Question

php获取input text输入的值，为上传文件改名，请直接给出怎么写

html代码：
号码：<input type="text" id="title" name='title'/>
<div style="width:100%;margin:10px auto; border:solid 1px #ddd; overflow:hidden; ">
<input type="file" id="uploadphoto" name="litpic" value="请点击上传图片" onchange="imgPreview(this)" style="display:none;" />
<div class="imglist"></div>
<a href="javascript:void(0);" onclick="uploadphoto.click()" class="uploadbtn">点击上传文件</a>
</div>
<script type="text/javascript">
$(document).ready(function(e) {
$('#uploadphoto').localResizeIMG({
//width: 400,
quality: 1,
success: function (result) {
var submitData={
base64_string:result.clearBase64,
};
$.ajax({
type: "POST",
url: "/member/templets/test/upload.php",
data: submitData,
dataType:"json",
success: function(data){
if (0 == data.status) {
alert(data.content);
return false;
}else{
alert(data.content);
var attstr= '<img src="'+data.url+'">';
$(".imglist").append(attstr);
return false;
}
},
complete :function(XMLHttpRequest, textStatus){
},
error:function(XMLHttpRequest, textStatus, errorThrown){ //上传失败
alert(XMLHttpRequest.status);
alert(XMLHttpRequest.readyState);
alert(textStatus);
}
});
}
});

});
</script>

php代码：
$base64_string = $_POST['base64_string'];

$savename = uniqid().'.jpeg';//localResizeIMG压缩后的图片都是jpeg格式

$savepath = 'images/'.$savename;

$image = base64_to_img( $base64_string, $savepath );

if($image){
echo '{"status":1,"content":"上传成功","url":"'.$image.'"}';
}else{
echo '{"status":0,"content":"上传失败"}';
}

function base64_to_img( $base64_string, $output_file ) {
$ifp = fopen( $output_file, "wb" );
fwrite( $ifp, base64_decode( $base64_string) );
fclose( $ifp );
return( $output_file );

Answer 1

ajax 提交的数据里增加一个参数，submitData.file_name=$("#title").val();
后台获取的时候 $file_name = $_POST['file_name'];

Answer 2

$savename = $_POST['title'].'.jpeg';

追问

不行，获取的是空值

追答

连post参数都获取不到？ 确定你的表单 method 设置了 并且是post