我是从 Excel 文件导入到 Mysql，但结果是 1970-01-01

Question

我是从 Excel 文件导入到 Mysql，但结果是 1970-01-01，这是我的代码：

// server info
$server = 'localhost';
$user = 'root';
$pass = '';
$db = 'agenda';

// connect to the database
$mysqli = new mysqli($server, $user, $pass, $db);

// show errors (remove this line if on a live site)
mysqli_report(MYSQLI_REPORT_ERROR);

?>

<?php
require 'classes/phpexcel.php';
require_once 'classes/phpexcel/iofactory.php';
$inputFileName  = 'database/sql/agenda.xlsx';
$inputFileType  = 'Excel2007';
$objReader      = PHPExcel_IOFactory::createReader("$inputFileType");
$objReader->setReadDataOnly(true);
$objPHPExcel    = $objReader->load("$inputFileName");
$objWorksheet   = $objPHPExcel->getActiveSheet();
$highestRow     = $objWorksheet->getHighestRow();
$highestColumn  = $objWorksheet->getHighestColumn();
$highestColumnIndex = PHPExcel_Cell::columnIndexFromString($highestColumn);

// view in table
echo '<table>' . "\n";
for ($row = 1; $row <= $highestRow; ++$row) {
    echo '<tr>' . "\n";
    for ($col = 0; $col <= $highestColumnIndex; ++$col) {
        echo '<td>' . $objWorksheet->getCellByColumnAndRow($col, $row)->getValue() . '</td>' . "\n";
    }
    echo '</tr>' . "\n";
}
echo '</table>' . "\n";

//input to mysql
for($row = 2; $row <= $highestRow; ++$row) {
    for($col = 0; $col < $highestColumnIndex; ++$col) {  
        $rows[$col] = $objWorksheet->getCellByColumnAndRow($col, $row); 
    }
    $id   ="$rows[0]";
    $ts   ="$rows[1]";
    $pd   ="$rows[2]";

    $get_data1  = "$rows[3]";
    $st   = date("Y-m-d", strtotime ($get_data1));

    $stm  ="$rows[4]";

    $get_data2  ="$rows[5]";
    $et   = date("Y-m-d", strtotime ($get_data2));

    $etm  ="$rows[6]";
    $g    ="$rows[7]";
    $w    ="$rows[8]";
    // Prepared Statement
    $stmt = $mysqli->prepare("INSERT INTO `keyin` VALUES ('$id', '$ts', '$pd', '$st', '$stm', '$et', '$etm', '$g', '$w')");
    //Prepared Statement Bound$stmt->bind_param('sssssssss', $id, $ts, $pd, $st, $stm, $et, $etm, $g, $w);     


    //Prepared Statement Executed 
    $stmt->execute();
    printf("%s Row Inserted.\n", $stmt->affected_rows);
    //Prepared Statement Closed
    $stmt->close();

    // If you don't want to use prepared statement, you can use this one

    $mysqli->query("INSERT INTO users (Username,Email,Gender,Country) VALUES ('$rows[0]', '$rows[1]', '$rows[2]', '$rows[3]')");
}
$mysqli->close();

Answer 1

由于 excel 计算自 1900-01-01 以来的天数，因此您不能使用 strtotime()。

您需要做的是将 excel 天数转换为秒数。

$unix = ($get_data1 - 25569) * 86400; 
$st = date("Y-m-d", $unix);

您的 ~42000 的 excel 值小于以秒为单位的一天 (86400)，因此您在 php 中得到 1970-01-01。
如果您在 php 代码中添加时间，您会看到时间大约是 1970 年 1 月 1 日中午。

https://3v4l.org/teI1F

Answer 2

试试这个：

$st = date('Y-m-d', PHPExcel_Shared_Date::ExcelToPHP($get_data1));