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View Indexed Field in another table

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View Indexed Field in another table

 
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  #1
Feb 11th, 2009
I'm having trouble figuring out how to display data from another indexed table. What I have is one table (assets) with a majority of my data in it. I use a simple select statement to display the data. Within that data is a field of "typeid" which is an indexed field to my "assettype" table. With my simple select statement I see only the indexed field "assets.typeid". I want to see the "assettype.type" field from the assettype table.

In a nutshell, i want to change the typeid of "1" to be "printer", but "printer" is in a different table that is indexed by "typeid" number. Here is a snippet, hope it helps.
PHP Syntax (Toggle Plain Text) 
  1. $query = "SELECT * FROM assets WHERE tag = $tag";
  2. $result = mysql_query($query) or die('Sorry, could not find tag requested');
  3.  
  4. $row = mysql_fetch_array($result, MYSQL_ASSOC) or die('No records retrieved');
  5.  
  6. $tag=$row['tag'];
  7. $typeid = $row['typeid'];
Further down the script
PHP Syntax (Toggle Plain Text) 
  1. echo "<h3><u>Tag#</u><font color=\"#663300\" size=\"2\"> $tag</font></h3>"; 
  2. echo "<h3><u>Type</u><font color=\"#663300\" size=\"2\"> $typeid</font></h3>\n";
Where $typeid is echoed, i want to echo $type instead, but it is in a different table that is indexed with $typeid.
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Re: View Indexed Field in another table

 
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  #2
Feb 11th, 2009
You'll have to query both tables with a WHERE clause and set the indexes equal to each other:

PHP Syntax (Toggle Plain Text) 
  1. SELECT table1.blah, table2.blahz FROM table1, table2 WHERE table1.tag = table2.tag1;

Sorry, I didn't pay attention to your column names, etc. but you get the idea.
Ken Hess
Linux DaniWeb Staff Writer/Columnist/Host
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Re: View Indexed Field in another table

 
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  #3
Feb 11th, 2009
Thank you for the reply. I did fail to leave out some information that, as I see, is now pertinant. This page is coming from another page where its using $tag as the id in a $_GET function.
PHP Syntax (Toggle Plain Text) 
  1. $tag = $_GET['id'];
  2.  
  3. $query = "SELECT * FROM assets WHERE tag = $tag";
  4. $result = mysql_query($query) or die('Sorry, could not find tag requested');
  5.  
  6. $row = mysql_fetch_array($result, MYSQL_ASSOC) or die('No records retrieved');
  7.  
  8. $tag=$row['tag'];
  9. $typeid = $row['typeid'];

So would i then have to alter my WHERE something like
PHP Syntax (Toggle Plain Text) 
  1. $query = "SELECT assets.*, assettype.* FROM assets WHERE assets.tag = "$tag" AND assettype.typeid = "$typeid"
  2.  

That didn't work but that's where my mind is going. Or create a $query2 and put them together somehow?
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Re: View Indexed Field in another table

 
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  #4
Feb 11th, 2009
You need to use a join I think

PHP Syntax (Toggle Plain Text) 
  1. $q = @mysql_query("SELECT a.tag, a.field2, at.assettype FROM assets AS a INNER JOIN assettype AS at
  2. ON a.assettypeid = at.assettypeid
  3.  WHERE a.tag = $tag);
  4.  
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Re: View Indexed Field in another table

 
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  #5
Feb 11th, 2009
Thanks TommyBs, I'll give it a try. Other work is pressing so I'll have to get back to this a little later, but really do appreciate the help.
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Re: View Indexed Field in another table

 
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  #6
Feb 11th, 2009
Ok, this is what I tried.
PHP Syntax (Toggle Plain Text) 
  1. $query = @mysql_query("SELECT assets.*, assettype.* FROM assets AS a INNER JOIN assettype AS at ON assets.typeid = assettype.typeid WHERE assets.tag = $tag");

I have an "or die" on my $result = and recieved an error of "sorry, could not find tag requested"
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Re: View Indexed Field in another table

 
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  #7
Feb 13th, 2009
Does anyone have any ideas?
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Re: View Indexed Field in another table

 
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  #8
Feb 13th, 2009
Well doesn't that mean the tag you passed in doesn't exist?
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Re: View Indexed Field in another table

 
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  #9
Feb 18th, 2009
No, the tag exsists. If I use the original query
PHP Syntax (Toggle Plain Text) 
  1. $query = "SELECT * FROM assets WHERE tag = $tag";
The script continues to run and is successful. Only when I try to pull data from 2 different tables in this query is when I'm running into problems.
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