【发布时间】:2017-06-30 14:52:29
【问题描述】:
我正在使用帮助 ajax 上传图像并显示它。现在我想要实现的是,一旦用户选择图像,它应该在 div 中可见,但是当我第一次尝试上传图像时当数据库为空时,它什么也没显示 ajax 正在运行 数据正在保存到数据库中,但我看不到图像 你能告诉我我在做什么错吗?
<div id="timelineBackground" class="">
<?php
{
if(!$timeline_image=="")
{
$image_properties = array('src' => base_url("uploads/".$timeline_image),'width' => '900px','height'=> '400px','id'=>'coverimg', 'title' => 'That was quite a night','rel' => 'lightbox');
echo img($image_properties);
}else{
$timeline_image = array(
'src' => base_url("assets/img/Facebook-cover.jpg"),
'width' => '900px',
'height'=> '400px',
);
echo img($timeline_image);
}
?>
<div id="timelineselector" class="grow">
<?php echo form_open_multipart('',["id"=>"form_cover"]); ?>
<input type="hidden" name="id" value="<?php echo $id ;?>">
<label class="icons-camera " for="timelineimage">
<p class="cover_upload"> <i class="fa fa-camera cover_upload" style="color:white;"> Upload Your Cover</p></i>
</label>
<?php echo form_upload(["name"=>"timelineimage","id"=>"timelineimage" ]); ?>
<?php echo form_close(); ?>
</div>
<?php
}
?>
</div>
jQuery 代码
jQuery('#timelineimage').on('change',function(e) {
e.preventDefault();
var form=jQuery('#form_cover')[0];
var formData = new FormData(form);
var url = '<?php echo base_url("user/coverimage"); ?>';
formData.value
jQuery.ajax({
type: "POST",
url: url,
data: formData,
cache: false,
contentType: false,
processData: false,
success: function(data) {
console.log(data);
$('#coverimg').attr('src', data);
},
error: function(data) {
//error function
}
});
});
这是我的代码,当第一次数据库为空时我该怎么办?
这是我的控制器
public function coverimage()
{
$config = [
'upload_path' => './uploads/',
'allowed_types' => 'jpg|gif|png|jpeg',
'max_size' => 10000000000000,
'max_width' => 1024000000,
'max_height' => 7680000000,
];
$this->load->library('upload', $config);
$this->upload->initialize($config);
$timelineimage="timelineimage";
if(!$this->upload->do_upload($timelineimage))
{
$upload_error = $this->upload->display_errors();
$this->load->view('dashboard/profile',compact('upload_error'));
}
else
{
$post = $this->input->post();
unset($post['submit']);
$upload_data = $this->upload->data();
$file_name=$_FILES['timelineimage'];
$this->load->model('Pmodel');
$this->Pmodel->timeline_upload_model($post,$file_name);
echo $image_path= base_url("uploads/".$upload_data['raw_name'].$upload_data['file_ext']);
}
}
型号代码
public function timeline_upload_model($arr,$arra)
{
$id=$arr['id'];
$image=$arra['name'];
$data=array('timelineimage'=>$image,'user_id'=>$id);
$query=$this->db->where('id',$id)->update('user_data',$data);
//print_r($query);
return $query;
}
现在让我告诉你问题是什么
[![First Image][1]][1]
[![Second Image][2]][2]
[![Third Image][3]][3]
[1]: https://i.stack.imgur.com/C2Rvu.png
[2]: https://i.stack.imgur.com/4fYXJ.png
[3]: https://i.stack.imgur.com/DG4NH.png
【问题讨论】:
-
在 ajax 成功中获得了什么数据
-
这样的图片的url "localhost/P_Display/uploads/1-logo2.png" @shafiq
-
你在使用任何框架吗? $timeline_image 是第一次为空
-
codeigniter @shafiq
-
第一次 $timeline_image 的输出是什么?你能分享你的控制器代码吗?
标签: php jquery html ajax codeigniter