[PHP]変数のみを参照渡しする必要があります

A PHP Error was encountered  Severity: Runtime Notice  Message: Only variables should be passed by reference  $file_name = $_FILES[$upload_name]['name'];  $file_extension = end(explode('.', $file_name)); //ERROR ON THIS LINE  原因：  获取后缀名时，使用end函数，将数组内部指针指向最后一个元素，而end是引用传值。  The problem is, that end requires a reference, because it modifies the internal representation of the array (i.e. it makes the current element pointer point to the last element).The result of explode('.', $file_name) cannot be turned into a reference. This is a restriction in the PHP language, that probably exists for simplicity reasons.The array. This array is passed by reference because it is modified by the function. This means you must pass it a real variable and not a function returning an array because only actual variables may be passed by reference.解决办法：(1) 定义一个变量获取explode后的值，再使用end函数：  $parts = explode('.', $file_name);  $file_extension = end($parts);  (2) 使用substr和strrchr函数来取文件后缀名：  $ext = substr( strrchr($file_name, '.'), 1);  (3) 使用pathinfo函数来取文件后缀名：  $file_ext = pathinfo($file_name, PATHINFO_EXTENSION);  end() 函数将数组内部指针指向最后一个元素，并返回该元素的值（如果成功）。  pathinfo() 函数以数组的形式返回文件路径的信息。  参考链接：http://www.w3school.com.cn/php/func_array_end.asp  http://www.w3school.com.cn/php/func_filesystem_pathinfo.asp  http://stackoverflow.com/questions/4636166/only-variables-should-be-passed-by-reference