Question

我喜欢插入我的数据，如下面的结构。

    +------+----+
    |id|JAN||FEB|
    +------+----+
    |1 |  1||  5|
    +------+----+
    |2 |  8|| 12|
    +------+----+
    |3 | 15|| 19|
    +------+----+
    |4 | 22|| 26|
    +------+----+


<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "epi";

$conn = new mysqli($servername, $username, $password, $dbname);

if (!$conn) 
{
  die('Could not connect: ' . mysql_error());
}

$number_of_dates = 4;
$startDate_Jan = strtotime("2018-01-01");
for ($i = 0; $i <= $number_of_dates; $i++)
{
    $date = strtotime("Monday +" . ($i * 1) . ' weeks', $startDate_Jan);
    $month=date('m', $date).PHP_EOL;
    $datex=date('d', $date).PHP_EOL;
    $intmonth = intval($month);
    $intdate = intval($datex);

    echo $intdate;
    echo "<br>";

    $sql="INSERT INTO testdata (ID_DATE, EPI_DATE_JAN) VALUES ('','$intdate')";
    $result = mysqli_query($conn, $sql);
}

$number_of_dates = 4;
$startDate_Feb = strtotime("2018-02-01");
for ($i = 0; $i <= $number_of_dates; $i++)
{
    $date = strtotime("Monday +" . ($i * 1) . ' weeks', $startDate_Feb);
    $month=date('m', $date).PHP_EOL;
    $datex=date('d', $date).PHP_EOL;
    $intmonth = intval($month);
    $intdate = intval($datex);

    echo $intdate;
    echo "<br>";

    $sql="INSERT INTO testdata (ID_DATE, EPI_DATE_FEB) VALUES ('','$intdate')";
    $result = mysqli_query($conn, $sql);
}
$conn->close();
?>

当我喜欢上面的时候，总共创建了8个id。所以现在我想解决我的问题。告诉我一些解决方案如何解决我的问题。如果你还有困惑那么问我，我也会提供一些细节。

Answer 1

如果你的表中只有4行且ID_DATE是星期数，你可以在第二个循环中使用UPDATE，如下所示：

// second loop:
$week = $i+1;
$sql="UPDATE testdata set EPI_DATE_FEB = $intdate WHERE ID_DATE=$week";
$result = mysqli_query($conn, $sql);

所以第二个循环可能是：

$number_of_dates = 4;
$startDate_Feb = strtotime("2018-02-01");
for ($i = 0; $i <= $number_of_dates; $i++)
{
    $date = strtotime("Monday +" . ($i * 1) . ' weeks', $startDate_Feb);
    $month=date('m', $date).PHP_EOL;
    $datex=date('d', $date).PHP_EOL;
    $intmonth = intval($month);
    $intdate = intval($datex);

    echo $intdate;
    echo "<br>";

    $week = $i+1;
    $sql="UPDATE testdata SET EPI_DATE_FEB=$intdate WHERE ID_DATE=$week";
    $result = mysqli_query($conn, $sql);
}

但使用prepared queries更好。

如何在相同的id但不同的列值下插入值？

1 个答案: