解析短语和关键字的搜索字符串

Question

例如,我需要在php中解析关键字和短语的搜索字符串

字符串1: value of "measured response" detect goal "method valuation" study

会产生: value,of,measured reponse,detect,goal,method valuation,study

如果字符串有:我也需要它工作:

1. 没有用引号括起来的短语,
2. 任何数量的短语都用引号括起来,引号外有任意数量的关键字,
3. 只有引号中的短语,
4. 只有空格分隔的关键字.

我倾向于使用preg_match模式'/(\".*\")/'将短语放入数组中,然后从字符串中删除短语,最后将关键字放入数组中.我不能把所有东西拉到一起!

我也在考虑用逗号替换引号之外的空格.然后将它们分解为数组.如果这是一个更好的选择,我该怎么做preg_replace？

有没有更好的方法来解决这个问题？救命!非常感谢大家

Answer 1

preg_match_all('/(?<!")\b\w+\b|(?<=")\b[^"]+/', $subject, $result, PREG_PATTERN_ORDER);
for ($i = 0; $i < count($result[0]); $i++) {
    # Matched text = $result[0][$i];
}

这应该产生您正在寻找的结果.

说明:

# (?<!")\b\w+\b|(?<=")\b[^"]+
# 
# Match either the regular expression below (attempting the next alternative only if this one fails) «(?<!")\b\w+\b»
#    Assert that it is impossible to match the regex below with the match ending at this position (negative lookbehind) «(?<!")»
#       Match the character “"” literally «"»
#    Assert position at a word boundary «\b»
#    Match a single character that is a “word character” (letters, digits, etc.) «\w+»
#       Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
#    Assert position at a word boundary «\b»
# Or match regular expression number 2 below (the entire match attempt fails if this one fails to match) «(?<=")\b[^"]+»
#    Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=")»
#       Match the character “"” literally «"»
#    Assert position at a word boundary «\b»
#    Match any character that is NOT a “"” «[^"]+»
#       Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»