Question

这是我的测试查询

PFQuery *lookingForSomething = [PFQuery queryWithClassName:@"AprilMessage"];
[lookingForSomething whereKey:@"user" equalTo:[PFUser currentUser]];
[lookingForSomething findObjectsInBackgroundWithBlock:^(NSArray *objects, NSError *error) {

    // nothing here

 for (PFObject *objectFound in objects) {

       NSLog(@" this is my objectFound", objectFound);

到目前为止一切顺利，我的NSLog（objectFound）的结果是：

{
message = hello world ! ;
reciever = "<PFUser:WNOmhmrO3L>";
user = "<PFUser:WNOmhmrO3L>";

我要做的是获取消息字符串“hello world！”并在屏幕上显示，在这里我做了什么，但由于某种原因它不起作用

 // same code from before just adding the last lines down there

PFQuery *lookingForSomething = [PFQuery queryWithClassName:@"AprilMessage"];
[lookingForSomething whereKey:@"user" equalTo:[PFUser currentUser]];
[lookingForSomething findObjectsInBackgroundWithBlock:^(NSArray *objects, NSError *error) {

// nothing here

   for (PFObject *objectFound in objects) {

   NSLog(@" this is my objectFound", objectFound);

   // adding this new line

 objectFound[@"message"] = _textChat.text;  <-- textChat is a UITextView !!!

任何想法我哪里出错了？

Answer 1

你的最后一行是倒退。它应该是，

_textChat.text = objectFound[@"message"];

这应该可以假设_textChat和objectFound [@“message”]都不是nil。

parse.com |如何在UITextView上显示结果？

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