use*_*579 3 cocoa-touch objective-c uitableview ios unrecognized-selector
当我从谷歌寻找答案时,大多数答案都说你试图在NSArray上使用不受支持的长度.
这里的问题是我甚至没有在我的代码中使用任何NSArray或长度.
我有一个
NSMutableArray *filteredContent;
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哪里filteredContent将包含来自plist的字典.
一切都很好,直到写cell.textLabel.text在牢房上tableView.
选中NSLog并且内容确实是一个数组.
这是我尝试编写单元格文本的方法:
cell.textLabel.text=[[self.filteredContent objectAtIndex:indexPath.row] valueForKey:@"recipeName"];
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但它给了我错误所以我改为:
NSString *myValue = [self.filteredContent valueForKey:@"recipeName"];
cell.textLabel.text=myValue;
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但结果是一样的.我不知道我得到了什么错误.
更多详情:
[results addObject:@[recipe]];
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这是我创建主数组的地方,而不是将其传递给PageViewwith seguefilteredContent = results
编辑:
recipe = arrayOfPlist[i]; 其中arrayOfPlist
NSArray *arrayOfPlist = [[NSArray alloc] initWithContentsOfFile:path];
//Output to NSLog(@"FIltered Content : %@", self.filteredContent);
FIltered Content : (
(
{
category = main;
difficultyLevel = "3/5";
numberOfPerson = 5;
recipeDetail = "Bulguru koy su koy beklet pisir ye ";
recipeImage = "nohutlu-pilav.jpg";
recipeIngredients = "pirinc,tereyag tuz,bulgur";
recipeName = "Bulgurlu Pilav";
time = "25 dk";
}
),
(
{
category = main;
difficultyLevel = "3/5";
numberOfPerson = 5;
recipeDetail = "Bulguru koy su koy beklet pisir ye ";
recipeImage = "nohutlu-pilav.jpg";
recipeIngredients = "pirinc,tereyag tuz,bulgur";
recipeName = "Bulgurlu Pilav";
time = "25 dk";
}
)
)
2014-04-27 02:47:41.704 deneme2[19820:60b] VALUE IN FILTERED TABLE is (
"Bulgurlu Pilav" // this is what i want to write to the cell label and i get it with myValue-look a bit above
)
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根据您的数据输出,您有一个额外的数组.所以你想要这个:
cell.textLabel.text = self.filteredContent[indexPath.row][0][@"recipeName"];
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filteredContentArray的每个元素都是另一个数组.每个内部数组都有包含所需数据的字典.
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