当我们不更新 Codeigniter 2.2.6 中多图像形式的图像时如何设置或检索旧路径图像
How to Set or retrieve old Path image when we not update the image in the form multiple images in Codeigniter 2.2.6
我在 codeigniter 中以那种形式编辑表单我还有 15 个图像文件,同时我更新它在数据库中更新的图像,但我什至没有更新它去的形式的图像并保存为 null 或空路径。但是我需要解决方案,当我不更新所有 15 个文件时,应该检索存储在 database.Please 中的相同旧图像路径,你们能为此提供更好的解决方案吗?
我的编辑表单
控制器:
if($this->input->post('submit'))
{
if($_FILES['file']['name']!='')
{
$filename = $_FILES['file']['name'];
$file_name1 = "upload_".$this->get_random_name()."_".$filename;
$_SESSION['file_upload']=$file_name1;
$file ="uploads/images/".$file_name1;
}
if($_FILES['file1']['name']!='')
{
$filename = $_FILES['file1']['name'];
$file_name2 = "upload_".$this->get_random_name()."_".$filename;
$_SESSION['file_upload']=$file_name2;
$file ="uploads/images/".$file_name2;
}
$query = $this->scener_model->popular_upload($file_name1,$file_name2)
}
如果你是空的,那么传递隐藏的输入值或者像这样传递新上传的值..
if(!empty($_FILES()){
//your uploaded value here
} else {
//hidden attribute value
}
if($this->input->post('submit1')){
$titlemain = $this->input->post('title_main');
$price = $this->input->post('price');
$package = $this->input->post('package');
$titleinner = $this->input->post('title_inner');
$city_package = $this->input->post('city_packge');
if(!empty($count =count($_FILES['file1']['name']))){;
for($i=0; $i<$count;$i++){
$filename1=$_FILES['file1']['name'][$i];
//print_r($filename);exit;
$filetmp=$_FILES['file1']['tmp_name'][$i];
$filetype=$_FILES['file1']['type'][$i];
$images = $_FILES['file1']['name'];
$filepath="uploads/images/".$filename1;
move_uploaded_file($filetmp,$filepath);
}
$filename1 = implode(',',$images);
}
else{
$filename1=$this->input->get('iti_image2') ;
//print_r( $data['newz1']);exit;
}
我的浏览页:
<div class="col-sm-8">
<input type="button" id="get_file" value="Grab file">
<?php
$imss= $result->iti_image2;
?>
<input type="file" id="my_file" name="file3[]" value="<?php echo $imss ?>" multiple/>
<div id="customfileupload">Select a file</div>
<input type="hidden" class="btn btn info" id="image" name="iti_image2" accept="image" value="<?php echo $result->iti_image2; ?>" multiple/ >
</div>
<script>
document.getElementById('get_file').onclick = function() {
document.getElementById('my_file').click();
};
$('input[type=file]').change(function (e) {
$('#customfileupload').html($(this).val());
});
</script>
</div>
我的模特:
function itinerary_updte($filename4,$filename5){
$update_itinerarys = array(
'iti_image2' =>$filename4,
'iti_image3' =>$filename5
);
$this->db->where('id', $id);
$result = $this->db->update('sg_itinerary', $update_itinerarys);
return $结果;
}
最后,伙计们,我在一些教程之后找到了图像更新的解决方案,请参阅 link
“http://www.2my4edge.com/2016/04/multiple-image-upload-with-view-edit.html”
感谢那些对我的 post 给出逻辑和评论的朋友,非常感谢@ankit singh @Lll
我在 codeigniter 中以那种形式编辑表单我还有 15 个图像文件,同时我更新它在数据库中更新的图像,但我什至没有更新它去的形式的图像并保存为 null 或空路径。但是我需要解决方案,当我不更新所有 15 个文件时,应该检索存储在 database.Please 中的相同旧图像路径,你们能为此提供更好的解决方案吗?
我的编辑表单
控制器:
if($this->input->post('submit'))
{
if($_FILES['file']['name']!='')
{
$filename = $_FILES['file']['name'];
$file_name1 = "upload_".$this->get_random_name()."_".$filename;
$_SESSION['file_upload']=$file_name1;
$file ="uploads/images/".$file_name1;
}
if($_FILES['file1']['name']!='')
{
$filename = $_FILES['file1']['name'];
$file_name2 = "upload_".$this->get_random_name()."_".$filename;
$_SESSION['file_upload']=$file_name2;
$file ="uploads/images/".$file_name2;
}
$query = $this->scener_model->popular_upload($file_name1,$file_name2)
}
如果你是空的,那么传递隐藏的输入值或者像这样传递新上传的值..
if(!empty($_FILES()){
//your uploaded value here
} else {
//hidden attribute value
}
if($this->input->post('submit1')){
$titlemain = $this->input->post('title_main');
$price = $this->input->post('price');
$package = $this->input->post('package');
$titleinner = $this->input->post('title_inner');
$city_package = $this->input->post('city_packge');
if(!empty($count =count($_FILES['file1']['name']))){;
for($i=0; $i<$count;$i++){
$filename1=$_FILES['file1']['name'][$i];
//print_r($filename);exit;
$filetmp=$_FILES['file1']['tmp_name'][$i];
$filetype=$_FILES['file1']['type'][$i];
$images = $_FILES['file1']['name'];
$filepath="uploads/images/".$filename1;
move_uploaded_file($filetmp,$filepath);
}
$filename1 = implode(',',$images);
}
else{
$filename1=$this->input->get('iti_image2') ;
//print_r( $data['newz1']);exit;
}
我的浏览页:
<div class="col-sm-8">
<input type="button" id="get_file" value="Grab file">
<?php
$imss= $result->iti_image2;
?>
<input type="file" id="my_file" name="file3[]" value="<?php echo $imss ?>" multiple/>
<div id="customfileupload">Select a file</div>
<input type="hidden" class="btn btn info" id="image" name="iti_image2" accept="image" value="<?php echo $result->iti_image2; ?>" multiple/ >
</div>
<script>
document.getElementById('get_file').onclick = function() {
document.getElementById('my_file').click();
};
$('input[type=file]').change(function (e) {
$('#customfileupload').html($(this).val());
});
</script>
</div>
我的模特:
function itinerary_updte($filename4,$filename5){
$update_itinerarys = array(
'iti_image2' =>$filename4,
'iti_image3' =>$filename5
);
$this->db->where('id', $id);
$result = $this->db->update('sg_itinerary', $update_itinerarys);
return $结果;
}
最后,伙计们,我在一些教程之后找到了图像更新的解决方案,请参阅 link “http://www.2my4edge.com/2016/04/multiple-image-upload-with-view-edit.html”
感谢那些对我的 post 给出逻辑和评论的朋友,非常感谢@ankit singh @Lll