JS实现浏览上传文件的代码
作者:MoleKing
本文通过实例代码给大家介绍了基于js实现浏览上传文件的代码,非常不错,代码简单易懂,具有参考借鉴价值,需要的朋友参考下吧
废话不多说了,直接给大家贴代码了,具体代码如下所示:
<div style="position:relative;width:380px;"> <iframe id="t_load" name="t_load" style="display: none"></iframe> <form action="" method="post" enctype="multipart/form-data" name="form1" target="t_load"> <input type='text' name='textfield' id='textfield' style="height:22px; border:2px solid #cdcdcd; width:230px; border-radius:50px;" /> <input type='button' value='浏览...' style="display: inline-block; margin-bottom: 0; font-size: 14px; border-radius: 4px; padding: 6px 12px; margin: 2px 2px; border: 1px solid #357ebd; background: #1b9ad5; color: #fff;" /> <input type="file" name="file1" accept="image/jpeg, image/gif" id="file1" onchange="document.getElementById('textfield').value=this.value" style="position:absolute; top:5px; right:74px; height:27px; filter:alpha(opacity:0);opacity: 0;width:300px" /> <input type="button" name="button" onclick="mysubmit()" value="上传" style="display: inline-block; margin-bottom: 0; font-size: 14px; border-radius: 4px; padding: 6px 12px; margin: 2px 2px; border: 1px solid #357ebd; background: #1b9ad5; color: #fff;"> </form> </div> <div style="display:none;color:#6D6D72;font-size: 20px;" id="res"></div> <script> function mysubmit() { var a = document.getElementById("textfield").value; if(a == "") { alert("请上传文件"); return; } document.form1.submit(); ajaxsend(); } function ajaxsend() { var xmlHttp; if(window.ActiveXObject) { xmlHttp = new ActiveXObject("Msxml2.XMLHTTP"); } else { xmlHttp = new XMLHttpRequest(); } xmlHttp.open("post", "upload", true); xmlHttp.onreadystatechange = function() { xmlHttp.readyState; if(xmlHttp.readyState == 4) { res.innerHTML = "上传成功!"; setTimeout(function() { res.innerHTML = ""; }, 2000); } else { document.getElementById("res").style.display = "inline"; } } xmlHttp.send(); var obj = document.getElementById('textfield'); obj.outerHTML = obj.outerHTML; } </script>
总结
以上所述是小编给大家介绍的JS实现浏览上传文件的代码,希望对大家有所帮助,如果大家有任何疑问请给我留言,小编会及时回复大家的。在此也非常感谢大家对脚本之家网站的支持!