如何避免JQuery衝突？

Question

對於我的ajax操作時，ajax正常工作表功能不加載。

我用我的表操作：

<script type="text/javascript" language="javascript" 
     src="Dia_tables/media/js/jquery.js"></script> 
<script type="text/javascript" language="javascript" 
     src="Dia_tables/examples_support/jquery.jeditable.js"></script> 
<script type="text/javascript" language="javascript" 
     src="Dia_tables/media/js/jquery.dataTables.js"></script> 
<script type="text/javascript" charset="utf-8"> 

的Javascript：

<script src=http://code.jquery.com/jquery-latest.min.js></script> 
<script type="javascript"> 
    jQuery17 = jQuery.noConfict(true); 
</script> 

<script> 
$(document).ready(function() { 
    $('#submit').click(function (event) { 
     var username = $('#user').val(); 
     var userk = $('#phone').val(); 
     $.get('test', { 
      user: username, 
      phone: userk 
     }, function (responseText) { 
      $('textarea#welcometext').text(responseText); 
     }); 
    }); 
}); 
</script> 

Answer 1

試試這個

var jq = jQuery.noConflict(); 
jq(document).ready(function() {       
       jq('#submit').click(function(event) { 
        var username=jq('#user').val(); 
        var userk=jq('#phone').val(); 
       jq.get('test',{user:username,phone:userk},function(responseText) { 
         jq('textarea#welcometext').text(responseText);   
        }); 
       }); 
      });