Question

这里我有一个数组，在这个数组中我还有一个数组名studentAbsentId，现在我想在数据库中插入studentAbsentId，就像这里studentAbsentId 2和{{1} } 2不存在所以我需要插入两行

我的更新代码 print_r（$ params）;

studentAbsentId

现在我想只接受studentAbsentId，该怎么办？

预期结果

 Array
(
    [studentAbsentId] => Array
        (
            [0] => 2
            [1] => 3
        )

    [schoolId] => 2
    [classId] => 1
    [sectionId] => 1
    [studentAbsentDate] => 2017-04-18
    [reg_date] => 2017-04-18 05:20:17
    [created_by] => 1
)

Answer 1

试试这有助于：）

<?php

$params=array(
        'studentAbsentId' => array
            ( 2,3
            ),
        'studentAbsentDate' => '2017-04-18',
        'reg_date' => '2017-04-18 04:41:21',
        'created_by' => '1',
    );


foreach($params['studentAbsentId'] as $ff)
{
        $atnEntry = array(
        "studentAbsentId" => $ff,
        "studentAbsentDate" =>$params['studentAbsentDate'],
        "morning" => "1"

        );

        $this->db->insert("student_absent_list" , $atnEntry);
}        
        $return = array("status" => "Success" );
        echo json_encode($return);  

?>

Answer 2

试试这个

$test = array('studentAbsentId' => array('2','3'),'studentAbsentDate' => '2017-04-18','reg_date' => '2017-04-18 04:41:21','created_by' => '1');

    foreach($test['studentAbsentId'] AS $tet)
    {
        $array_val = array('studentAbsentId' => $tet,'studentAbsentDate' => $test['studentAbsentDate'],'created_by' => $test['created_by']);

        // echo "<pre>";print_r($array_val);

        $this->db->insert('table_name',$array_val);
    }

如何获取数组中的值以及如何使用codeiniter在数据库中插入

2 个答案: