PHP:变量在函数内部不起作用? [英] PHP: variable not working inside of function?

问题描述

echo $path; //working
function createList($retval) {
    echo $path; //not working
    print "<form method='POST' action='' enctype='multipart/form-data'>";
    foreach ($retval as $value) {
            print "<input type='checkbox' name='deletefiles[]' id='$value' value='$value'>$value<br>";
    }
    print "<input class='submit' name='deleteBtn' type='submit' value='Datei(en) löschen'>";
    print "</form>";    
}

我做错了什么?为什么 $path 在 createList 函数外正确打印，但在函数内无法访问?

what am I doing wrong? why is $path printed correctly outside of the createList function, but it's not accessible inside the function?

推荐答案

因为没有在函数中定义.

有几种方法可以解决这个问题:

Because it's not defined in the function.

There are a few ways to go about this:

1) 使用 Alex 所说的，告诉函数它是一个全局变量:

echo $path; // working

function createList($retval) {
  global $path;

  echo $path; // working
}

2) 将其定义为常量:

define(PATH, "/my/test/path"); // You can put this in an include file as well.
  
echo PATH; // working

function createList($retval) {
  echo PATH; // working
}

3) 如果它特定于该函数，则将其传递给函数:

echo $path; // working

function createList($retval, $path) {
  echo $path; // working
}

根据该功能的实际工作方式，其中之一可以.

Based on how the function really works, one of those will do ya.

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