Is this Possible - PHP/MySQL

[Phillip T. Murphy]

Well, after half pulling my hair out messing with this, I am thinking it is
not possible. I did research and found references to "sub-queries" not
possible in MySQL (I am using 4.0.18-32 with PHP 4). But, not sure if I am
breaking that rule or not.

Psuedo code of what I am trying to do...

<?php

$sql = "select * from table1";
$db_result = mysql_query($sq l);

while ($row = mysql_fetch_arr ay($db_result)) {
$table_key = $row("table_key ");


$sql2 = "select * from table2 where tabel_key='$tab le_key'";
$db_result2 = mysql_query($sq l2);
$row2 = mysql_fetch_arr ay($db_result2) ;

if record exists in table2
do something with table 2;
else
do something with table 1;
}
?>

The second query seems to always return an empty array. Is this nested
query allowed in PHP/MySQL?

Thanks in advance for any help,

PTM

[Geoff Berrow]

Re: Is this Possible - PHP/MySQL

I noticed that Message-ID: <bg8kd.753$z3.2 79@bignews5.bel lsouth.net>
from Phillip T. Murphy contained the following:
[color=blue]
>The second query seems to always return an empty array.[/color]

tabel_key?
--
Geoff Berrow (put thecat out to email)
It's only Usenet, no one dies.
My opinions, not the committee's, mine.
Simple RFDs http://www.ckdog.co.uk/rfdmaker/

[Michael Fesser]

Re: Is this Possible - PHP/MySQL

.oO(Phillip T. Murphy)
[color=blue]
>Well, after half pulling my hair out messing with this, I am thinking it is
>not possible. I did research and found references to "sub-queries" not
>possible in MySQL (I am using 4.0.18-32 with PHP 4). But, not sure if I am
>breaking that rule or not.[/color]

A subquery (or subselect) is a SELECT statement nested inside another:

SELECT something FROM somewhere WHERE foo = (SELECT somethingelse ...)

Subqueries are supported as of MySQL 4.1.
[color=blue]
>The second query seems to always return an empty array. Is this nested
>query allowed in PHP/MySQL?[/color]

No problem with that.

But you have no error checking in your code, you won't notice when a
query fails. Check if mysql_query() returns a resource ID, if not have a
look at mysql_error() to see if MySQL encountered a problem.

http://www.php.net/mysql_query

http://www.php.net/mysql_error

Micha

[Phillip T. Murphy]

Re: Is this Possible - PHP/MySQL


"Michael Fesser" <netizen@gmx.ne t> wrote in message
news:cp62p01egp cntcr210fu10g3b bm0igjbf1@4ax.c om...[color=blue]
> .oO(Phillip T. Murphy)
>[color=green]
>>Well, after half pulling my hair out messing with this, I am thinking it
>>is
>>not possible. I did research and found references to "sub-queries" not
>>possible in MySQL (I am using 4.0.18-32 with PHP 4). But, not sure if I
>>am
>>breaking that rule or not.[/color]
>
> A subquery (or subselect) is a SELECT statement nested inside another:
>
> SELECT something FROM somewhere WHERE foo = (SELECT somethingelse ...)
>
> Subqueries are supported as of MySQL 4.1.
>[color=green]
>>The second query seems to always return an empty array. Is this nested
>>query allowed in PHP/MySQL?[/color]
>
> No problem with that.
>
> But you have no error checking in your code, you won't notice when a
> query fails. Check if mysql_query() returns a resource ID, if not have a
> look at mysql_error() to see if MySQL encountered a problem.
>
> http://www.php.net/mysql_query
> http://www.php.net/mysql_error
>
> Micha[/color]

Thanks for the quick reply. in the interest of space, I cut that out. I
have a function defined that replaces the mysql_query() call and does error
checking.

At least I know what I am trying to do is possible now. I am starting over.
I would like to be able to just use a LEFT JOIN, but the second table needs
a where clause on a field that does not exist in the first table, so too
many records are elimiated.

I will keep trying. Thanks for the input.

[kingofkolt]

Re: Is this Possible - PHP/MySQL

"Phillip T. Murphy" <pStPmAuMrphy@b ellsouth.net> wrote in message
news:bg8kd.753$ z3.279@bignews5 .bellsouth.net. ..[color=blue]
> Well, after half pulling my hair out messing with this, I am thinking it[/color]
is[color=blue]
> not possible. I did research and found references to "sub-queries" not
> possible in MySQL (I am using 4.0.18-32 with PHP 4). But, not sure if I[/color]
am[color=blue]
> breaking that rule or not.
>
> Psuedo code of what I am trying to do...
>
> <?php
>
> $sql = "select * from table1";
> $db_result = mysql_query($sq l);
>
> while ($row = mysql_fetch_arr ay($db_result)) {
> $table_key = $row("table_key ");[/color]
[...]

Not sure if/what this will fix, but in that last line ($table_key =
$row("table_key ");), the parantheses in $row("table_key ") should be changed
to brackets: $row["table_key"]

- JP

[Phillip T. Murphy]

Re: Is this Possible - PHP/MySQL


"kingofkolt " <jessepNOSPAM@c omcast.net> wrote in message
news:B4ckd.2145 7$5K2.20313@att bi_s03...[color=blue]
> "Phillip T. Murphy" <pStPmAuMrphy@b ellsouth.net> wrote in message
> news:bg8kd.753$ z3.279@bignews5 .bellsouth.net. ..[color=green]
>> Well, after half pulling my hair out messing with this, I am thinking it[/color]
> is[color=green]
>> not possible. I did research and found references to "sub-queries" not
>> possible in MySQL (I am using 4.0.18-32 with PHP 4). But, not sure if I[/color]
> am[color=green]
>> breaking that rule or not.
>>
>> Psuedo code of what I am trying to do...
>>
>> <?php
>>
>> $sql = "select * from table1";
>> $db_result = mysql_query($sq l);
>>
>> while ($row = mysql_fetch_arr ay($db_result)) {
>> $table_key = $row("table_key ");[/color]
> [...]
>
> Not sure if/what this will fix, but in that last line ($table_key =
> $row("table_key ");), the parantheses in $row("table_key ") should be
> changed
> to brackets: $row["table_key"]
>
> - JP
>
>[/color]

Thanks for the responses everyone. Thanks to suggestions here and assuring
me that it is possible to do nested queries, I got it to work.

Come to find out, it had to do with the way I was checking if the second
query returned anything or not. When it didn't return anything, I was
checking one of the array variables to see if it was NULL or Blank. That
did not work. I backed up and now check what the mysql_query returns (FALSE
or an Array).

All is now well. As always, coming to a group like this to get answers fast
solved my problem.