CODEWITHSUNDEEP
php
primero
primero

Reputation: 591

php undefined variable in option value

i'm sorry for asking this question because i think it has a really simple answer, but i really don't know php at all. I want to install bandwidthd for network monitoring, and i wanted the option which lets you run the bandwidthd daemon on the gateway and all the info is exported to a postgres database on another machine. Also on that machine, the one with the database i've set up an apache server for displaying the webapp of bandwidthd. The web up is a bunch of php files which don't seem to work at all so i'm just spending my time corecting them. I've been able to get a few of the right but now i'm stuck. I have the following code:

<td><SELECT name="sensor_name">

<OPTION value="none">--Select A Sensor--
<?php
$dbh = pg_pconnect("host=localhost dbname=BandwidthD user=postgres");
$sql = "SELECT sensor_name from sensors order by sensor_name;";
$result = pg_query($dbh, $sql);
while ($r = pg_fetch_array($result))
    echo "<option value=\"".$r[0]."\" ".($sensor_name==$r[0]?"SELECTED":"").">".$r[0]."\n";
?>

And when the page executes it is saying that sensor_name is undefined, but from what i see, it looks to me that the variable should be initialized there right at that line. If you have any suggestions please let me know but bear in mind that i don't know any php, i'm just venturing guesses because it's a programming language and it's supposed to be logical.

I was thinking that $sensor_name would be initialized in that peace of code because i also have this code:

 <td><SELECT name="limit">
 <OPTION value="none">--How Many Results--
 <OPTION value=20 <?php=$limit==20?"SELECTED":""?>>20
 <OPTION value=50 <?php=$limit==50?"SELECTED":""?>>50
 <OPTION value=100 <?php=$limit==100?"SELECTED":""?>>100
 <OPTION value=all <?php=$limit=="all"?"SELECTED":""?>>All
 </select>

and limit is not declared anywhere else and doesn't give an error.

Upvotes: 0

Views: 1194

Answers (2)

Marcus
Marcus

Reputation: 8669

It seems to me that $sensor_name does not exist. You fetch $results from the database which is an enumeration of database results in which you have an index of 'sensor_name'.

Thus, I think you should replace this line (not tested):

echo "<option value=\"".$r[0]."\" ".($sensor_name==$r[0]?"SELECTED":"").">".$r[0]."\n";

With this

echo "<option value=\"".$r[0]."\" ".($result['sensor_name']==$r[0]?"SELECTED":"").">".$r[0]."\n";

Upvotes: 2

Shaun Moss
Shaun Moss

Reputation: 623

The code, $sensor_name == $r[0] does not assign a value to $sensor_name, but compares it for equality with $r[0].

The code $sensor_name==$r[0]?"SELECTED":"" means that if $sensor_name is equal to $r[0] (the sensor_name field in the row), then evaluate the operation to SELECTED, otherwise evaluate to ''.

$sensor_name isn't defined anywhere in your code, which is why you're getting the notice.

Upvotes: 0

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