不能下标类型为[String：AnyObject]的值？索引类型为String

Question

我知道有很多相同的问题，但仍然找不到修复错误的方法。请参阅图片了解更多详情。我使用了Xcode 7和swift 2.0

编辑：搞砸Swift的警告。 finnaly <?php include "config.php"; extract($_POST); $file_exts = array("jpg", "bmp", "jpeg", "gif", "png"); $upload_exts = end(explode(".", $_FILES["img"]["name"])); if ((($_FILES["img"]["type"] == "img/gif") || ($_FILES["img"]["type"] == "img/jpeg") || ($_FILES["img"]["type"] == "img/png") || ($_FILES["img"]["type"] == "img/pjpeg")) && ($_FILES["img"]["size"] < 2000000) && in_array($upload_exts, $file_exts)) { if ($_FILES["img"]["error"] > 0) { header("Location: Add_Decoration_Item_form.php?err=imgpro1"); //echo "Return Code: " . $_FILES["img"]["error"] . "<br>"; } else { $tempupname = time().$_FILES["img"]["name"]; $imgpathtostore="imgs/".$tempupname; $imgpathtostoreDB="imgs/".$tempupname; // Enter your path to upload file here move_uploaded_file($_FILES["img"]["tmp_name"], $imgpathtostore); $sql= "INSERT INTO user_regi.u_reg (u_id, u_name, u_gender, u_mail, u_pwd, u_dob,u_hobbies,u_image,u_about) VALUES (NULL, '$uname', '$gender', '$email', '$pwd', '$DOB',$team, $img,'$about')"; mysqli_query($con,$sql); if(mysqli_errno($con)) { echo mysqli_errno(); /*header("Location: Form.php?msg=err");*/ } else { header("Location: Form.php?msg=ok"); } } } else { header("Location: Form.php?err=imgpro"); } ?> 修正了错误

Answer 1

change是可选的。打开可选的

let isCaptureStillImage = change![NSKeyValueChangeNewKey]!.boolValue

或使用可选绑定

if let changeNewKey = change?[NSKeyValueChangeNewKey] {
  let isCaptureStillImage = changeNewKey.boolValue

...