Hello again

[lenbell]

Help again

This works

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <=
$session->userleve)";

I seem to have missed something about concatenation

Len Bell

[Heiko Richler]

lenbell wrote:

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <=
$session->userleve)";

To use $session->userleve ?
".... <= {$session->userleve})";

Heiko

--
http://portal.richler.de/ Namensportal zu Richler
http://www.richler.de/ Heiko Richler: Computer - Know How!
http://www.richler.info/ private Homepage

[Paul Lautman]

lenbell wrote:

Help again

This works

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <=
$session->userleve)";

I seem to have missed something about concatenation

Len Bell

I would tend to write that as a JOIN rather than a correalted sub-query.

However, regardless of that, telling us that something "doesn't work" is
pointless. In precisely what way does it not work?

[sheldonlg]

lenbell wrote:

Help again

This works

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <=
$session->userleve)";

I seem to have missed something about concatenation

Hint: When you have a query that doesn't work, put a print of the
query. You will then readily find what is wrong in building the query
(missing ' or something else), or when you try it using phpMyAdmin
directly, it will tell you where the error is.

[Jerry Stuckle]

lenbell wrote:

Help again

This works

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <=
$session->userleve)";

I seem to have missed something about concatenation

Len Bell

Maybe ask in a SQL newsgroup (or one for your database)? This isn't a
PHP question, after all.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
js*******@attglobal.net
==================

[Roger]

On May 23, 9:30*pm, "lenbell" <[e-mail]:

Help again

This works

* *$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenetsWHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

* * *$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <=
$session->userleve)";

I seem to have missed something about concatenation

Len Bell

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <= '" .
$session->userleve . "')";

or

$q = "SELECT * FROM groupnames LEFT JOIN tenets ON groupnames.groupnum
= tenents.groupnum AND tenets.level <= '" .
$session->userleve . "')";

[sheldonlg]

Jerry Stuckle wrote:

lenbell wrote:

>Help again

This works

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <=
$session->userleve)";

I seem to have missed something about concatenation

Len Bell

Maybe ask in a SQL newsgroup (or one for your database)? This isn't a
PHP question, after all.

Actually, Jerry, it is. He is having trouble building the query when he
uses a php variable rather than a constant. That kind of makes it a php
question.

For the OP, are you sure you meant userleve and not userlevel?

[Jerry Stuckle]

sheldonlg wrote:

Jerry Stuckle wrote:

>lenbell wrote:

>>Help again

This works

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM
tenets WHERE groupnames.groupnum = tenets.groupnum AND tenets.level
<= $session->userleve)";

I seem to have missed something about concatenation

Len Bell

Maybe ask in a SQL newsgroup (or one for your database)? This isn't a
PHP question, after all.

Actually, Jerry, it is. He is having trouble building the query when he
uses a php variable rather than a constant. That kind of makes it a php
question.

For the OP, are you sure you meant userleve and not userlevel?

Then why are all of the answers SQL statements?

He needs to simply echo $q and fix his problem.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
js*******@attglobal.net
==================

[lenbell]

Heres how I have it NOW in the php file
----------------------------------------------------
function displayGroups(){
global $database;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '" .
$session->userlevel . "')";

echo $q;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

echo $q;

$result = $database->query($q);
__________________________________

Here's what I get when it runs (both echos run together - small but pisses
me off)

SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '')SELECT * FROM
groupnames WHERE EXISTS (select * FROM tenets WHERE groupnames.groupnum =
tenets.groupnum AND tenets.level <= '1')

Also why can't I get those 2 echos to play on separate lines?

Thanks for your support

Len Bell

"Roger" <le*********@natpro.comwrote in message
news:7f**********************************@d45g2000 hsc.googlegroups.com...
On May 23, 9:30 pm, "lenbell" <[e-mail]:

Help again

This works

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <=
$session->userleve)";

I seem to have missed something about concatenation

Len Bell

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <= '" .
$session->userleve . "')";

or

$q = "SELECT * FROM groupnames LEFT JOIN tenets ON groupnames.groupnum
= tenents.groupnum AND tenets.level <= '" .
$session->userleve . "')";

[sheldonlg]

Jerry Stuckle wrote:

sheldonlg wrote:

>Jerry Stuckle wrote:

>>lenbell wrote:
Help again

This works

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

This doesn't

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM
tenets WHERE groupnames.groupnum = tenets.groupnum AND tenets.level
<= $session->userleve)";

I seem to have missed something about concatenation

Len Bell
Maybe ask in a SQL newsgroup (or one for your database)? This isn't
a PHP question, after all.

Actually, Jerry, it is. He is having trouble building the query when
he uses a php variable rather than a constant. That kind of makes it
a php question.

For the OP, are you sure you meant userleve and not userlevel?

Then why are all of the answers SQL statements?

He needs to simply echo $q and fix his problem.

That's what I told him a few replies ago.

