Question

标题可能混淆不清,但我认为我怀疑清楚。

我要解释一下。我在此问:

 $sql =  SELECT * FROM calendar WHERE day = " .$day. " AND month = " .$month. " AND year = " .$year. " AND realizada = 0 AND colaborador = "What to put here?!" ;

但是,实地“实验室”是一个序列式阵列。

一例:Iprint_r 我的阵列在进行空中勘测后的价值就是如此:

Array (0] => l3g iii 页: 1



我想在前几句中寻求“l3g”的魔ine,我如何能够这样做?

谢谢。

l3g

Answer 1


如果你需要从你的阵列中询问个别内容,则不储存序列号。 在每个儿童桌上储存每个单元,并将其与贵日历表的主要价值联系起来:

CREATE TABLE colaboradoras (
  calendar_id INT NOT NULL,
  colaborador VARCHAR(20) NOT NULL,
  FOREIGN KEY (calendar_id) REFERENCES calendar(calendar_id)
);

INSERT INTO colaboradoras VALUES
(1234,  l3gion ),
(1234,  Someone ),
(1234,  teste );

$sql = "SELECT * FROM calendar AS c JOIN colaboradoras AS o 
              ON c.calendar_id = o.calendar_id
        WHERE c.day = $day AND c.month = $month AND c.year = $year 
          AND c.realizada = 0 AND o.colaborador =  l3gion ";


这是正常的做法。

如果你必须储存序列式阵列,你可以检查HowvantFeed Uses MySQL,以指数化“schemaless”数据。 但这也涉及为指数编制新的表格。

如果你真的能够创建任何新的表格或索引,你可以尝试使用<条码>LIKE或<条码>,但这两种解决办法都将非常低效和容易发生错误。

SELECT ... WHERE ... AND colaborador REGEXP  [[:<:]]s:6:"l3gion";

你再说一遍。

Answer 2

简短的答复: 将数据编为数据库领域。 http://en.wikipedia.org/wiki/Database_normalization”rel=“nofollow noreferer”>normalization ......

长久的答案是很难做到。页: 1 你也可以写出处理该问题的法规(Andre use MySQL s >>>:/code> > 表述。

但最佳解决办法是不将序列化数据储存在数据库中......

Answer 3

Colaborador = “What to put here?

该法典将处理序列化和原始数据以及阵列和便衣价值。

if (is_string($colaborador)) {
    // prevent errors from showing
    ob_start();
    $data = unserialize($colaborador);
    // $data is now false or output buffer is non-empty if $colaborador was
    // not serialized
    $data = ob_get_length() > 0 || false === $data ? $colaborador : $data;
    // lose the possible error
    ob_end_clean();
} else {
    $data = $colaborador;
}
// use the first value as colaborador if an array
$sql =  SELECT * FROM calendar WHERE day = " .$day. " AND month = " .$month. "
AND year = " .$year. " AND realizada = 0 AND colaborador =    .
(is_array($data) ? (empty($data) ?    : $data[0]) : $data) .    ;`

或者,如果你想找他们的话:

$sql =  SELECT * FROM calendar WHERE day = " .$day. " AND month = " .$month. "
AND year = " .$year. " AND realizada = 0 AND colaborador = IN (   .
(is_array($data) ? (empty($data) ?    : implode(" ,  ", $data[0])) : $data) .   ) ;`

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