change the name of a variable in a for() cycle

marcnyc

Hello I would like to get some help/input on how to change the name of a variable within a for() cycle because the things I have tried don't work...

My piece of code is:

for($y=2004;$y<="2010";$y++) {
if ( ereg($y,$r->work_title) ) { $year[$y] = "selected"; }
}

What I want basically is that at every cycle the name (not the value) of the variable changes from $year2004 to $year2004 to $year2010...

I have tried $year[$y], $year($y), $year$y, $year.$y and $year{$y} and none of them seem to work... Any suggestions?

As you might gather what I am trying to do is to have different names of variables so that the one that will match the ereg() case will be assigned the value "selected" so that I can use it in a drop down menu like this:

		<select class="text" name="Title_year" id="year">
		  <option $year[$y] value="">year</option>
		  <option $year[$y] value="2004">2004</option>
		  <option $year[$y] value="2005">2005</option>
		  <option $year[$y] value="2006">2006</option>
		  <option $year[$y] value="2007">2007</option>
		  <option $year[$y] value="2008">2008</option>
		  <option $year[$y] value="2009">2009</option>
		  <option $year[$y] value="2010">2010</option>
		</select>

yelvington

It can be done (look up "variable variables" in the manual), but why aren't you using an associative array for this?

lazzerous

if $year[$y] == "2004"

then you can create the variable $2004 by doing this:

${$year[$y]} = "whatever value you want this new var to have";

or to quote your code:

<select class="text" name="Title_year" id="year"> 
<option value="">year</option>

<?
for ($i=0; $i<sizeof($year); $i++)
    {
    echo "<option value='" . ${$year[$i]} . "'>" . ${$year[$i]} . "</option>\n";
    }
?>

</select>

-=Lazz=-

marcnyc

My problem is that I want to create variables whose name will be $year2004, $year2005, $year2006 etc... How can I do that?

Also I don't know what an associative array is but I can tell you that the piece of code you provided won't do what I need, which is add a "selected" to the html of the drop down menu for when a condition is met...

yelvington

Actually, this is easier than you're making it out to be, and you don't need variable variables, nor arrays.

Assuming $r->work_title is a string containing a year:


for($y=2004; $y<=2010; $y++){
        echo "<option ";
        if ( ereg($y,$r->work_title) ) echo " selected ";
        echo "value='$y'>$y</option>\r\n";
        }

If $r->work_title is '2008' the output will be

Which is what you asked for.

However ... you do need to learn about arrays before getting entangled in any ill-considered variable-naming nightmares in the future.

PHP does support variable variables (see the appropriate section of the manual) but any time you have a collection of similar things, you should use an array with a numeric index (a normal array) or one with strings as the index value (associative arrays). There are many tutorials on the subject; just hit Google.

marcnyc

You're rigth, your code worked like a charm!

The only mistake in it (and I say this for other beginners like me, not to point out a small mistake in a code I wouldn't have been able to come up with myself) is that it needed to escape the variables with '. and .', like this:

for($y=2004; $y<=2010; $y++){
echo '<option ';
if ( ereg($y,$r->work_title) ) { echo ' selected '; }
echo 'value="'.$y.'">'.$y.'</option>\r\n';
}

You are right I should know more about PHP, the thing is I am not a programmer, I just do this for fun and never have time to study it the way it should be studied... I learn from doing what I need to do at a specific moment in time and by learning and modifying other people's code... Believe it or not I have actually learned more by modifying and looking at third party code than not by reading the manual...

yelvington

Actually you don't need to chop up the code and use concatenation in order to change the single quotes to double quotes in the output. Instead, escape double quotes within a double-quoted string:

$foo = 'selected';

echo "the variable is \"$foo\"";