Question

我正在将多个图像上传到我的sql数据库，但是我的爆破方法不起作用...

Private Sub Worksheet_Change(ByVal Target As Range)
    If Not Intersect(Target, Range("A2", Range("A" & Rows.Count).End(xlUp))) 
        Is Nothing Then
            Application.EnableEvents=False
            If Target="Closed" Then
                Target.EntireRow.Copy Sheet2.Range("A1").End(xlDown)(2)
                Target.EntireRow.Delete
            End If
        End If
        Application.EnableEvents=True
End Sub

Answer 1

$filename应该是一个数组。您正在将字符串传递给内爆函数。做这样的事情。

 if(isset($_POST['submit'])) {
    $id     = $_POST['id'];
    $filename= [];
    for($i = 0; $i < count($_FILES['file_upload']['name']); $i++){
      $filetmp  = $_FILES['file_upload']['tmp_name'][$i];
      $filename[] = basename($_FILES['file_upload']['name'][$i]);
      $filetype = $_FILES['file_upload']['type'][$i];
      $filepath = "../images/".$filename;
      move_uploaded_file($filetmp,$filepath);
    }
      $mainfiles = implode(", ", $filename);

      $sql  = "INSERT INTO pictures2 (";
      $sql .= "image, photograph_id";
      $sql .= ") VALUES ('";
      $sql .= $database->escape_character($mainfiles) ."', '";  
      $sql .= $id ."')";
      $result = $database->query($sql);
        if($result){
        $session->message('<div class="success-msg">Pictures uploaded sucessfully.</div>');
        }

}

希望这会有所帮助。

Answer 2

进行了这样的更改，我正在考虑$id是您的photograph_id

if (isset($_POST['submit'])) {
    $id = $_POST['id'];
    for ($i = 0; $i < count($_FILES['file_upload']['name']); $i++) {
        $filetmp    = $_FILES['file_upload']['tmp_name'][$i];
        $filename[] = [$id, basename($_FILES['file_upload']['name'][$i])]; // here changes
        $filetype   = $_FILES['file_upload']['type'][$i];
        $filepath   = "../images/" . $filename;
        move_uploaded_file($filetmp, $filepath);
    }
    // I made changes at below code
    $valueStr = array_map(function ($item) {
        return "('" . implode("','", $item) . "'),"; // to convert into `,` separated values
    }, $filename);
    $mainfiles = rtrim(implode("", $valueStr), ','); // removing right training `,`

    $sql = "INSERT INTO pictures2 (";
    $sql .= "image, photograph_id";
    $sql .= ") VALUES ('";
    $sql .= $database->escape_character($mainfiles) . "', '";
    $sql .= $id . "')";
    $result = $database->query($sql);
    if ($result) {
        $session->message('<div class="success-msg">Pictures uploaded sucessfully.</div>');
    }

}

如何将图像文件名内推到mysql数据库中

2 个答案: