Question

任何人都可以看到我使用这个类做了什么incorreclty它应该改变之前和之后的单词，但它只是坐在那里什么都不做。第二个php pgm是我正在使用的实际类..

<?php

include("parse.inc.php");

$bob = new parseClass();

$tag = "{Before}";
$replacement = "After";
$content = "My animal runs around all over the place and his name is {Before}.";
$bob->ReplaceTag($tag, $replacement, $content);
echo $bob;

?>




parse.inc.php (file name)
***************************

<?php
Class parseClass {

    function ReplaceTag($tag, $replacement, $content) {
        $content = str_replace($tag, $replacement, $content);
        return $content;
    }

}

// END Class parseClass
?>

Answer 1

在您的情况下，您将内容传递给类的函数因此，您应该将返回值存储到变量第一个

中

$result = $bob->ReplaceTag($tag, $replacement, $content);

echo $result;

Answer 2

您只需将函数的结果存储在局部变量中，然后打印出来。

$result = $bob->ReplaceTag($x, $y, $z);
print($result);

Answer 3

我想你要打印要更换的字符串，所以你应该：

$result = $bob->ReplaceTag($tag, $replacement, $content);
echo $result;

为什么使用echo来打印类？如果你想这样做，请改用“var_dump”或“print_r”;

Answer 4

如果硬性要求是您希望echo parseClass的实例，请尝试以下操作：

class parseClass 
{
    private $content;

    public function ReplaceTag($tag, $replacement, $content) 
    {
        $this->content = str_replace($tag, $replacement, $content);
    }

    public function __toString()
    {
        return $this->content;
    }
}

$bob = new parseClass();

$tag = "{Before}";
$replacement = "After";
$content = "My animal runs around all over the place and his name is {Before}.";

$bob->ReplaceTag($tag, $replacement, $content);

echo $bob;

供参考，见：

http://php.net/manual/en/language.oop5.magic.php#object.tostring

Answer 5

只需更换这两行

即可

$bob->ReplaceTag($tag, $replacement, $content);

echo $bob;

与

$content = $bob->ReplaceTag($tag, $replacement, $content);

echo $content;

请让这个php类按预期工作

5 个答案: