连接引用jQuery中的php和javascript函数

时间:2013-03-22 18:17:55

标签: php jquery ajax

$.ajax({
    type: "POST",
    url: <?php echo base_url().'index.php/user_management/manage_users/ViewProfile/'?>"+JSON.parse(item.UserID),
    success: function(output_string){
        $('.second_column_content_container').html(output_string);
    }
});

我希望将url的结果括在单引号或双引号中; 例如

'url: <?php echo base_url().'index.php/user_management/manage_users/ViewProfile/'?>"+JSON.parse(item.UserID)' or 
"url: <?php echo base_url().'index.php/user_management/manage_users/ViewProfile/'?>"+JSON.parse(item.UserID)"

2 个答案:

答案 0 :(得分:0)

试试这个

url: "<?php echo base_url().'index.php/user_management/manage_users/ViewProfile/'?>"+JSON.parse(item.UserID),
//---^^--here

url: "<?php echo base_url() ?>"+'index.php/user_management/manage_users/ViewProfile/'+JSON.parse(item.UserID),

答案 1 :(得分:0)

试试这个:

$.ajax({
    type: "POST",
    url: '<?php echo base_url(); ?>'+'index.php/user_management/manage_users/ViewProfile/'+JSON.parse(item.UserID),
    success: function(output_string){
       $('.second_column_content_container').html(output_string);
    }
});

而不是php连接.使用javascript concatenation +