檢查沒有複選框調用DestroyeAll功能

Question

如果沒有選擇複選框警報前應檢查顯示沒有記錄中選擇如果複選框選中，那麼就應該到控制器，否則只是警惕複選框前刪除 這裏是我的方法

function delete() { 
    if(confirm('Are you SURE!')) { 
     var checkedValues = $('input:checkbox:checked').map(function() { 
      return this.value; 
     }).get(); 
     $.ajax({ 
      type : "POST", 
      url  : "{{ url('/school/destroyAll') }}", 
      data : {ids: checkedValues, _token: "{{ csrf_token() }}"}, 
      success: function(result) { 
       if(result == 'delete') { 
        window.location.href = "/school"; 
       } 
      } 
     }); 
    } 
} 

只是我想在這裏添加新的東西，如果沒有複選框sleeted應該沒有顯示警報應該去URL

Answer 1

試試這個：

function delete() { 
     if(!$('input:checkbox:checked').length){ 
      alert("No checkbox has selected"); 
      return false; 
     } 
     if(confirm('Are you SURE!')) { 
      var checkedValues = $('input:checkbox:checked').map(function() { 
      return this.value; 
      }).get(); 
      $.ajax({ 
       type : "POST", 
       url  : "{{ url('/school/destroyAll') }}", 
       data : {ids: checkedValues, _token: "{{ csrf_token() }}"}, 
       success: function(result) { 
           if(result == 'delete') { 
         window.location.href = "/school"; 
        } 
       } 
     }); 
    } 
} 

Answer 2

你可以試試這個

function delete(){ 

if($('input[type=checkbox]:checked').length != 0){ 
      if(confirm('Are you SURE!')) { 
     var checkedValues = $('input[type=checkbox]:checked').map(function() { 
      return this.value; 
     }).get(); 
     $.ajax({ 
     type : "POST", 
     url  : "{{ url('/school/destroyAll') }}", 
     data : {ids: checkedValues, _token: "{{ csrf_token() }}"}, 
     success: function(result) { 
         if(result == 'delete') { 
          window.location.href = "/school"; 
         } 
        } 
}); 
}else{ 
alert('no checkbox checked') 
} 

}