css - Jquery .click and .mouseover not working on css3 display:table ,row, cell worked on html table? -

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I had a table that was generated from standard HTML and JQuery was used. MCOOVER event listener as well as .click listener Everything was working fine. I have taken steps to collect data from the Facebook API and then used divs and css, making notifications on the page. The breakdown of the code is as follows.

This is jquery function 

  $ (document). Mouseover (function (.ready (function () {$ ( '.fbLikeImage')) {$ likeID = this .id; document.getElementById ( 'uiFBULikeName') innerHTML = this.alt ;. document.getElementById ( 'uiFBLikeImage') innerHTML = ". & lt; img width = '70 'height =' 70 'class =' ​​fbUserImage 'src = "+ this.src +" & gt; ";}); $ ( '. fbLikeImage') function (click (.) {showDiv ( 'uiCommentTextInputWrapper'); showDiv ( 'likeButtonWrapper');});   
});  
 Then the HTML presented here  
    
  
 There are several more rows, but this is the basic structure.  
 Here is the code from my CSS. Table {display: Table; }  
  .tableRow {display: table-row; } .tableCell {display: table-cell; Vertical-align: top; }   
 Before I changed it into CSS based table layout was working fine, if any idea is great, then I can not solve it.  
 Thanks Neil   
 
 
  $ (function () {$ ( ' .fbLikeImage '). Mouseover (function () $ (' # UiFBLikeImage ') .HTML ("and lt; img width = '70' height = $ (this) .ID; $ ('# uiFBULikeName'). Html ($ (This) .alt); $ ('# uiFBLikeImage' '70' class = 'fbUserImage' src = '+ $ (this) .src +' & gt;)}); $ $ ('.fbLikeImage') Click on (function () {$ ('div uiCommentTextInputWrapper') show (); $ ('Use the button like Div'). Show ();});}); I will suggest such a thing, but I'm not sure That click function works like this .. and try it by passing the class or id T

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