PHP class initialization, you can not use variable parameters, how to change it?

In the making of Gay Bay, and then ran into problems.

I just want to be used as a function is enough, the direct double quotation mark is not a problem, change the variable will not respond. Amount

$entxt =$_get["Entxt"];
Echo $entxt;
$x = new Fanyi ($entxt, 1);
echo $x->go ();
Class Fanyi
{

Reply to discussion (solution)

Direct double quotation marks are not a problem, but they do not respond to variables.
I don't understand.

Direct double quotation marks are not a problem, but they do not respond to variables.
I don't understand.
This can:

 
  Go (), class Fanyi{var $cilentID, Var $clientSecret, var $authUrl, var $scopeUrl, var $grantType; var $inputstring; var $ Tolanguage;var $fromLanguage; function __construct ($inputstring, $biao = ' 0 ') {

Not this one.
$entxt =$_get["Entxt"];
Echo $entxt;
$x = new Fanyi ("Hello Baby", 1);
echo $x->go ();
Class Fanyi
{
var $cilentID;
var $clientSecret;
var $authUrl;
var $scopeUrl;
var $grantType;
var $inputstring;
var $toLanguage;
var $fromLanguage;
function __construct ($inputstring, $biao = ' 0 ')
{

Direct double quotation marks are not a problem, but they do not respond to variables.
I don't understand.
Said the reverse, the above can not, the following May

New Fanyi ($entxt ;, 1);
More than a semicolon, resulting in grammatical errors can not be resolved, and you blocked the error prompt, so did not find this problem

Upstairs, with a semicolon.

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