php:在"echo"中执行数学运算.还将变量名称传递给函数

Question

我正在构建一个表(不使用<table>)并用mysql中的数据填充它.我有以下代码来构建"表":

$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>";

我遇到了一个问题" . $NYC_min-$NYC . ":它打破了行(该行被解释为刚刚结束" . $NYC_min-$NYC . ".但是,如果我有<span>$NYC_min-$NYC</span>(不作为回声的附加组件),则打印单元格的值300-500而不是200.

另外我不确定我是否正确设置了我的功能(但这并没有破坏表格/行).从cityMinMet('$NYC')我想要$NYC发送到函数的文字字符串(而不是它的值).在函数内部,我需要追加_min到$NYC,然后调用$NYC_min,并与它的价值回报.

编辑:我改变的顺序$NYC和$NYC_min在方程.

Answer 1

用括号括起来:

" . ($NYC - $NYC_min) . "