PHP and sql problem

[Bartosz Wegrzyn]

Please look at the code and if you can find my mistake.

Please tell me why
customerid is saved with wrong value the workorder table

Instead of the correct value I get '0"

When I try to display the $customerid value before the querry
I have the correct value for example 1 but even that in the
database I still have 0.

Whats wrong???

Thanks

<?php


$addcustomer = "insert into addressindex(na me,address,city ,state,zipcode, email,phone,ss, cc,exp,notes) values ('$name','$addr ess','$city','$ state','$zipcod e','$email','$p hone','$ss','$c c','$exp','$not es')";
$addworkorder = "insert into workorder(custo merid,date,time ,service,due,sp ecial,extrawork ,due2,company,i nstaller,total) values ('$customerid', '$date','$time' ,'$service','$d ue','$special', '$extrawork','$ due2','$company ','$installer', '$total')";
$getdata = "select * from addressindex where name = '$name' or customerid = '$customerid'";

$rs = mysql_query($ge tdata);

if(( $rs ) && ( mysql_errno() == 0 )) {

// query was successful
if( mysql_num_rows( $rs ) > 0 ) {

$result2 = mysql_query($ad dworkorder);

if ($result2) {
include '../login/print.php';
} else {
include '../login/error-broken.php';}

} else {

$result = mysql_query($ad dcustomer);

echo $customerid;

$getid = "select customerid from addressindex where name = '$name'";
$sqgetid = mysql_query($ge tid);
$customerid = mysql_result($s qgetid,0);

echo $customerid;

//now i do a querry and even $customerid is 1 here now
it is saved under 0


$result2 = mysql_query($ad dworkorder);

if ($result && $result2) {
include '../login/print.php';
} else {
include '../login/error-broken.php';}

}
} else {
include '../login/error-broken.php';}

//Close connection with MySQL
MySQL_close()


?>

[Tom Thackrey]

Re: PHP and sql problem


On 29-Aug-2003, "Bartosz Wegrzyn" <blwegrzyn@lexo n.ws> wrote:
[color=blue]
> Please look at the code and if you can find my mistake.
>
> Please tell me why
> customerid is saved with wrong value the workorder table
>
> Instead of the correct value I get '0"
>
> When I try to display the $customerid value before the querry
> I have the correct value for example 1 but even that in the
> database I still have 0.
>
> Whats wrong???
>
> Thanks
>
> <?php
>
>
> $addcustomer = "insert into
> addressindex(na me,address,city ,state,zipcode, email,phone,ss, cc,exp,notes)
> values
> ('$name','$addr ess','$city','$ state','$zipcod e','$email','$p hone','$ss','$c c','$exp','$not es')";
> $addworkorder = "insert into
> workorder(custo merid,date,time ,service,due,sp ecial,extrawork ,due2,company,i nstaller,total)
> values
> ('$customerid', '$date','$time' ,'$service','$d ue','$special', '$extrawork','$ due2','$company ','$installer', '$total')";
> $getdata = "select * from addressindex where name = '$name' or customerid
> = '$customerid'";[/color]

Please cross-post!!!

It looks like you are inserting customerid into workorder but trying
retrieving it from addressindex.



--
Tom Thackrey

https://www.creative-light.com

[Bartosz Wegrzyn]

Re: PHP and sql problem

Thats is correct !

Thats what I want to do.

On my main form I do enter all the data for customer and work order.

Customer is is created automaticaly.

I do not know how to add that id to my work order so I
query what is that id and than I try to add it.

On Fri, 29 Aug 2003 19:31:12 +0000, Tom Thackrey wrote:
[color=blue]
>
> On 29-Aug-2003, "Bartosz Wegrzyn" <blwegrzyn@lexo n.ws> wrote:
>[color=green]
>> Please look at the code and if you can find my mistake.
>>
>> Please tell me why
>> customerid is saved with wrong value the workorder table
>>
>> Instead of the correct value I get '0"
>>
>> When I try to display the $customerid value before the querry
>> I have the correct value for example 1 but even that in the
>> database I still have 0.
>>
>> Whats wrong???
>>
>> Thanks
>>
>> <?php
>>
>>
>> $addcustomer = "insert into
>> addressindex(na me,address,city ,state,zipcode, email,phone,ss, cc,exp,notes)
>> values
>> ('$name','$addr ess','$city','$ state','$zipcod e','$email','$p hone','$ss','$c c','$exp','$not es')";
>> $addworkorder = "insert into
>> workorder(custo merid,date,time ,service,due,sp ecial,extrawork ,due2,company,i nstaller,total)
>> values
>> ('$customerid', '$date','$time' ,'$service','$d ue','$special', '$extrawork','$ due2','$company ','$installer', '$total')";
>> $getdata = "select * from addressindex where name = '$name' or customerid
>> = '$customerid'";[/color]
>
> Please cross-post!!!
>
> It looks like you are inserting customerid into workorder but trying
> retrieving it from addressindex.[/color]

[Matthew Vickers]

Re: PHP and sql problem

On Fri, 29 Aug 2003 18:00:41 GMT
"Bartosz Wegrzyn" <blwegrzyn@lexo n.ws> wrote:
[color=blue]
> Please look at the code and if you can find my mistake.
>
> Please tell me why
> customerid is saved with wrong value the workorder table
>
> Instead of the correct value I get '0"
>
> When I try to display the $customerid value before the querry
> I have the correct value for example 1 but even that in the
> database I still have 0.
>
> Whats wrong???
>
> Thanks
>
> <?php
>
>
> $addcustomer = "insert into
> addressindex(na me,address,city ,state,zipcode, email,phone,ss, cc,exp,no
> tes) values
> ('$name','$addr ess','$city','$ state','$zipcod e','$email','$p hone','$s
> s','$cc','$exp' ,'$notes')";$ad dworkorder = "insert into
> workorder(custo merid,date,time ,service,due,sp ecial,extrawork ,due2,com
> pany,installer, total) values
> ('$customerid', '$date','$time' ,'$service','$d ue','$special', '$extrawo
> rk','$due2','$c ompany','$insta ller','$total') ";$getdata = "select *
> from addressindex where name = '$name' or customerid = '$customerid'";[/color]

<SNIP>

Just a thought:

Is customerid declared as an integer in the MySQL table ?
You are inserting it as a char value.

Matt


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