《PHP实例:PHP中ajax无刷新上传图片与图片下载功能》要点:
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PHP ajax无刷新上传图片与图片下载功能的实现代码如下所示:PHP实战
<Meta charset="utf-8" > <form id= "uploadForm"> <p >指定文件名: <input type="text" name="filename" value= ""/></p > <p> 上传文件: <input type="file" name="photo" onchange="showPreview(this)" class="file" /> <img id="portrait" src="" width="70" height="75"> </p> <input type="button" value="上传" onclick="doUpload()" /> </form> <script src="http://www.haoyunyun.cn/jquery.js"></script> <script> function doUpload() { var formData = new FormData($( "#uploadForm" )[0]); $.ajax({ url: 'submit.PHP',type: 'POST',data: formData,async: false,cache: false,contentType: false,processData: false,success: function (returndata) { alert(returndata); },error: function (returndata) { alert(returndata); } }); } </script> <script type="text/javascript"> function showPreview(source) { var file = source.files[0]; if (window.FileReader) { var fr = new FileReader(); fr.onloadend = function(e) { document.getElementById("portrait").src = e.target.result; }; fr.readAsDataURL(file); } } </script>
<?PHP if($_FILES['photo']['error']>0){ echo "上传文件失败"; die; } $dir='./photo/'; $type=substr($_FILES['photo']['name'],strrpos($_FILES['photo']['name'],'.')); $filename=time().rand(1000,9999).$type; if(is_uploaded_file($_FILES['photo']['tmp_name'])){ move_uploaded_file($_FILES['photo']['tmp_name'],$dir.$filename); echo "上传成功"; }else{ echo "上传文件失败"; }