using functions to query mysql database in php

Question

This is one thing I can't understand. I'm using a function which fetches records from the database. And if that record doesn't exist. It performs an insert query. I even echoed out to ensure that those 2 parameters contains something. And it really contain something. What I can't understand is that it doesn't perform the insert query. I also tested the insert query below on another script which doesn't use function. And it worked.

function fetch_customer($cust, $credit){

$getcnum=query_database("SELECT Cust_Name, CUSID FROM customer_credit WHERE Cust_Name='$cust'","onstor",$link);

if(mysql_num_rows($getcnum)==0){
    $fullname=explode(",", $cust);
    $lname= $fullname[0];
    $fname= $fullname[1];


    query_database("INSERT INTO customer_credit(Cust_Name, CREDIT) VALUES('$cust','$credit')",'onstor' ,$link);

    echo "customer: ".$cust."<br/>";

    echo "credit: ".$credit;
    query_database("INSERT INTO customer_table(CLNAME, CFNAME) VALUES('$lname', '$fname')",'onstor',$link);

}

Then I would call the function later on.

fetch_customer($customer, $custcred);

Where do I start if I want to debug this one?

Where does your $link come from ? if it's defined out of the function it may be the problem :) — tsadiq
– tsadiq, Commented Mar 4, 2011 at 15:53
whats this query_database ? ,if you want to query or insert it should be mysql_query — sush
– sush, Commented Mar 4, 2011 at 15:53

Community · Accepted Answer · 2023-11-17 20:42:18Z

2

I'm thinking that your insert queries aren't working because, to wit, there is no query_database function defined (unless it's one that you've defined yourself).

The function that (I believe) you are looking for is mysql_query.

Here's a link to the docs: mysql_query

edited Nov 17, 2023 at 20:42

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answered Mar 4, 2011 at 15:54

Brian Driscoll

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1 Comment

user225269 Over a year ago

seems like the problem is with using functions. I have an include statement which contains the function query_database. And I think the function cannot see the include. That's why it doesn't work. But thanks

7dr3am7 · Accepted Answer · 2011-03-04 15:55:28Z

1

Are you sure that it goes into the if statement? mysql_num_rows($getcnum)==0 echo mysql_num_rows($getcnum) and see how many rows contains, I have also notice the single quotes into 'onstor' , could that be a problem in your function?

answered Mar 4, 2011 at 15:55

7dr3am7

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Comments

Jemaclus · Accepted Answer · 2011-03-04 15:56:20Z

1

It's hard to answer without seeing the query_database() function. I'm not sure why you're doing that instead of just using mysql_query("INSERT INTO...") or whatever. It's possible that your query_database() function isn't returning the proper query variable.

The only thing I can tell you is that in the code above, you're missing a closing bracket. The if statement is closed, but the function isn't. That might be the problem.

Good luck.

answered Mar 4, 2011 at 15:56

Jemaclus

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sush · Accepted Answer · 2011-03-04 16:03:56Z

0

Here is php code to query mysql

function fetch_customer($cust, $credit){


    $link = mysql_connect( 'localhost' , 'username' , 'password' );

    mysql_select_db( 'onstor' , $link ); 

    $getcnum=mysql_query("SELECT Cust_Name, CUSID FROM customer_credit WHERE Cust_Name='$cust'");

        if(mysql_num_rows($getcnum)==0){
            $fullname=explode(",", $cust);
            $lname= $fullname[0];
            $fname= $fullname[1];


            mysq_query("INSERT INTO customer_credit(Cust_Name, CREDIT) VALUES('$cust','$credit')");

            echo "customer: ".$cust."<br/>";

            echo "credit: ".$credit;
            mysql_query("INSERT INTO customer_table(CLNAME, CFNAME) VALUES('$lname', '$fname')");

        }

answered Mar 4, 2011 at 16:03

sush

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1 Comment

tsadiq Over a year ago

$link = mysql_connect( 'localhost' , 'username' , 'password' ); here you have the link. Previously, you didn't have any link. You were passing undefined var as second parameter

Carlos Campderrós · Accepted Answer · 2011-03-04 16:49:33Z

0

You are passing the $link to the database in your query_database($sql, $db, $link), but it is not defined anywhere. So it must be a global variable, that you forget to declare. So, declare $link as being global:

function fetch_customer($cust, $credit)
{
     global $link;
     // rest of your function
}

answered Mar 4, 2011 at 16:49

Carlos Campderrós

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