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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 = L Код:
<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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