How to upload image and move in folder using jquery ?












1















View:



<script>
$("#team_image").change(function(){
var file_data = $('#team_image').prop('files')[0];
var member_name = $("#member_name").val();
var form_data = new FormData();
form_data.append('file', file_data);
$('#imgs').html("<img src='<?php echo base_url(); ?>resource/loading.gif' />");
$.ajax({
url: '<?php echo base_url(); ?>upload',
dataType: 'text',
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'post',
success: function(php_script_response){
$("#imgs").html(php_script_response);
}
});
});
</script>
<input id="team_image" type="file" name="team_image" />
<input id="member_name" type="text" name="member_name" />


Controller:



public function upload()
{
if ( 0 < $_FILES['file']['error'] )
{
echo 'Error: ' . $_FILES['file']['error'] . '<br>';
}
else
{
$filename = $_FILES['file']['name'];
$ext = pathinfo($filename, PATHINFO_EXTENSION);
$allowed = array("jpg", "jpeg", "png");
if(!in_array($ext, $allowed))
{
echo '<p id="red">Only jpg, jpeg, and png files are allowed.</p>';
}
else
{
$data = array(
'member_name'=>$this->input->post('member_name'),
'team'=>$filename
);
$sql = $this->db->insert('team',$data);
move_uploaded_file($_FILES['file']['tmp_name'], ''.FCPATH.'image/team_image/'.$_FILES['file']['name']);
if($sql==true)
{
echo '<p style="color:green;font-weight:bold;">File uploaded Successfully</p>';
}
else
{
echo '<p id="red">Unable to proceed!</p>';
}
}
}
}


In this code I have simple two file i.e. type="file" and type="text". Now, What I actually want when I click on type=" file" image and member_name will insert into database and image should be moved into a folder. I don't know where am I doing wrong? So, How can I do this? Please help me.



Thank You










share|improve this question























  • Could you also explain what is currently happening? Do you get an error?

    – Sven Hakvoort
    Nov 23 '18 at 6:39
















1















View:



<script>
$("#team_image").change(function(){
var file_data = $('#team_image').prop('files')[0];
var member_name = $("#member_name").val();
var form_data = new FormData();
form_data.append('file', file_data);
$('#imgs').html("<img src='<?php echo base_url(); ?>resource/loading.gif' />");
$.ajax({
url: '<?php echo base_url(); ?>upload',
dataType: 'text',
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'post',
success: function(php_script_response){
$("#imgs").html(php_script_response);
}
});
});
</script>
<input id="team_image" type="file" name="team_image" />
<input id="member_name" type="text" name="member_name" />


Controller:



public function upload()
{
if ( 0 < $_FILES['file']['error'] )
{
echo 'Error: ' . $_FILES['file']['error'] . '<br>';
}
else
{
$filename = $_FILES['file']['name'];
$ext = pathinfo($filename, PATHINFO_EXTENSION);
$allowed = array("jpg", "jpeg", "png");
if(!in_array($ext, $allowed))
{
echo '<p id="red">Only jpg, jpeg, and png files are allowed.</p>';
}
else
{
$data = array(
'member_name'=>$this->input->post('member_name'),
'team'=>$filename
);
$sql = $this->db->insert('team',$data);
move_uploaded_file($_FILES['file']['tmp_name'], ''.FCPATH.'image/team_image/'.$_FILES['file']['name']);
if($sql==true)
{
echo '<p style="color:green;font-weight:bold;">File uploaded Successfully</p>';
}
else
{
echo '<p id="red">Unable to proceed!</p>';
}
}
}
}


In this code I have simple two file i.e. type="file" and type="text". Now, What I actually want when I click on type=" file" image and member_name will insert into database and image should be moved into a folder. I don't know where am I doing wrong? So, How can I do this? Please help me.



Thank You










share|improve this question























  • Could you also explain what is currently happening? Do you get an error?

