$.ajax({ type: "POST",url: "post.PHP",data: dataString,success: function(data) { console.log('OK: ' + data); },error: function(jqXHR,textStatus,errorThrown) { console.log('Error: ' + errorThrown + ' ' + textStatus + ' ' + jqXHR); } });借助于 console.log()函数 原文链接:https://www.f2er.com/ajax/166046.html