我在表格视图中的输出未正确显示输出

Question

Even after giving constrain, My output is not as per my expectation, The labels are getting overlapped and my image is shifted to left.

My code snippet -

- (void)viewDidLoad {
    [super viewDidLoad];
  //  [self.emplyoyeeTableView registerNib:[UINib nibWithNibName:@"SimpleTableCell" bundle:nil] forCellReuseIdentifier:@"SimpleTableCell"];

    NSManagedObjectContext *manageobjectcontext = [self manageobjectcontext];
    NSFetchRequest *fetchrequest = [[NSFetchRequest alloc]initWithEntityName:@"Employee"];
    self.employeeArray = [[manageobjectcontext executeFetchRequest:fetchrequest error:nil]mutableCopy];
    [self.emplyoyeeTableView reloadData];
}

#pragma mark - <UITableViewDataSource>
-(NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section{
    return self.employeeArray.count;

}

-(NSInteger)numberOfSectionsInTableView:(UITableView *)tableView{
    return 1;
}

- (CGFloat)tableView:(UITableView *)tableView heightForRowAtIndexPath:(NSIndexPath *)indexPath{
    return 110.0f;
}

-(UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{

    SimpleTableCell *cell = [tableView dequeueReusableCellWithIdentifier:@"SimpleTableCell"];
    NSManagedObject *emp = [self.employeeArray objectAtIndex:indexPath.row];
    [cell.nameLabel setText:[NSString stringWithFormat:@"%@",[emp valueForKey:@"name"]]];
    [cell.emailLabel setText:[NSString stringWithFormat:@"%@",[emp valueForKey:@"email"]]];
    [cell.designationLabel setText:[NSString stringWithFormat:@"%@",[emp valueForKey:@"designation"]]];

    [cell.picImageView setImage:[UIImage imageNamed:@"skinage 2.png"]];

    return cell;
}

the output is like this-

Answer 1

我认为您需要像这样在viewController中设置tableview数据源。

tableView.dataSource = self

Answer 2

The problem is, you got an instance of UITableViewCell instead of SimpleTableCell when dequeuing the cell.

Just a guess. Try Replace this snippet:

[self.emplyoyeeTableView registerNib:[UINib nibWithNibName:@"SimpleTableCell" bundle:nil] forCellReuseIdentifier:@"SimpleTableCell"];

SimpleTableCell *cell = (SimpleTableCell *) [tableView dequeueReusableCellWithIdentifier:@"SimpleTableCell"];

if (cell == nil) {
    NSArray *nib = [[NSBundle mainBundle] loadNibNamed:@"SimpleTableCell" owner:self options:nil];
    cell = [nib objectAtIndex:0];
}

with:

SimpleTableCell *cell = [tableView dequeueReusableCellWithIdentifier:@"SimpleTableCell"];

When using storyboard designs, the prototyping cell can be accessible without register.

Also, if above not working, try restarting XCode. Sometimes i got mess with XCode file indexing.

Answer 3

You should remove this line:

 [self.emplyoyeeTableView registerNib:[UINib nibWithNibName:@"SimpleTableCell" bundle:nil] forCellReuseIdentifier:@"SimpleTableCell"];

Because you create the SimpleTableCell inside your viewController already, so there is no need to registerNib.

Answer 4

Try This Code:-

- (void)viewDidLoad {
    [super viewDidLoad];

[self.emplyoyeeTableView registerNib:[UINib nibWithNibName:@"SimpleTableCell" bundle:nil] forCellReuseIdentifier:@"SimpleTableCell"];
}
-(UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
    // static NSString *simpleTableIdentifier = @"SimpleTableCell";



   SimpleTableCell *cell = (SimpleTableCell *) [tableView dequeueReusableCellWithIdentifier:@"SimpleTableCell"];


    if (cell == nil)
    {
        NSArray *nib = [[NSBundle mainBundle] loadNibNamed:@"SimpleTableCell" owner:self options:nil];
        cell = [nib objectAtIndex:0];
    }
  cell.clipsToBounds=YES;

    NSManagedObject *emp = [self.employeeArray objectAtIndex:indexPath.row];
    [cell.nameLabel setText:[NSString stringWithFormat:@"%@",[emp valueForKey:@"name"]]];
    [cell.emailLabel setText:[NSString stringWithFormat:@"%@",[emp valueForKey:@"email"]]];
    [cell.designationLabel setText:[NSString stringWithFormat:@"%@",[emp valueForKey:@"designation"]]];

    [cell.picImageView setImage:[UIImage imageNamed:@"skinage 2.png"]];

    return cell;
}