PHP从表单中插入日期时间

Question

我有一个表单，用户需要选择两个日期，一个是日期，一个是日期和时间。我有一个基本表单，其中包含两个字段，用户必须输入日期（如2014-02-02）和时间日期时间（如2014-02-02 10:20:00），此表单验证效果很好。

但是，将值插入数据库时​​会出现问题。这就是我所拥有的：

<?php
class Quote
{
  public $job_deadline = null;
  public $job_dispatchdate = null;
}

public function __construct( $data=array() ) {
    if ( isset( $data['job_deadline'] ) ) $this->job_deadline = (int) $data['job_deadline'];
    if ( isset( $data['job_dispatchdate'] ) ) $this->job_dispatchdate = (int) $data['job_dispatchdate'];
    //if ( isset( $data['job_dispatchdate'] ) ) $this->job_dispatchdate = date('Y-m-d H:i:s',strtotime($data['job_dispatchdate']));
}

public function storeFormValues ( $params ) {
    $this->__construct( $params );

       if ( isset($params['job_deadline']) ) {
      $job_deadline = explode ( '-', $params['job_deadline'] );

      if ( count($job_deadline) == 3 ) {
        list ( $y, $m, $d ) = $job_deadline;
        $this->job_deadline = gmmktime ( 0, 0, 0, $m, $d, $y );
      }
    }

    if ( isset($params['job_dispatchdate']) ) {
        $job_deadline = $params['job_dispatchdate'];
        list ( $y, $m, $d, $h, $i, $s ) = $job_dispatchdate;
        $this->job_dispatchdate = gmmktime ( 0, 0, 0, 0, 0, 0, $y, $m, $d, $h, $i, $s );
    }

    /*
    if ( isset($params['job_dispatchdate']) ) {
        $datetime = date('Y-m-d H:i:s', strtotime($params['job_dispatchdate']));
    }
    */
}

public function insertjob() {
    $conn = new PDO( DB_DSN, DB_USERNAME, DB_PASSWORD );

    //$datetime = date('Y-m-d H:i:s', strtotime($params['job_dispatchdate']));  USING $datetime inplace of FROM_UNIXTIME(:job_dispatchdate) and removing st for job_dispatchdate

    $sql = "INSERT INTO tbl1 (job_deadline, job_dispatchdate) 
    VALUES (FROM_UNIXTIME(:job_deadline), FROM_UNIXTIME(:job_dispatchdate))";               
    $st = $conn->prepare ( $sql );
    $st->bindValue( ":job_deadline", $this->job_deadline, PDO::PARAM_INT );
    $st->bindValue( ":job_dispatchdate", $this->job_dispatchdate, PDO::PARAM_INT );
    $st->execute();
    $this->job_id = $conn->lastInsertId();
    $inserted_id = $this->id = $conn->lastInsertId();
    $conn = null;
}
?>

job_deadline 插入正常，问题出在job_dispatchdate。注释掉的部分是我尝试过的东西，但也是不同的工作。我从来没有得到任何插入，所以该字段显示为 NULL 或像 1970-01-01 00:00:00 这样的日期。

谁能帮我插入日期时间。

提前致谢。

伊恩

---编辑---

<?php
class Quote
{
  public $job_deadline = null;
  public $job_dispatchdate = null;
}

public function __construct( $data=array() ) {
    if ( isset( $data['job_deadline'] ) ) $this->job_deadline = (int) $data['job_deadline'];
    if ( isset( $data['job_dispatchdate'] ) ) $this->job_dispatchdate = preg_replace ( "/[^\.\,\-\_\'\|\+\#\"\@\%\?\!\&\:\;\£\$\/\\\(\n) a-zA-Z0-9()]/", "", $data['job_dispatchdate'] );
}

public function storeFormValues ( $params ) {
    $this->__construct( $params );

       if ( isset($params['job_deadline']) ) {
      $job_deadline = explode ( '-', $params['job_deadline'] );

      if ( count($job_deadline) == 3 ) {
        list ( $y, $m, $d ) = $job_deadline;
        $this->job_deadline = gmmktime ( 0, 0, 0, $m, $d, $y );
      }
    }

    if ( isset($params['job_dispatchdate']) ) {
        $job_deadline = $params['job_dispatchdate'];
        list ( $y, $m, $d, $h, $i, $s ) = $job_dispatchdate;
        $this->job_dispatchdate = gmmktime ( 0, 0, 0, 0, 0, 0, $y, $m, $d, $h, $i, $s );
    }
}

public function insertjob() {
    $conn = new PDO( DB_DSN, DB_USERNAME, DB_PASSWORD );
    //$datetime = date('Y-m-d H:i:s', strtotime($params['job_dispatchdate']));  USING $datetime inplace of FROM_UNIXTIME(:job_dispatchdate) and removing st for job_dispatchdate
    $sql = "INSERT INTO tbl1 (job_deadline, job_dispatchdate) 
    VALUES (FROM_UNIXTIME(:job_deadline), FROM_UNIXTIME(:job_dispatchdate))";               
    $st = $conn->prepare ( $sql );
    $st->bindValue( ":job_deadline", $this->job_deadline, PDO::PARAM_INT );
    $st->bindValue( ":job_dispatchdate", $this->job_dispatchdate, PDO::PARAM_STR );
    $st->execute();
    $this->job_id = $conn->lastInsertId();
    $inserted_id = $this->id = $conn->lastInsertId();
    $conn = null;
}
?>

基于 cmets，我已将代码更改为此。并将 DB 中的字段设置为字符串 Varchar 但无济于事。

Answer 1

if ( isset($params['job_dispatchdate']) ) {
        $job_deadline = $params['job_dispatchdate'];
        $this->job_dispatchdate = date('Y-m-d H:i:s', strtotime( $job_deadline ));
}

这似乎解决了它。谢谢大家

Answer 2

 $st->bindValue( ":job_dispatchdate", $this->job_dispatchdate, PDO::PARAM_INT)

应该是

 $st->bindValue( ":job_dispatchdate", $this->job_dispatchdate, PDO::PARAM_STR)

建议：最好将所有日期以 unix 时间戳格式存储在 mysql 中。

EDIT 2 ：将您的 job_dispatch 列类型更改为 STRING，如果它是 INT 类型