    – Sven Hakvoort
    Nov 23 '18 at 6:39














1












1








1








View:



<script>
$("#team_image").change(function(){
var file_data = $('#team_image').prop('files')[0];
var member_name = $("#member_name").val();
var form_data = new FormData();
form_data.append('file', file_data);
$('#imgs').html("<img src='<?php echo base_url(); ?>resource/loading.gif' />");
$.ajax({
url: '<?php echo base_url(); ?>upload',
dataType: 'text',
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'post',
success: function(php_script_response){
$("#imgs").html(php_script_response);
}
});
});
</script>
<input id="team_image" type="file" name="team_image" />
<input id="member_name" type="text" name="member_name" />


Controller:



public function upload()
{
if ( 0 < $_FILES['file']['error'] )
{
echo 'Error: ' . $_FILES['file']['error'] . '<br>';
}
else
{
$filename = $_FILES['file']['name'];
$ext = pathinfo($filename, PATHINFO_EXTENSION);
$allowed = array("jpg", "jpeg", "png");
if(!in_array($ext, $allowed))
{
echo '<p id="red">Only jpg, jpeg, and png files are allowed.</p>';
}
else
{
$data = array(
'member_name'=>$this->input->post('member_name'),
'team'=>$filename
);
$sql = $this->db->insert('team',$data);
move_uploaded_file($_FILES['file']['tmp_name'], ''.FCPATH.'image/team_image/'.$_FILES['file']['name']);
if($sql==true)
{
echo '<p style="color:green;font-weight:bold;">File uploaded Successfully</p>';
}
else
{
echo '<p id="red">Unable to proceed!</p>';
}
}
}
}


In this code I have simple two file i.e. type="file" and type="text". Now, What I actually want when I click on type=" file" image and member_name will insert into database and image should be moved into a folder. I don't know where am I doing wrong? So, How can I do this? Please help me.



Thank You










share|improve this question














View:



<script>
$("#team_image").change(function(){
var file_data = $('#team_image').prop('files')[0];
var member_name = $("#member_name").val();
var form_data = new FormData();
form_data.append('file', file_data);
$('#imgs').html("<img src='<?php echo base_url(); ?>resource/loading.gif' />");
$.ajax({
url: '<?php echo base_url(); ?>upload',
dataType: 'text',
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'post',
success: function(php_script_response){
$("#imgs").html(php_script_response);
}
});
});
</script>
<input id="team_image" type="file" name="team_image" />
<input id="member_name" type="text" name="member_name" />


Controller:



public function upload()
{
if ( 0 < $_FILES['file']['error'] )
{
echo 'Error: ' . $_FILES['file']['error'] . '<br>';
}
else
{
$filename = $_FILES['file']['name'];
$ext = pathinfo($filename, PATHINFO_EXTENSION);
$allowed = array("jpg", "jpeg", "png");
if(!in_array($ext, $allowed))
{
echo '<p id="red">Only jpg, jpeg, and png files are allowed.</p>';
}
else
{
$data = array(
'member_name'=>$this->input->post('member_name'),
'team'=>$filename
);
$sql = $this->db->insert('team',$data);
move_uploaded_file($_FILES['file']['tmp_name'], ''.FCPATH.'image/team_image/'.$_FILES['file']['name']);
if($sql==true)
{
echo '<p style="color:green;font-weight:bold;">File uploaded Successfully</p>';
}
else
{
echo '<p id="red">Unable to proceed!</p>';
}
}
}
}


In this code I have simple two file i.e. type="file" and type="text". Now, What I actually want when I click on type=" file" image and member_name will insert into database and image should be moved into a folder. I don't know where am I doing wrong? So, How can I do this? Please help me.



Thank You







php jquery ajax codeigniter






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 23 '18 at 6:32









BruceBruce

235




235













  • Could you also explain what is currently happening? Do you get an error?

    – Sven Hakvoort
    Nov 23 '18 at 6:39



















  • Could you also explain what is currently happening? Do you get an error?

    – Sven Hakvoort
    Nov 23 '18 at 6:39

















Could you also explain what is currently happening? Do you get an error?

– Sven Hakvoort
Nov 23 '18 at 6:39





Could you also explain what is currently happening? Do you get an error?

– Sven Hakvoort
Nov 23 '18 at 6:39












1 Answer
1






active

oldest

votes


















0














Please change AJAX code, don't need to use prop method to get files data, with FormData() method, it will send $_POST & $_FILES data on the server. Also, use form tag.



