php 尝试将正则表达式匹配传递给函数时出错

Error trying to pass regex match to function

提问by Giri

I'm getting Syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$'

我越来越 Syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$'

This is the code i'm using

这是我正在使用的代码

function wpse44503_filter_content( $content ) {
    $regex = '#src=("|\')'.
        '(/images/(19|20)(0-9){2}/(0|1)(0-9)/[^.]+\.(jpg|png|gif|bmp|jpeg))'.
        '("|\')#';
    $replace = 'src="'.get_site_url(  ).'"';

    $output = preg_replace( $regex, $replace, $content );

    return $output;
}

This is the line where i'm getting that error $replace = 'src="'.get_site_url( $2 ).'"';

这是我收到错误的那一行 $replace = 'src="'.get_site_url( $2 ).'"';

Can anyone help me to fix it? Thanks

任何人都可以帮我修复它吗？谢谢

采纳答案by Joachim Isaksson

What you're trying to do (ie replacing the matched string with the result of a function call) can't be done using preg_replace, you'll need to use preg_replace_callbackinstead to get a function called for every match.

您尝试执行的操作（即用函数调用的结果替换匹配的字符串）无法使用 完成preg_replace，您需要使用preg_replace_callback来代替为每个匹配调用一个函数。

A short example of preg_replace_callback;

preg_replace_callback 的一个简短示例；

$get_site_url =                    // Returns replacement
  function($row) { 
    return '!'.$row[1].'!';        // row[1] is first "backref"
  };                                                     

$str = 'olle';
$regex = '/(ll)/';                 // String to match

$output = preg_replace_callback(   // Match, calling get_site_url for replacement
    $regex,
    $get_site_url,
    $str);

var_dump($output);                 // output "o!ll!e"

回答by Farray

You can't have '$2' as a variable name. It must start with a letter or underscore.

您不能将 '$2' 作为变量名。它必须以字母或下划线开头。

http://php.net/manual/en/language.variables.basics.php

http://php.net/manual/en/language.variables.basics.php

Variable names follow the same rules as other labels in PHP. A valid variable name starts with a letter or underscore, followed by any number of letters, numbers, or underscores. As a regular expression, it would be expressed thus: '[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'

变量名遵循与 PHP 中其他标签相同的规则。有效的变量名以字母或下划线开头，后跟任意数量的字母、数字或下划线。作为正则表达式，它将被表示为：'[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'

EditAbove was my original answer and is the correct answer to the simple "syntax error" question. More in-depth answer below...

上面的编辑是我的原始答案，是对简单的“语法错误”问题的正确答案。下面有更深入的回答...

You are trying to use $2 to represent "the second capture group", but you haven't done anything at that point to match your regex. Even if $2 was a valid PHP variable name, it still wouldn't be set at that point in your script. Because of this, you can determine that you are using preg_replaceimproperly and that it may not suit your actual needs.

您正在尝试使用 $2 来表示“第二个捕获组”，但此时您还没有做任何事情来匹配您的正则表达式。即使 $2 是一个有效的 PHP 变量名，它仍然不会在脚本中的那个点被设置。因此，您可以确定您使用preg_replace不当，并且它可能不适合您的实际需要。

Note that the preg_replacedocumentation doesn't support using $n as a separate variable outside of the replacement operation. In other words, 'foo' . $1 . 'bar'is not a valid replacement string, but 'foo$1bar'is.

请注意，preg_replace文档不支持将 $n 用作替换操作之外的单独变量。换句话说，'foo' . $1 . 'bar'不是有效的替换字符串，而是'foo$1bar'。

Depending on the complexity of get_site_url, you have 2 options:

根据 的复杂程度get_site_url，您有两种选择：

1. If get_site_urlis simply adding a root directory or server name, you could change your replacement string to src="/myotherlocation$2". This will effectively replace "/image/..." with "/myotherlocation/image/..." in the img src. This will not work if get_site_urlis doing something more complex.

2. If get_site_urlis complex, you should use preg_replace_callbackper other answers. Give the documentation a read and post a new question (or I guess update this question?) if you have trouble with the implementation.

1. 如果get_site_url只是添加根目录或服务器名称，您可以将替换字符串更改为src="/myotherlocation$2". 这将有效地将 img src 中的“/image/...”替换为“/myotherlocation/image/...”。如果get_site_url正在做一些更复杂的事情，这将不起作用。

2. 如果get_site_url很复杂，你应该使用preg_replace_callbackper other answers。如果您在实施时遇到问题，请阅读文档并发布一个新问题（或者我想更新这个问题？）。

回答by prodigitalson

PHP variable names cant begin with a number.

PHP 变量名不能以数字开头。

回答by Ignacio Vazquez-Abrams

$2is not a valid PHP variable. If you meant the second group in the regex then you want to put \2in a string. However, since you're passing it to a function then you'll need to use preg_replace_callback()instead and substitute appropriately in the callback.

$2不是有效的 PHP 变量。如果您指的是正则表达式中的第二组，那么您想放入\2一个字符串。但是，由于您将它传递给函数，因此您需要preg_replace_callback()在回调中使用并适当替换。

回答by Krishnamoorthy Acharya

if PHP variable begins with number use following:

如果 PHP 变量以数字开头，请使用以下内容：

when I was getting the following as the result set from thrid party API

当我从第三方 API 获得以下结果集时

Code Works

代码工作

$stockInfo->original->data[0]->close_yesterday

Code Failed

代码失败

$stockInfo->original->data[0]->52_week_low

Solution

解决方案

$stockInfo->original->data[0]->{'52_week_high'}