Dynamic Variable Name Issue

[brokencode]

Necessary info:

 

$svar is equal to "DOG"

 

The code that is causing problems for me is:

 

$subid = mysql_query("SELECT sourcevar1 FROM sources WHERE id='$tsid'");

$row = mysql_fetch_array($subid);

$svar = $row['sourcevar1'];

$tvar = $_GET[$svar];

print "$svar - $tvar";

 

When loading the code with a url matching: http://www.site.com/?DOG=poodle I would expect the print statement to read:

 

DOG - poodle

 

but instead, the $tvar variable comes out blank and returns:

 

DOG -

 

Thanks for your time and help!

 

 

[wildteen88]

Where is the variable $tsid defined?

[kenrbnsn]

If you put

<?php
echo '<pre>' . print_r($_GET,true) . '</pre>';
?>

just before

<?php
$tvar = $_GET[$svar];
?>

what prints?

 

Ken

[PFMaBiSmAd]

The code blindly assumes that your query worked and returned a row from your database. Since you have no error checking logic in your code to make sure you got a value from your query, there is no guarantee that $_GET[$svar] will produce anything.

 

 

[brokencode]

Where is the variable $tsid defined?

 

$tsid is defined by a separate $_GET variable. In actuallity there is another variable in the url ($tsid=3)

 

If you put

<?php
echo '<pre>' . print_r($_GET,true) . '</pre>';
?>

just before

<?php
$tvar = $_GET[$svar];
?>

what prints?

 

Ken

 

I modified the code with the echo line you suggested:

 

$subid = mysql_query("SELECT sourcevar1 FROM sources WHERE id='$tsid'");

$row = mysql_fetch_array($subid);

$svar = $row['sourcevar1'];

echo '<pre>' . print_r($_GET,true) . '</pre>';

$tvar = $_GET[$svar];

print "$svar - $tvar";

 

The code now prints the following:

 

Array

(

    [tsid] => 3

    [OVKEY] => hello

)

 

OKVEY -

 

Thanks for your help guys, don't know what I would do without forums like this and people like you. And also, as I'm sure you noticed my inital post replaced OVKEY with DOG

 

 

[wildteen88]

If $tsid is comming from the url you'll need to grab this values using $_GET['tsid'].

// check to to see if tsid set in the url and ensure it holds a numeric value
if(isset($_GET['tsid']) && is_numeric($_GET['tsid']))
{
    // grab tsid from the url
    $tsid = (int) $_GET['tsid'];

    $subid = mysql_query("SELECT sourcevar1 FROM sources WHERE id='$tsid'");
    $row = mysql_fetch_array($subid);
    $svar = $row['sourcevar1'];
    echo '<pre>' . print_r($_GET,true) . '</pre>';
    $tvar = $_GET[$svar];
    print "$svar - $tvar";
}

[brokencode]

Sorry, I should have copied more of the function to begin with and not altered anything:

 

I am pulling the $tsid from $_GET['tsid'] and using it's value to pull $row['sourcevar1'] (which is OVKEY)

 

The url is http://www.site.com/?tsid=3&OVKEY=hello

 

The code I have at this point is:

 

///// GET SITE ID

if(isset($_GET['tsid']) && is_numeric($_GET['tsid'])){

$tsid = (int) $_GET['tsid'];

 

///// GET SUBID1 FOR ABOVE TSID

$subid = mysql_query("SELECT sourcevar1 FROM sources WHERE id='$tsid'");

$row = mysql_fetch_array($subid);

$svar = $row['sourcevar1'];

echo '<pre>' . print_r($_GET,true) . '</pre>';

$tvar = $_GET[$svar];

print "$svar - $tvar";

}

 

 

 

And the code prints the same thus far:

 

Array

(

    [tsid] => 3

    [OVKEY] => hello

)

 

OKVEY -

[wildteen88]

Make sure your query is actually returning a result.

$subid = mysql_query("SELECT sourcevar1 FROM sources WHERE id='$tsid'");

// check to see if the query ran fine and results have been returned
if($subid && mysql_num_rows($subid) > 0)
{
    $row = mysql_fetch_array($subid);
    $svar = $row['sourcevar1'];
    echo '<pre>' . print_r($_GET,true) . '</pre>';
    $tvar = $_GET[$svar];
    print "$svar - $tvar";
}
// something is wrong or there was no results returned
else
{
    // check that no error codes where returned
    if(!mysql_errno())
    {
        echo 'No results where returned';
    }
    // Error code returned, lets find out why
    else
    {
        echo 'There is an error! Which is... ' . mysql_error();
    }
}

 

Also ensure you are connected to mysql too.

