I have a form to upload a picture like below :
<head>
<title> Image Upload </title>
</head>
<body>
<div id="container">
<?php echo form_open_multipart('upload/uploadImage')?>
<input type="file" name="userfile" />
<p>
<input type="submit" name="submit" value="submit" />
</p>
<?php echo form_close();?>
</div>
</body>
</html>
How can I show a display of choose file in the form?
Seems you need javascript for this:
<div class="col-lg-6">
<input id="userfile" name="userfile" type="file" onchange="showMyImage(this)">
<img id="dummyImage" src="#" alt="your image" style="display: none; width: 200px;" />
</div>
<script>
function showMyImage(fileInput) {
var files = fileInput.files;
$('#dummyImage').show();
for (var i = 0; i < files.length; i++) {
var file = files[i];
var imageType = /image.*/;
if (!file.type.match(imageType)) {
continue;
}
var img = document.getElementById("dummyImage");
img.file = file;
var reader = new FileReader();
reader.onload = (function (aImg) {
return function (e) {
aImg.src = e.target.result;
};
})(img);
reader.readAsDataURL(file);
}
}
</script>
You can also call the showMyImage function on document.ready
You can do that using the javascript FileReader(), example below:
function readURL(input) {
if (input.files && input.files[0]) {
var reader = new FileReader();
reader.onload = function (e) {
$('.image_preview').html('<img src="'+e.target.result+'" alt="'+reader.name+'" class="img-thumbnail" width="304" height="236"/>');
}
reader.readAsDataURL(input.files[0]);
}
}
Use the function to your like this - I also have a github repo for this using codeigniter, check the link below:
https://github.com/adrianrios25/CI_upload
I hope this will help you.