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 How to get grab content from certain id+column? 
		
		
		Kak mozhno vziat content so stranizi i assing it to a some value to output it into template?  
	Est naprimer straniza s id 223. Nado vziat s nee colonky 0(normal), s language id 2, i assing it to temp.langbar please help:) or suggest syntax/keywords to explore  | 
		
 Chto zdes ne tak? :) Ya prosto pitaius vziat content s id 223, default language, i column 0. tam stoit plugin na languages. potom pipe it to template. 
	#typoscrpt setup temp.langMenu = CONTENT temp.langMenu { select { pidInList = 233 # uid of the page languageField=sys_language_uid where=colPos=0 } table = tt_content } A vot polnostiu ves moi setup - Код: 
	config.linkVars = L | 
		
 Код: 
	temp.langMenu = RECORDS | 
		
 Ne rabotaet:( 
	Mozhet ya ne tak ego vstavliaiu v setup(langmenu) Код: 
	config.linkVars = LКод: 
	<div id="langMenu"> | 
		
 In case anyone will stumble upon the same problem here is the solution 
	temp.whatever = CONTENT temp.whatever { table = tt_content select { pidInList = HERE_GOES_PID_OF_PAGE orderBy = sorting andWhere = colPos=0 } }  | 
		
 <div id="langMenu"> 
	<!--###langMenu### start--> Some text <!--###langMenu### end--> </div> Примерно так должен выгядеть кусок кода, куда вы хотите вставить ваше языковое меню.  | 
		
 A zachem <!--###langMenu### start-->:confused: ? 
	U menia prosto s <div id="langMenu"> rabotaet. Tolko kogda ne rabotet v outpute vidaet <!--###langMenu### start--> :confused: Ya videl na drugih saitah tozhe pishut <!--###someitme### start-->.. Tak vkakih slychiah nado isplozivat <!--###someitme### start-->?  | 
		
 Цитата: 
	
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 Since I am new to typo3 snachalo nado  razobratsa s basics, a potom migrate to WYSWIG if needed:) 
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