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php: performing math inside of an "echo". also passing variable name to a function

开发者 https://www.devze.com 2023-04-05 17:38 出处:网络
I\'m building a table (tho not using <table>) and populating it with data from mysql. I have the following code to build the \"table\":

I'm building a table (tho not using <table>) and populating it with data from mysql. I have the following code to build the "table":

$NYC = // data from mysql, working fine
$NYC_min = 300;
switch($cou_mun){
    case "New York":
        $NYC++;
        break;
    //etc
}

function cityMinMet($city){
    $city_min= "$city" . "_min";
    if($city>$city_min){return "yes";}
    else{return "no";};
}

echo "<h3>Table 6: Recruitment by target area</h3>";
echo "<ul>\n<li><span>Target Area</span><span>Number Recruited</span><span>Amount under minimum</span><span>Minimum</span><span>Minimum met?</span></li>";
echo "<li><span>12 Jurisdictions with Highest AIDS Prevalance</span><span></span><span></span><span></span><span></span></li>";
echo "<li><span>New York</span><span>$NYC</span><span>" . $NYC_min-$NYC . "</span><span>$NYC_min</span><span>"; cityMinMet('$NYC'); echo "</span></li>";

I encounter a problem with " . $NYC_min-$NYC . ": it breaks t开发者_JAVA百科he row (the row gets interpreted as ending just before " . $NYC_min-$NYC . ". However, if I have <span>$NYC_min-$NYC</span> (not as an additional component of the echo), the value of the cell is printed as 300-500 instead of 200.

Also I'm not sure I've setup my function properly (but this is not breaking the table/row). From cityMinMet('$NYC') I want the literal string $NYC (and not its value) sent to the function. Inside the function I need to append _min to $NYC and then call $NYC_min and it return with its value.

EDIT: I changed the order of $NYC and $NYC_min in the equation.


Wrap it in parenthesis:

" . ($NYC - $NYC_min) . "


For the " . $NYC-$NYC_min . ", change to " . ($NYC-$NYC_min) . "

To pass the literal string $NYC use "\$NYC". Note, you need to use double quotes " instead of single quotes ' here, otherwise it will pass the string literal \$NYC

As @webbiedave points out in his comment, passing it as '$NYC' should already be sending the string literal $NYC


Your first problem: $NYC - $NYC_min is being interpreted as a string (subtly cast as such) because you are concatenating it with a string. You should perform the maths before the echo and store it in a variable which you write in there (clean) or you can put it in brackets so its interpreted before the php parser considers the string concatenation context.


From cityMinMet('$NYC') I want the literal string $NYC (and not its value) sent to the function.

The way you're calling the function (with single quotes) will pass the literal string.

However, if you need the value of $NYC_min in function cityMinMet then just pass the value to the function:

function cityMinMet($city, $city_min)
{
    if ($city > $city_min) {
        return "yes";
    } else {
        return "no";
    }
}

In your calling code do:

echo cityMinMet($NYC, $NYC_min);

Or you can forgo the function all together and just do:

echo ($city > $city_min) ? 'yes' : 'no';


Might be easiest to make a quick function to do what you need to do:

function calculate($nyc, $nyc_min) {
  return $nyc - $nyc_min;
}

then replace what you have with:

echo "<li><span>New York</span><span>$NYC</span><span>" . calculate($NYC,$NYC_min) . "</span>...";
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