Please follow below code:-



<script>
$("#team_image").change(function(){
var form_data = new FormData();
$('#imgs').html("<img src='<?php echo base_url(); ?>resource/loading.gif' />");
$.ajax({
url: '<?php echo base_url(); ?>upload',
dataType: 'text',
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'post',
success: function(php_script_response){
$("#imgs").html(php_script_response);
}
});
});
</script>
<form method="post" enctype="multipart/form-data">
<input id="team_image" type="file" name="team_image" />
<input id="member_name" type="text" name="member_name" />
</form>


On server side (on PHP script) you can get data by using $_FILES (Array) & $_POST (Array).



Please use this.






share|improve this answer























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    bindNavPrevention: true,
    postfix: "",
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    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Please change AJAX code, don't need to use prop method to get files data, with FormData() method, it will send $_POST & $_FILES data on the server. Also, use form tag.



    Please follow below code:-



    <script>
    $("#team_image").change(function(){
    var form_data = new FormData();
    $('#imgs').html("<img src='<?php echo base_url(); ?>resource/loading.gif' />");
    $.ajax({
    url: '<?php echo base_url(); ?>upload',
    dataType: 'text',
    cache: false,
    contentType: false,
    processData: false,
    data: form_data,
    type: 'post',
    success: function(php_script_response){
    $("#imgs").html(php_script_response);
    }
    });
    });
    </script>
    <form method="post" enctype="multipart/form-data">
    <input id="team_image" type="file" name="team_image" />
    <input id="member_name" type="text" name="member_name" />
    </form>


    On server side (on PHP script) you can get data by using $_FILES (Array) & $_POST (Array).



    Please use this.






    share|improve this answer




























      0














      Please change AJAX code, don't need to use prop method to get files data, with FormData() method, it will send $_POST & $_FILES data on the server. Also, use form tag.



      Please follow below code:-



      <script>
      $("#team_image").change(function(){
      var form_data = new FormData();
      $('#imgs').html("<img src='<?php echo base_url(); ?>resource/loading.gif' />");
      $.ajax({
      url: '<?php echo base_url(); ?>upload',
      dataType: 'text',
      cache: false,
      contentType: false,
      processData: false,
      data: form_data,
      type: 'post',
      success: function(php_script_response){
      $("#imgs").html(php_script_response);
      }
      });
      });
      </script>
      <form method="post" enctype="multipart/form-data">
      <input id="team_image" type="file" name="team_image" />
      <input id="member_name" type="text" name="member_name" />
      </form>


      On server side (on PHP script) you can get data by using $_FILES (Array) & $_POST (Array).



      Please use this.






      share|improve this answer


























        0












        0








        0







        Please change AJAX code, don't need to use prop method to get files data, with FormData() method, it will send $_POST & $_FILES data on the server. Also, use form tag.



        Please follow below code:-



        <script>
        $("#team_image").change(function(){
        var form_data = new FormData();
        $('#imgs').html("<img src='<?php echo base_url(); ?>resource/loading.gif' />");
        $.ajax({
        url: '<?php echo base_url(); ?>upload',
        dataType: 'text',
        cache: false,
        contentType: false,
        processData: false,
        data: form_data,
        type: 'post',
        success: function(php_script_response){
        $("#imgs").html(php_script_response);
        }
        });
        });
        </script>
        <form method="post" enctype="multipart/form-data">
        <input id="team_image" type="file" name="team_image" />
        <input id="member_name" type="text" name="member_name" />
        </form>


        On server side (on PHP script) you can get data by using $_FILES (Array) & $_POST (Array).



        Please use this.






        share|improve this answer













        Please change AJAX code, don't need to use prop method to get files data, with FormData() method, it will send $_POST & $_FILES data on the server. Also, use form tag.



        Please follow below code:-



        <script>
        $("#team_image").change(function(){
        var form_data = new FormData();
        $('#imgs').html("<img src='<?php echo base_url(); ?>resource/loading.gif' />");
        $.ajax({
        url: '<?php echo base_url(); ?>upload',
        dataType: 'text',
        cache: false,
        contentType: false,
        processData: false,
        data: form_data,
        type: 'post',
        success: function(php_script_response){
        $("#imgs").html(php_script_response);
        }
        });
        });
        </script>
        <form method="post" enctype="multipart/form-data">
        <input id="team_image" type="file" name="team_image" />
        <input id="member_name" type="text" name="member_name" />
        </form>


        On server side (on PHP script) you can get data by using $_FILES (Array) & $_POST (Array).



        Please use this.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 24 '18 at 13:36









        Sorav GargSorav Garg

        8611720




        8611720
































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