从函数PHP中更改全局变量

I am trying to change a variable that is outside of a function, from within a function. Because if the date that the function is checking is over a certain amount I need it to change the year for the date in the beginning of the code.

我试图改变一个函数之外的变量，从函数内部。因为如果函数正在检查的日期超过一定数量，我需要它在代码开始时更改日期的年份。

$var = "01-01-10";
function checkdate(){
     if("Condition"){
            $var = "01-01-11";
      }
}

4 个解决方案

#1

A. Use the global keyword to import from the application scope.

使用全局关键字从应用程序范围导入。

$var = "01-01-10";
function checkdate(){
    global $var;  
    if("Condition"){
        $var = "01-01-11";
    }
}
checkdate();

B. Use the $GLOBALS array.

使用$GLOBALS数组。

$var = "01-01-10";
function checkdate(){
    if("Condition"){
        $GLOBALS['var'] = "01-01-11";
    }
}
checkdate();

C. Pass the variable by reference.

通过引用传递变量。

$var = "01-01-10";
function checkdate(&$funcVar){  
    if("Condition"){
        $funcVar = "01-01-11";
    }
}
checkdate($var);

#2

Just use the global keyword like so:

使用全局关键字，比如:

$var = "01-01-10";
function checkdate(){
     global $var;

     if("Condition"){
            $var = "01-01-11";
      }
}

Any reference to that variable will be to the global one then.

对该变量的任何引用都将指向全局变量。

#3

All the answers here are good, but... are you sure you want to do this?

这里所有的答案都很好，但是……你确定你要这么做吗?

Changing global variables from within functions is generally a bad idea, because it can very easily cause spaghetti code to happen, wherein variables are being changed all over the system, functions are interdependent on each other, etc. It's a real mess.

从函数内部更改全局变量通常不是一个好主意，因为它很容易导致通心粉代码的发生，其中变量在整个系统中都被更改，函数相互依赖等等。

Please allow me to suggest a few alternatives:

请允许我提出几个选择:

1) Object-oriented programming

1)面向对象编程

2) Having the function return a value, which is assigned by the caller.

2)让函数返回由调用者分配的值。

e.g. $var = checkdate();

例如:$ var = checkdate();

3) Having the value stored in an array that is passed into the function by reference

将值存储在数组中，并通过引用传递给函数

function checkdate(&$values) { if (condition) { $values["date"] = "01-01-11"; } }

函数检验日期(&$values) {if (condition) {$values["date"] = "01-01-11";} }

Hope this helps.

希望这个有帮助。

#4

Try this pass by reference

试试这个推荐信

  $var = "01-01-10";
    function checkdate(&$funcVar){  
        if("Condition"){
            $funcVar = "01-01-11";
        }
    }
    checkdate($var);

or Try this same as the above, keeping the function as same.

或者和上面一样，保持函数不变。

 $var = "01-01-10";
    function checkdate($funcVar){  
        if("Condition"){
            $funcVar = "01-01-11";
        }
    }
    checkdate(&$var);

#1