I'm stuck and hope somebody on this forum can help me.
In my MySQL table (named Mytable), I have made a field named "obsolete", details: varchar (10)
I have made a page, on which I want to show obsolete items. If the "obsolete" field of my MySQL table contains "true", I want to show the item on the obsolete page. On this obsolete page I have made a grid, named mytable. With Visual Query Builder I have added this query to the grid:
SELECT *
FROM mytable
WHERE Obsolete LIKE '%{true}%'
However, it doesn't work. The page shows all items in my database.
What do I do wrong. Details of my query:
field = Obsolete (Text)
condition = contains (like '%...%' )
parameter = true (url)
Markie
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wkempees
Posts: 1679
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| Posted: 08/06/2008, 6:41 AM |
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field = Obsolete (Text)
condition = equals(= )
parameter = "true" (expression)
if you actualy test for obsolete to contain the string "true"
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Markie
Posts: 251
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| Posted: 08/06/2008, 7:45 AM |
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Dank je Walter !
I indeed had to use the quotes ("...") and expression instead of url
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The Netherlands, GMT+1
Tools: CCS 5.1, Windows 7, Navicat, Ultraedit
Local server: XAMPP with Apache, php and MySQL
Webserver: Windows 2008 IIS 7, php and MySQL |
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