php - 如何从查询中获取结果并将结果插入到其他表中？

Question

我查询了需要结果的数据库，然后将其存储在一个变量中。然后我会将变量传递给 INSERT INTO 语句，但由于某种原因我的代码不起作用。这是我的代码/

$query = "SELECT * from animals where old= 1 AND user_id=".$_SESSION['user_id'];
$result = mysqli_query(mysqli_connect("","","", ""), $query);


while ($row = mysqli_fetch_array($result))
{
    $variable[] = $row['number'];
}

//现在我将把 $variable 传递给 INSERT INTO 语句

if(isset($_POST['submit_d']))
{    
    foreach($variable as $var)
    {
        $query="INSERT INTO selectedanimals(number) VALUES ({$var},2)";
        mysqli_query($con, $query) or die (mysql_error());
    }
?>
<script>
    alert("Animal added.");
    self.location="chooseAnimals.php";
</script>
<?php
    }
?>

Answer 1

您可以在单个查询中为此目的使用 INSERT INTO ... SELECT:

INSERT INTO selectedanimals (number)
SELECT number
FROM animals
WHERE old = 1 AND user_id = some_id

PHP代码:

$query = "INSERT INTO selectedanimals (number) ";
$query.= "SELECT number FROM animals WHERE old = 1 AND user_id = ".$_SESSION['user_id'];
mysqli_query($con, $query) or die (mysql_error());