Extremely simple function not working, help?

  • Resolved phpchick

    (@phpchick)


    13 years, 6 months ago

    function getstockstats($stocksymbol)
  {
 if ( $stocksymbol == "aapl" ) {
	$result = "Your name is someguy!<br />";
}
return $result;
  }

getstockstats('aapl');

    Anyone have any suggestions? super simple

Viewing 9 replies - 1 through 9 (of 9 total)
  • Mark

    (@taart)

    13 years, 6 months ago

    What are you trying to do?

    Shouldn’t you add:

    $baboon = getstockstats(‘aapl’);

    echo $baboon;

    ?

    Andrew Nevins

    (@anevins)

    WCLDN 2018 Contributor | Volunteer support

    13 years, 6 months ago

    What isn’t working, in which line?
    What, if any, are the errors shown?

    Thread Starter phpchick

    (@phpchick)

    13 years, 6 months ago

    whats the difference between echo and return ?

    All I wanted to do was print “yourname is some guy”

    Thread Starter phpchick

    (@phpchick)

    13 years, 6 months ago

    I changed to this, 

    function getstockstats($stocksymbol)
  {
 if ( $stocksymbol == "aapl" ) {
	$result = "Test!<br />";
}
echo $result;
  }

    and it worked. But shouldnt returning $result also print it out?

    Michael

    (@alchymyth)

    13 years, 6 months ago

    http://php.net/manual/en/function.echo.php
    http://php.net/manual/en/function.return.php

    however simple functions you are trying to code, you need to learn the basics.

    Thread Starter phpchick

    (@phpchick)

    13 years, 6 months ago

    One last question, 

    function widget($args, $instance)
  {

  function getstockstats($stocksymbol)
  {
	include_once('class.yahoostock.php');
	$objYahooStock = new YahooStock;
	$objYahooStock->addFormat("snl1d1t1cvc1p2");
	$objYahooStock->addStock("msft");

	foreach( $objYahooStock->getQuotes() as $code => $stock)
{
    ?>
    Code: <?php echo $stock[0]; ?> <br />
    Name: <?php echo $stock[1]; ?> <br />
    Last Trade Price: <?php echo $stock[2]; ?> <br />
    Last Trade Date: <?php echo $stock[3]; ?> <br />
    Last Trade Time: <?php echo $stock[4]; ?> <br />
    Change and Percent Change: <?php echo $stock[5]; ?> <br />
    Volume: <?php echo $stock[6]; ?> <br /><br />
    <?php
}

 if ( $stocksymbol == "aapl" ) {
	$result = "Test!<br />";
	echo $result;
}
  }

getstockstats("aapl");
    echo $after_widget;
  }

    Why doesn’t my function work when I add the echos inside that foreach loop? I have a feeling it is some sort of permissions issue but am not sure

    Thread Starter phpchick

    (@phpchick)

    13 years, 6 months ago

    UGH.

    The directory starts in the plugin root and NOT the directory that the file is located in.

    I changed
    include_once(‘class.yahoostock.php’);
    to
    include_once(‘prediction-widget/class.yahoostock.php’);

    and now it works. ARGH

    Thread Starter phpchick

    (@phpchick)

    13 years, 6 months ago

    One more issue

    $objYahooStock->addStock(“$stocksymbol”);

    That line is getting read as $stocksymbol
    What do I have to do change in order to make it read the variable, as opposed to $stocksymbol?

    I’ve tried wrapping in braces like

    $objYahooStock->addStock(“{$stocksymbol}”);

    but it did not work

    Thread Starter phpchick

    (@phpchick)

    13 years, 6 months ago

    Solved!

    I removed the quotes.

    $objYahooStock->addStock($stocksymbol);

    Thank you everybody for so much help 🙂

Viewing 9 replies - 1 through 9 (of 9 total)

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