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。我有两个字段的基本表单，其中用户必须输入2014-02-02之类的日期和2014-02-02 10:20:00之类的时间日期时间，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，问题在于job_dispatchdate。注释掉的部分是我尝试过的，但也有不同的工作。我没有插入任何内容，因此字段显示为NULL或某个日期显示为1970-01-01 00:00:00。

任何人都可以帮我插入日期时间。

提前致谢。

伊恩

--- --- EDIT

<?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;
}
?>

根据评论，我已将代码更改为此。并将DB中的字段设置为字符串Varchar但无效。

Answer 1

 $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：最好以unix时间戳格式将所有日期存储在mysql中。

编辑2 ：如果 INT 类型

，则将您的job_dispatche列类型更改为 STRING

Answer 2

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

这似乎解决了这个问题。谢谢大家

PHP从表单插入日期时间

2 个答案: