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在PHP中解析属性/值列表

parsing regex php

3

dreeves · 技术社区 · 15 年前

给定一个具有属性/值对的字符串,例如

attr1="some text" attr2 = "some other text" attr3= "some weird !@'#$\"=+ text"

目标是解析它并输出一个关联数组,在这种情况下:

array('attr1' => 'some text',
      'attr2' => 'some other text',
      'attr3' => 'some weird !@\'#$\"=+ text')

注意等号周围的间距不一致,输入中有转义双引号,输出中有转义单引号。

2 回复 | 直到 15 年前

1

6

Bart Kiers 15 年前

尝试如下操作:

$text = "attr1=\"some text\" attr2 = \"some other text\" attr3= \"some weird !@'#$\\\"=+ text\"";
echo $text;
preg_match_all('/(\S+)\s*=\s*"((?:\\\\.|[^\\"])*)"/', $text, $matches, PREG_SET_ORDER);
print_r($matches);

产生:

attr1="some text" attr2 = "some other text" attr3= "some weird !@'#$\"=+ text"

Array
(
    [0] => Array
        (
            [0] => attr1="some text"
            [1] => attr1
            [2] => some text
        )

    [1] => Array
        (
            [0] => attr2 = "some other text"
            [1] => attr2
            [2] => some other text
        )

    [2] => Array
        (
            [0] => attr3= "some weird !@'#$\"=+ text"
            [1] => attr3
            [2] => some weird !@'#$\"=+ text
        )

)

简单解释一下:

(\S+)               // match one or more characters other than white space characters
                    // > and store it in group 1
\s*=\s*             // match a '=' surrounded by zero or more white space characters 
"                   // match a double quote
(                   // open group 2
  (?:\\\\.|[^\\"])* //   match zero or more sub strings that are either a backslash
                    //   > followed by any character, or any character other than a
                    //   > backslash
)                   // close group 2
"                   // match a double quote

2

Amarghosh 15 年前

编辑:如果值以反斜杠结尾,则此regex将失败,如 attr4="something\\"

我不知道php,但是因为regex在任何语言中基本上都是相同的,所以我在actionscript中就是这样做的:

var text:String = "attr1=\"some text\" attr2 = \"some other text\" attr3= \"some weird !@'#$\\\"=+ text\"";

var regex:RegExp = /\s*(\w+)\s*=\s*(?:"(.*?)(?<!\\)")\s*/g;

var result:Object;
while(result = regex.exec(text))
    trace(result[1] + " is " + result[2]);

我得出了以下结论:

ATTR1是一些文本
ATTR2是其他文本
ATTR3有些奇怪!@ '$$\ ' = +文本