Question

以下查询会因@（符号）而导致错误。删除后，查询将正常工作。我试图逃避@角色，但没有运气。

SELECT * FROM clients WHERE MATCH (form) AGAINST ('test@test.com' IN BOOLEAN MODE);

产生的错误是：

#1064 - syntax error, unexpected '@', expecting $end

请注意，我正在phpMyAdmin SQL控制台区域中测试这些查询，因此我的其他编程不会出现转义错误。

MySQL服务器版本为5.6.17。

有什么想法吗？感谢。

Answer 1

这与INNODB FULLTEXT索引相关联。

它是作为以下组合引入的：

InnoDB全文搜索不支持在单个搜索词上使用多个运算符



@距离   此运算符仅适用于InnoDB表。它测试两个或多个单词是否都在相互指定的距离内开始，用单词测量。

http://dev.mysql.com/doc/refman/5.6/en/fulltext-boolean.html

# Running a test search for MATCH('+test{$ascii}test' IN BOOLEAN MODE) on all ASCII chars returns errors on:
40 (
41 )
64 @

MYSQL似乎将这些符号视为wordbreaks并且我发现无法逃脱并将其包含在实际查询中，因此我的解决方案是对符号进行拆分并将它们作为一个组包含在内。 “test @ bar”==（+ test + bar）

# As a further test, running a search for MATCH('+{$ascii}' IN BOOLEAN MODE) returns errors for:
40 (
41 )
42 *
43 +
45 -
60 <
62 >
64 @
126 ~

这与MYSQL文档中的预期一样，是特殊的BOOLEAN修饰符

# As a testcase (Requires MYSQL 5.6+): 
CREATE TABLE `fulltext_innodb` (
`id` int(11) NOT NULL AUTO_INCREMENT,
 `text` text COLLATE utf8_unicode_ci,
 PRIMARY KEY (`id`),
 FULLTEXT KEY `text` (`text`)
) ENGINE=InnoDB
INSERT INTO `fulltext_innodb` (`id`, `text`) VALUES (1, 'test@bar');

SELECT * FROM `fulltext_innodb` WHERE MATCH (`text`) AGAINST( '+test@bar’ IN BOOLEAN MODE )

#1064 - syntax error, unexpected '@', expecting $end

Answer 2

似乎没有办法用任何其他字符替换“ @”。到目前为止，发现“ @”并在搜索后添加字符串是最好的解决方法。这意味着

$mail = 'test@test.com';
$mail = str_replace("@", " +", $mail); //replace the @ that causes the problem
$query = "SELECT * FROM clients WHERE MATCH (form) AGAINST ('$mail' IN BOOLEAN MODE)'; //query with replaced mail address

应该带来所需的结果。

另一种方法是像在this post中那样处理它，它包含一个类似的问题，但又有一个很好的解决方法。

Answer 3

不是直接的答案，但是如果有人在寻找PHP代码来处理用户输入的搜索字符串以进行全文搜索的标记化，则可以使用以下代码：

/**
 * Method to take an input string and tokenize it into an array of words for Full Text Searching (FTS).
 *
 * This method is used when an input string can be made up of multiple words (let's say, separated by space characters),
 * and we need to use different Boolean operators on each of the words. The tokenizing process is similar to extraction
 * of words by FTS parser in MySQL. The operators used for matching in Boolean condition are removed from the input $phrase.
 * These characters as of latest version of MySQL (8+) are: +-><()~*:""&|@  (@ is specific for InnoDB)
 * We can also execute the following query to get updated list: show variables like 'ft_boolean_syntax';
 * Afterwards, the modified string is split into individual words considering either space, comma, and, period (.) characters.
 * Details at: https://dev.mysql.com/doc/refman/8.0/en/fulltext-natural-language.html
 *
 * @param string $phrase Input statement/phrase consisting of words
 * @return array Tokenized words
 * @author Madhur, 2019
 */
function tokenizeStringIntoFTSWords(string $phrase) : array {
    $phrase_mod = trim(preg_replace('/[><()~*:"&|@+-]/', ' ', trim($phrase)));
    $words_arr = preg_split('/[\s,.]/', $phrase_mod, null, PREG_SPLIT_NO_EMPTY);

    // filter out the fulltext stop words and words whose length is less than 3.
    $fts_words = array();
    $fulltext_stop_words = array(
        'about','are','com','for','from','how','that','this','was','what',
        'when','where','who','will','with','und','the','www'
    );
    foreach($words_arr as $word) {
        // By default MySQL FULLTEXT index does not store words whose length is less than 3.
        // Check innodb_ft_min_token_size Ref: https://dev.mysql.com/doc/refman/8.0/en/innodb-parameters.html#sysvar_innodb_ft_min_token_size
        // So we need to ignore words whose length is less than 3.
        if(strlen($word) < 3) continue;

        // Ignore the fulltext stop words, whose length is greater than 3 or equal to 3.
        // Ref: https://dev.mysql.com/doc/refman/8.0/en/fulltext-stopwords.html
        if (in_array($word, $fulltext_stop_words)) continue;

        $fts_words[] = $word;
    }

    return $fts_words;
}

以上代码将处理停用词，最小字长限制以及布尔模式运算符。因此，例如，如果用户输入：Search@bar with in Javascript，它将返回(Search, bar, Javascript)的数组。然后，可以使用此数组编写全文查询。

使用@符号进行MySQL全文搜索会产生错误＆＃34;语法错误，意外＆＃39; @＆＃39;，期待$ end＆＃34;

3 个答案: