javascript - jquery - click, append, load, slideDown not displaying as expected -

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I'm trying to apply, I thought a simple click , Load , slidedown landscape but I can not get to show the slidedown part

I have the following two buttons: 

  & lt; Div & gt; & Lt; Fieldet id = "btn" & gt; & Lt; Input class = "databasebtn" type = "submit" name = "nameDatabaseBtn" id = "db1" Data-id = 1 "VALUE =" DB1 "/>  "Databasebtn" type = "submit" name = "nameDatabaseBtn" id = "db2" Data-id = "2" VALUE = "DB2" /> gt;  & lt; / fieldset & gt;

Then I have the following jQuery: 

  $ (document) .ready (function () {$ (' ('Click', function ($ {'(BTANLIST) is the idea that I click on the button, I attach it to  #btnlist   < / Pre> 
 / Code> button, and fill the new device with the contents of test7b.php A checkbox must be prepared. 
 
 All this works well, except that I can not see the checkbox if I see the code 
 
 If I use 'test78b.php' Separately, it is displayed as expected. 
 
 Am I missing something

You can not include a button in the div, in this case you can add div to the parent in the case of filtration.

Or you can use the insert before adding a code before you click with this code 

  & lt; Script type = "text / javascript" & gt; $ (Document) .ready (function () {$ ('databasebtn'). ('Click', function () {$ ("id id = 'btnlist' & gt; & lt; / div & gt;) Insert ($ (this)) $ ('# btnlist') Slide down ('slow', function () {$ ('# btnlist'). Load ("your page");})})}); & Lt; / Script & gt;

or attach the body tag to this body code with

And then, for a correct HTML code, you should not have multiple items with the same ID on the same page.
The Division ID to be added through the script should not be btnlist, but class = "btnlist"


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