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PHP Query, OUTPUT

Posted on 2011-03-17
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Last Modified: 2012-05-11
Hi Experts,

Ive got a query that returns results from a database but does not return all entries under the "last_spoken_to" field.. which is in the notes table..

the current query i have is as follows:

$sql = "select * from customer left join notes on customer.customer_name = notes.customer_name WHERE customer.customer_name = '" . $val . "'";

I have tried the following:

$sql = "select * from customer left join notes on customer.customer_name = notes.customer_name AND customer.customer_name = last_spoke_to.customer_name WHERE customer.customer_name = '" . $val . "'";

but that does not work, it returns an error, What i have got is attached in the picture, And i need it to produce the same output as the notes category.

Many thanks

AS you can see from the picture Notes repeats but last spoke to does not how would i get it to repeat like the notes? as i stated ive tried the above query but that did not work.


 
picture5.jpg
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Comment
Question by:NeoAshura
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5 Comments
 
LVL 6

Assisted Solution

by:brb6708
brb6708 earned 1000 total points
ID: 35155553
so far i can see is "last_spoke_to" a field in one of the tables "customer" or "notes".

The way you use it is wrong because "last_spoke_to.customer_name" combines two field names which is not correct.

So to answer your question it is necessary to get information abot the tables and fiellds in the database description.
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LVL 17

Accepted Solution

by:
Chris Harte earned 1000 total points
ID: 35155563
You say last_spoke_to is a field rather than a table, so should the second query not be

$sql = "
SELECT *
FROM customer
    LEFT JOIN notes
        ON customer.customer_name = notes.customer_name
        AND customer.customer_name = notes.last_spoke_to
WHERE customer.customer_name = '" . $val . "'";
0
 
LVL 6

Author Comment

by:NeoAshura
ID: 35155582
thanks for the feedback now my tables returns nothing they are blank...

Both notes is a table and contains "notes" and "last_spoke_to"

where customer is another table and customer_name is a field in that table which also exsits in the notes table as a referance for the notes and last spoke to. does that make sense?
0
 
LVL 6

Expert Comment

by:brb6708
ID: 35155587
@MunterMan:

that was my idea as well; but does it make sense selecting all customers who have talked to themselves?????

That's why I said that knowledge about the tables and the fields within those tables is necessary to answer NeoAshura's question.
0
 
LVL 6

Author Comment

by:NeoAshura
ID: 35155711
ok guys ive solved it but i appricate your input so points will be awrded equally. Ive outputed what i needed using the following:

$sql = "
SELECT *
FROM customer
    LEFT JOIN notes
        ON customer.customer_name = notes.customer_name
WHERE customer.customer_name = '" . $val . "'";

It is a notes page based on each customer, These notes needed to be logged.. it now looks as follows which is what i needed.
picture6.jpg
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