php form表单ajax上传图片方法

form表单ajax上传图片方法

先引用jquery.form.js

前台代码
<pre>
<form id="form1">
<input id="file_temporaryImage" type="file" name="file_temporaryImage" onchange="TemporaryMedia();"/>
<input type="hidden" id="userpicpath">
<img id="userpic" src=""/>
</form>
<script>
function TemporaryMedia() {
//var image0 = $("input[name='file_temporaryImage']").val();
//判断上传控件中是否选择了图片
var image = $("#file_temporaryImage").val();
if ($.trim(image) == "") {
alert("请选择图片!");
return;
}
//提交请求处理的url
var actionUrl = "/home/kelatoupiao/sctpic/";
//开始ajax操作
$("#form1").ajaxSubmit({
type: "POST",
dataType: "json",
url: actionUrl,
data: {},
success: function (data) {
if (data.success == 1) {
$('#userpicpath').val(data.data);
$('#userpic').attr('src',data.data);
} else {
alert(data.msg);
}


}


});


}
</script>
</pre>


PHP后台代码
<pre>
//上传图片
public function sctpic()
{
$openid = $this->_check_login();
$spath = __DIR__ . '/../../../Public/kelatoupiao/uploads/' . $openid . '_' . time() . '.jpg';
$webpath='/kelatoupiao/uploads/' . $openid . '_' . time() . '.jpg';
if (move_uploaded_file($_FILES["file_temporaryImage"]["tmp_name"], $spath)) {
echo json_encode(array('success' => 1, 'msg'=>'上传成功','data'=>$webpath));
exit();
} else {
echo json_encode(array('success' => 0, 'msg'=>'网络繁忙','data'=>''));
exit();
}
}
</pre>

posted @ 2019-11-16 17:38  newmiracle宇宙  阅读(548)  评论(0编辑  收藏  举报