jQuery load（）后提交不起作用

Question

我有这个脚本：

$('form').each(function validate() {  // attach to all form elements on page
  $(this).bootstrapValidator({
    // initialize plugin on each form
    message: 'This value is not valid',
    feedbackIcons: {
      valid: 'glyphicon glyphicon-ok',
      invalid: 'glyphicon glyphicon-remove',
      validating: ''
    },
    live: 'enabled',
    trigger: null,

    fields: {
      ServiceCategory: {
        validators: {
          notEmpty: {message: "<?php echo $Missing_Service_Category ?>"},
          regexp: {regexp: /^[A-Za-z]/, message: "<?php echo $Wrong_Service_Category ?>"},
        }
      },
    }
  })
  .on('success.form.bv', function(e, data) {
    // Prevent form submission
    e.preventDefault();
    // Get the form instance
    var $form = $(e.target);
    // Get the BootstrapValidator instance
    var bv = $form.data('bootstrapValidator');
    // Use Ajax to submit form data

    // Use Ajax to submit form data
    $.ajax({
      url: $form.attr('action'),
      type: 'POST',
      contentType: "application/x-www-form-urlencoded;charset=utf-8",
      data: $form.serialize(),
      success: function(data){
        var parts = data.split('|')
        var type = parts[0]
        var text = parts[1]
        if(type==1){
          // If Success type is 1, show success message with type success
          $('.top-right').notify({
            message: { text: text },
            type: 'info',
          }).show();
          $(".tab-content").load(" .tab-content > *", $(document).ready(function() {
            $.ajax({
              url: 'scripts.php',
              success: function(data) {
                $("#scripts").html(data);
              }
            });
          }));
        }
        else if(type==2){
          // If Success type is 1, show success message with type success
          $('.top-right').notify({
          message: { text: text },
          type: 'danger',
          }).show();
        }
        else if(type==3){
          // If Success type is 3, print messaage to #ResponseDIV
          $('#ResponseDiv').html(text);
        }else{
          // If Success type is 3, print messaage to #ResponseDIV
          $('#ResponseDiv').html(text);
        }
      },
      error: function(xhr){
        // If Success type is 1, show success message with type success
        $('.top-right').notify({
          message: { text: xhr.status + " " + xhr.statusText  },
          type: 'danger',
        }).show();
      },
    });
  })
});

完成此操作后，将使用实际数据重新加载div。

在div内有一个表格，但之后有

$(".tab-content").load(" .tab-content > *"，

表单提交不再起作用。当我刷新整个页面时，一切正常。

有什么办法可以使div加载（）并使表单上的提交按钮再次起作用吗？