0

大家好，我在执行后得到了数组我如何将该数组存储在数据库编码中

if (isset($_POST['submit'])) {

    $data_t1 = $_POST['t1'];

    foreach ($data_t1 as $key => $value) {

        echo 'T1: ' . $value . '<br />';
        echo 'T2: ' . $_POST['t2'][$key] . '<br />';
          echo 'T3: ' . $_POST['a1'][$key] . '<br />';
        echo 'Username: ' . $_POST['username'][$key] . '<br /><br />';


 echo 'T11: ' . $_POST['t11'] . '<br />';
          echo 'T22: ' . $_POST['t22'][$key] . '<br />';
              echo 'T33: ' . $_POST['a11'][$key] . '<br />';
        echo 'name: ' . $_POST['name'][$key] . '<br /><br />';


    }
      $data_t2 = $_POST['t2'];
      $data_t3=$_POST['a1'];
      $data_t4=$_POST['username'];
      $data_t11 = $_POST['t11'];
      $data_t22=$_POST['t22'];
      $data_t33=$_POST['a11'];
       $data_t44=$_POST['name'];
}
var_dump($data_t1);
var_dump($data_t2);
var_dump($data_t3);
var_dump($data_t4);
var_dump($data_t11);
var_dump($data_t22);
var_dump($data_t33);
var_dump($data_t44);

?>

数组以这种形式显示

 $data_t1( 12  12)
    $data_t2( 4    3)
    $data_t3( 5    6)
    $data_t4( 1.44  1.33)
    $data_t11( 12  12)
     $data_t22( 4    3)
     $data_t33( 5    6)
     $data_t44( 1.44  1.33)

我需要以 ths 形式将这些数组存储在数据库中

row1 12   4  5 1.44 12 4 5 1.44
row2 12   3  6 1.33 12 3 6 1.33

我编辑了我的问题

4

2 回答 2

1

这是一个片段，可以满足您的需求：

//database connection
$db = new PDO("mysql:host=$dbhost;dbname=$dbname",$dbuser,$dbpass);

//query
$sql = "INSERT INTO table (c1, c2, c3, c4) VALUES (:c1, :c2, :c3, :c4)";
$stmt = $db->prepare($sql);
$stmt->execute(array(
   ':c1'=>$data_t1[0],
   ':c2'=>$data_t2[0],
   ':c3'=>$data_t3[0],
   ':c4'=>$data_t4[0],
));

//this is for row1:
+--------------------+
| 12  | 4 | 6 | 1.33 |
+--------------------+   

+++++++++++++++++++++++++++++++++++++++++ 

//and for row2
$stmt->execute(array(
   ':c1'=>$data_t1[1],
   ':c2'=>$data_t2[1],
   ':c3'=>$data_t3[1],
   ':c4'=>$data_t4[1],
));

//this is for row2:
+----------------------+
| 12  |  3  | 5 | 1.44 |
+----------------------+

注意：我没有对此代码进行测试运行，因此您可能会遇到一些错误。

编辑

使用 Barmar 的想法只需添加额外的列c5, c6, c7, c8并将它们绑定到新值$data_t11[$i], $data_t22[$i], $data_t33[$i], $data_t44[$i]

$sql = "INSERT INTO table (c1, c2, c3, c4, c5, c6, c7, c8) VALUES (:c1, :c2, :c3, :c4, :c5, :c6, :c7, :c8)";
$stmt = $db->prepare($sql);
foreach ($data_t1 as $i => $value) {
    $stmt->execute(array(
       ':c1'=>$data_t1[$i],
       ':c2'=>$data_t2[$i],
       ':c3'=>$data_t3[$i],
       ':c4'=>$data_t4[$i],
       ':c5'=>$data_t11[$i],
       ':c6'=>$data_t22[$i],
       ':c7'=>$data_t33[$i],
       ':c8'=>$data_t44[$i],
    ));
}
于 2013-05-26T06:14:45.070 回答
1

这就像 Tifa 的答案，但使用for循环，因此它可以处理任意数量的值。

$db = new PDO("mysql:host=$dbhost;dbname=$dbname",$dbuser,$dbpass);

//query
$sql = "INSERT INTO table (c1, c2, c3, c4) VALUES (:c1, :c2, :c3, :c4)";
$stmt = $db->prepare($sql);
foreach ($data_t1 as $i => $value) {
    $stmt->execute(array(
       ':c1'=>$data_t1[$i],
       ':c2'=>$data_t2[$i],
       ':c3'=>$data_t3[$i],
       ':c4'=>$data_t4[$i],
    ));
}
于 2013-05-26T06:20:11.883 回答