[brokencode]

Make sure your query is actually returning a result.

$subid = mysql_query("SELECT sourcevar1 FROM sources WHERE id='$tsid'");

// check to see if the query ran fine and results have been returned
if($subid && mysql_num_rows($subid) > 0)
{
    $row = mysql_fetch_array($subid);
    $svar = $row['sourcevar1'];
    echo '<pre>' . print_r($_GET,true) . '</pre>';
    $tvar = $_GET[$svar];
    print "$svar - $tvar";
}
// something is wrong or there was no results returned
else
{
    // check that no error codes where returned
    if(!mysql_errno())
    {
        echo 'No results where returned';
    }
    // Error code returned, lets find out why
    else
    {
        echo 'There is an error! Which is... ' . mysql_error();
    }
}

 

Also ensure you are connected to mysql too.

 

I altered my code to read:

 

$tsid = $_GET['tsid'];

 

$subid = mysql_query("SELECT sourcevar1 FROM sources WHERE id='$tsid'");

// check to see if the query ran fine and results have been returned

if($subid && mysql_num_rows($subid) > 0){

    $row = mysql_fetch_array($subid);

    $svar = $row['sourcevar1'];

    echo '<pre>' . print_r($_GET,true) . '</pre>';

    $tvar = $_GET[$svar];

    print "$svar - $tvar";

}

// something is wrong or there was no results returned

else{

    // check that no error codes where returned

    if(!mysql_errno())

        echo 'No results';

    // Error code returned, lets find out why

    else

        echo 'There is an error! Which is... ' . mysql_error();

}

 

 

It is still printing:

 

Array

(

    [tsid] => 3

    [OVKEY] => hello

)

 

OKVEY -

 

 

 

Thanks for your continued help.

[PFMaBiSmAd]

I'm going to guess that you have some non-printing characters as part of your data and even though it displays as expected, it does not actually match the $_GET variable name.

 

Use the following to see exactly what $svar might contain -

var_dump($svar);

[brokencode]

I added the dump command - here is what the code looks like:

 

$tsid = $_GET['tsid'];

 

$subid = mysql_query("SELECT sourcevar1 FROM sources WHERE id='$tsid'");

// check to see if the query ran fine and results have been returned

if($subid && mysql_num_rows($subid) > 0){

    $row = mysql_fetch_array($subid);

    $svar = $row['sourcevar1'];

    echo '<pre>' . print_r($_GET,true) . '</pre>';

    $tvar = $_GET[$svar];

    print "$svar - $tvar<br />";

    var_dump($svar);

}

// something is wrong or there was no results returned

else{

    // check that no error codes where returned

    if(!mysql_errno())

        echo 'No results';

    // Error code returned, lets find out why

    else

        echo 'There is an error! Which is... ' . mysql_error();

}

 

The printed result is:

 

Array

(

    [tsid] => 3

    [OVKEY] => hello

)

 

OKVEY -

string(5) "OKVEY"

[PFMaBiSmAd]

Looks OK (the length matches the visible characters.)

 

How about -

var_dump($_GET);

[brokencode]

Okay, changed the code to the following:

 

$tsid = $_GET['tsid'];

 

$subid = mysql_query("SELECT sourcevar1 FROM sources WHERE id='$tsid'");

// check to see if the query ran fine and results have been returned

if($subid && mysql_num_rows($subid) > 0){

    $row = mysql_fetch_array($subid);

    $svar = $row['sourcevar1'];

    echo '<pre>' . print_r($_GET,true) . '</pre>';

    $tvar = $_GET[$svar];

    print "$svar - $tvar<br />";

    var_dump($svar);

    print "<br />";

    var_dump($_GET);

}

// something is wrong or there was no results returned

else{

    // check that no error codes where returned

    if(!mysql_errno())

        echo 'No results';

    // Error code returned, lets find out why

    else

        echo 'There is an error! Which is... ' . mysql_error();

}

 

 

 

And the printed result is:

 

Array

(

    [tsid] => 3

    [OVKEY] => hello

)

 

OKVEY -

string(5) "OKVEY"

array(2) { ["tsid"]=> string(1) "3" ["OVKEY"]=> string(5) "hello" }

 

 

Thank you all for trying to help me through this. It's still not working as expected. But I'm learning alot about error checking!

[PFMaBiSmAd]

I spy a spelling error.

[brokencode]

OKVEY

 

Thank you - thank you. The only record in the table with a spelling error and I pick it for test purposes.

 

Thank you, thank you, thank you all!

 

I learned alot from this thread, my error checking, among other things, is weak.