PHP5:如果我调用它，为什么包含的函数总是首先回显?

PHP5: How come an included function always echoes first if I call it?(PHP5:如果我调用它，为什么包含的函数总是首先回显?)

问题描述

我这里有两个文件:

ToBeIncludedFile.php

ToBeIncludedFile.php

<?php
function printOut(){
    echo "World!";
}
?>

MainFile.php

MainFile.php

<?php
include("ToBeIncludedFile.php");

echo "Hello ".printOut();
?>

我会期待Hello World！".相反，我得到了这个:世界！你好".

I would expect "Hello World!". Instead I get this: "World!Hello ".

我知道如果我写return而不是echo，那么一切都很好.是因为我正在回显一个已经回显字符串的函数吗?但是为什么它会打印出字符串World！"先不报错?

I know if I write return instead of echo, then everything is fine. Is it because I'm echoing a function that is already echoing a string? But then why would it print out the string "World!" first and not throw an error?

推荐答案

之所以先回显，是因为它被调用，而之后是字符串连接"(更多内容在第二):

The reason it echos first, is because it is called, and afterwards are the strings "concatenated" (more on that in a second):

你想要的 ToBeIncludedFile.php 是 return "World!";，而不是 echo.

What you want in ToBeIncludedFile.php is return "World!";, not echo.

现在是这样的:

• 您包含文件，它不打印任何内容，这是正确的.
• 您将字符串Hello"和printOut() 的返回值 串联起来.这意味着，首先调用该函数:
• printOut() 执行并打印World!"，不返回任何内容.
• 然后，您的主脚本将Hello"与任何内容连接起来并打印出来.

• You include the file, which doesn't print anything, this is correct.
• You do a concatenation of the string "Hello" and the return value of printOut(). That means, first that function is called:
• printOut() executes and prints "World!", returning nothing.
• Your main script then concatenates "Hello" with nothing and prints that.

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