prroblem of linking lists

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om.bitsian
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Posts: 36
Joined: Wed Sep 09, 2009 4:13 am

prroblem of linking lists

Post by om.bitsian »

Hello friends
i have 2 list in my form 1. Employee Type and 2. Designation in first
list (1. Employee Type) i have 3 valuse 2,3,4 ....for designation field
i ahve a table DESIG .....I want output in such a way that when i select
employee==2 .....it should show the related designation from table and when
i select employe_type==3 it should show the designation related to that types of employees

in DESIG table i have two columns
employ_type || designation
2 LECT
2 ASOP
3 LEST
3 PLOC
4 IPCK
etc........i am getting these designation values from database but now i do not know how to send these values to the designation list(i mean loop or somting else)
can anybody suggest me how to code for this so that i can link those two lists
Attachments
payentry.zip
(5.66 KiB) Downloaded 44 times
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Ollie Saunders
DevNet Master
Posts: 3179
Joined: Tue May 24, 2005 6:01 pm
Location: UK

Re: prroblem of linking lists

Post by Ollie Saunders »

Something like this: SELECT employeeType, employeeDesignation FROM employee, designation WHERE employeDesignation = designationId;

Or you could use a join.
om.bitsian
Forum Commoner
Posts: 36
Joined: Wed Sep 09, 2009 4:13 am

Re: prroblem of linking lists

Post by om.bitsian »

with quary i am ok....
$query = "SELECT DESIG from desig WHERE EMP_TYPE=2";
$result = mysql_query($query) or die ("Error in query: $query");
$row = mysql_result($result,0,"DESIG");


now this result i have to send to the other list disignation....
<label >9. Designation</label></select>

can you please tell me how can i post the result of quary into the designation list
(coding part.......so that only emplyee type -2 , designation appear in the list)
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Ollie Saunders
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Posts: 3179
Joined: Tue May 24, 2005 6:01 pm
Location: UK

Re: prroblem of linking lists

Post by Ollie Saunders »

You're very difficult to understand. What are these lists you speak of?
so that only emplyee type -2 , designation appear in the list
2 or -2? I think you mean 2.

Try:

Code: Select all

$q = "SELECT * FROM desig WHERE EMP_TYPE = 2";
$result = mysql_query($q) or die(mysql_error());
while ($row = mysql_fetch_assoc($result)) { 
   $rowDump = var_dump($row);
   echo "<pre>$rowDump</pre><br />";
}
om.bitsian
Forum Commoner
Posts: 36
Joined: Wed Sep 09, 2009 4:13 am

Re: prroblem of linking lists

Post by om.bitsian »

values are coming in the valriable $rowDump (last post) but still the designation list is blank.......please check the both list and suggest me how to solve this problem
last post: while($row = mysql_fetch_assoc($result)) {
$rowDump = var_dump($row);
echo "<pre>$rowDump</pre><br />";
$desg=($_POST['rowDump']); }

this is the form code:
<label>3. Employee Type</label>
<select name="emptyp">
<option value="2">2 (Teaching faculty)</option>
<option value="3">3 (Supporting staff)</option>
<option value="4">4 (Helping staff)</option>
</select>



<label >9. Designation</label>
<select name="desg" value="desg" > </select><br><br>
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Ollie Saunders
DevNet Master
Posts: 3179
Joined: Tue May 24, 2005 6:01 pm
Location: UK

Re: prroblem of linking lists

Post by Ollie Saunders »

You never answered me question on what a list is but I managed to glean it from your example. For future reference, you should describe these things as one of the following: combo-box, select field, or drop-down menu. Any of those work. List, on the other hand, is very generic and generally meaningless in the context of software.

Also, whilst I'm correcting your language; it's query, not quary.
but still the designation list is blank.......please check the both list and suggest me how to solve this problem
You need to generate the HTML. Replace my var_dump() with output of an option tag (using echo or something). I'm not going to do it for you but post back if you have difficulty.
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