[sheldonlg]

lenbell wrote:

Heres how I have it NOW in the php file
----------------------------------------------------
function displayGroups(){
global $database;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '" .
$session->userlevel . "')";

echo $q;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

echo $q;

$result = $database->query($q);
__________________________________

Here's what I get when it runs (both echos run together - small but pisses
me off)

SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '')SELECT * FROM
groupnames WHERE EXISTS (select * FROM tenets WHERE groupnames.groupnum =
tenets.groupnum AND tenets.level <= '1')

Also why can't I get those 2 echos to play on separate lines?

Well, your problem is now obvious. You have nothing in
$session->userlevel. It is yielding ''. Fix that error and you are all
set.

As for the two echos, add <brto the first string and it will display
the next one on a new line.

[lenbell]

sheldonlg
thanks for your help

Tried - echo $q;<br>
Tried - echo $q<br>;

both gave

Parse error: syntax error, unexpected '<' in
/home/host020/public_html/tenets/TenetGroupList.php on line 23

23 is the line above

There must be a scope problem with $session->userlevel and I'm looking

<sheldonlgwrote in message
news:_I******************************@giganews.com ...

lenbell wrote:

>Heres how I have it NOW in the php file
----------------------------------------------------
function displayGroups(){
global $database;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '" .
$session->userlevel . "')";

echo $q;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

echo $q;

$result = $database->query($q);
__________________________________

Here's what I get when it runs (both echos run together - small but
pisses me off)

SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '')SELECT *
FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')

Also why can't I get those 2 echos to play on separate lines?

Well, your problem is now obvious. You have nothing in
$session->userlevel. It is yielding ''. Fix that error and you are all
set.

As for the two echos, add <brto the first string and it will display the
next one on a new line.

[lenbell]

Found it - Thanks sheldonlg

It was a scope problem and now it's gone <smile>

Len

<sheldonlgwrote in message
news:_I******************************@giganews.com ...

lenbell wrote:

>Heres how I have it NOW in the php file
----------------------------------------------------
function displayGroups(){
global $database;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '" .
$session->userlevel . "')";

echo $q;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

echo $q;

$result = $database->query($q);
__________________________________

Here's what I get when it runs (both echos run together - small but
pisses me off)

SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '')SELECT *
FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')

Also why can't I get those 2 echos to play on separate lines?

Well, your problem is now obvious. You have nothing in
$session->userlevel. It is yielding ''. Fix that error and you are all
set.

As for the two echos, add <brto the first string and it will display the
next one on a new line.

[Spamm Trappe]

What a cleverly selected Subject:!

[Curtis]

lenbell wrote:

sheldonlg
thanks for your help

Tried - echo $q;<br>
Tried - echo $q<br>;

both gave

Parse error: syntax error, unexpected '<' in
/home/host020/public_html/tenets/TenetGroupList.php on line 23

23 is the line above

There must be a scope problem with $session->userlevel and I'm looking

<sheldonlgwrote in message
news:_I******************************@giganews.com ...

>lenbell wrote:

>>Heres how I have it NOW in the php file
----------------------------------------------------
function displayGroups(){
global $database;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '" .
$session->userlevel . "')";

echo $q;

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets
WHERE groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')";

echo $q;

$result = $database->query($q);
__________________________________

Here's what I get when it runs (both echos run together - small but
pisses me off)

SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '')SELECT *
FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= '1')

Also why can't I get those 2 echos to play on separate lines?

Well, your problem is now obvious. You have nothing in
$session->userlevel. It is yielding ''. Fix that error and you are all
set.

As for the two echos, add <brto the first string and it will display the
next one on a new line.

He meant:
echo "$q<br>";

It's HTML markup, not PHP.

--
Curtis

[sheldonlg]

lenbell wrote:

Found it - Thanks sheldonlg

It was a scope problem and now it's gone <smile>

Len

Glad to be of help. In the future, try to think of debugging as "what
is this line seeing?". In this case it was, "what is the sql engine
seeing?". To find out do a print or an echo and then you see exactly
what it sees. Very often that will tell it to you right away (as was
here). Other times, not. For example, the sql query may have looked
correct but not run correctly. Taking the output to phpMyAdmin and
running it interactively will tell you what line is in error.

Anyway, backtracking is one of the best ways of debugging. Since we
can't set breakpoints and examine values, printing is the next best thing.

[lenbell]

Yes and I sure do miss my break points

Thanks for the work around

<sheldonlgwrote in message
news:ta******************************@giganews.com ...

lenbell wrote:

>Found it - Thanks sheldonlg

It was a scope problem and now it's gone <smile>

Len

Glad to be of help. In the future, try to think of debugging as "what is
this line seeing?". In this case it was, "what is the sql engine
seeing?". To find out do a print or an echo and then you see exactly what
it sees. Very often that will tell it to you right away (as was here).
Other times, not. For example, the sql query may have looked correct but
not run correctly. Taking the output to phpMyAdmin and running it
interactively will tell you what line is in error.

Anyway, backtracking is one of the best ways of debugging. Since we can't
set breakpoints and examine values, printing is the next best thing.