如何在变量中使用变量来定义php函数,然后回显它
问题描述:
I am trying to use variable in variable function, then to do some calculations and later use that variable and print out answer:
<?php
$a($b)=function() {
if ($b == 10 ) {
return 10 ;
}else{
return 20 ;
}
}
$a(10);
echo $a(10);
?>
I am getting error:
Fatal error: Can't use function return value in write context
我试图在变量函数中使用变量,然后进行一些计算,然后使用该变量并打印出答案 : p>
&lt;?php
$ a($ b)= function(){
if($ b == 10){
return 10 ;
}其他{
返回20;
}
}
$ a(10);
echo $ a(10);
?&gt;
code> pre>
我收到错误: p>
致命错误:无法在写上下文中使用函数返回值 p> \ n blockquote>
div>
答
Here $a($b)
means invoking a function
instead of declaration
. you should define $a
and pass $b
as a variable to the function.
Change this to:
$a($b)=function() {
This:
$a=function($b) {
Whole PHPcode Try this code snippet here
<?php
ini_set('display_errors', 1);
$a = function($b)
{
if ($b == 10)
{
return 10;
} else
{
return 20;
}
};
echo $a(10);
?>
答
Change following line
$a($b)=function() {
to
$a=function($b) {