How to pass variable from a database query to a PHP function

Question

get_population($row['ID']) is causing error Variable undefined:

foreach($filterList as $row)
{
   echo '<tr>';
   echo '<td>' . $row['county_ID'] . '</td>';
   echo '<td>' . get_population($row['ID']) . '</td>'; 
   echo '<td>' . $row['ID'] . '</td>';     
   echo '</tr>';
}

When I use a database value in the function parameter I get the error :Undefined variable: It must be out of scope but I don't know how to get around it.

If I print $row['ID'] it shows the value of 1 for this example. If I hard code 1 as the parameter value it works. It is only when I try to pass $row['ID'] to the function that I get the error.

I am sure it is a simple answer but I am chasing my tail. Thanks

Can we see your query? Perhaps the table structure? The get_population() function? — Jay Blanchard
– Jay Blanchard, Commented May 29, 2014 at 19:38
Might be you're getting the error inside the get_population() function? Can you post that too? — Damien Pirsy
– Damien Pirsy, Commented May 29, 2014 at 19:38
Please post the whole error message. Undefined variable warnings contain the name of the variable that's undefined. — Barmar
– Barmar, Commented May 29, 2014 at 19:43
The error message also has a line number. Please point out the specific line that's getting the error. — Barmar
– Barmar, Commented May 29, 2014 at 19:44

Glauber Portella · Accepted Answer · 2014-05-29 19:41:25Z

1

In your code there's a mismatch in ''.['ID'].'';

The correct is $row['ID']:

foreach($filterList as $row)
{
   echo '<tr>';
   echo '<td>' . $row['county_ID'] . '</td>';
   echo '<td>' . get_population($row['ID']) . '</td>'; 
   // error here
   echo '<td>' . ['ID'] . '</td>';     
   // change with
   echo '<td>' . $row['ID'] . '</td>';     
   echo '</tr>';
}

answered May 29, 2014 at 19:41

Glauber Portella

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1 Comment

Barmar Over a year ago

I suspect that's a copying error, because that would cause a parse error, not undefined variable.

user1633947 · Accepted Answer · 2014-06-02 14:11:53Z

0

The code is correct, the problem was a null value in the database. Since there was no value to pass to the function it returned variable undefined.

answered Jun 2, 2014 at 14:11

user1633947

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