创建具有未知类型的图像格式是错误没有上传图像。迅速

时间:2017-07-02 05:45:42

标签: ios swift xcode firebase image-uploading

我试图让用户上传图片并创建一个帐户。然而,在选择图像之后,用户应该选择“创建帐户”。但是程序停止了,之后什么也没做。然后,此消息显示在编译器中,“创建具有未知类型的图像格式是一个错误,没有上传图像。”

这是代码。

<?php
    $sqlteams = "SELECT * FROM tournaments where status='Active'";
    $sqlquery  = mysqli_query($cnn,$sqlteams);
    while($rsres = mysqli_fetch_array($sqlquery))
    {

(line 47)   if($rsres["tournamentid"]== $_GET["tournamentid"])
        {
        echo "<option value='$rsres[tournamentid]' selected>$rsres[name]</option>";
        }
        else
        {
        echo "<option value='$rsres[tournamentid]'>$rsres[name]</option>";
        }
    }
?>

很抱歉,如果这太多了,但我不确定在哪里确定此错误,因为Xcode没有将其显示为错误。似乎解决方案可能在某个地方here,但我找不到它。提前谢谢。

1 个答案:

答案 0 :(得分:0)

您可以使用availableMediaTypes确保仅显示有效的图片类型。

        let type = kUTTypeImage as String
        let imagePicker = UIImagePickerController()

        if UIImagePickerController.isSourceTypeAvailable(.camera) {                
            if let availableTypes = UIImagePickerController.availableMediaTypes(for: .camera) {                    
                if availableTypes.contains(type) {                        
                    imagePicker.mediaTypes = [type]
                    imagePicker.sourceType = UIImagePickerControllerSourceType.camera
                }
            }
        } else {
            return
        }

        imagePicker.allowsEditing = true
        imagePicker.showsCameraControls = true
        imagePicker.delegate = self

然后您也可以查看UIImagePickerControllerOriginalImage

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]) {

        if let image = info[UIImagePickerControllerOriginalImage] as? UIImage ?? info[UIImagePickerControllerEditedImage] as? UIImage {
            userImagePicker.image = image
            imageSelected = true
        } else {
            print("image wasnt selected")
        }
        imagePicker.dismiss(animated: true, completion: nil)
    }