I have some hard times to save a blob (image file) to my sql-database.
I have three files: a html form (form.php), an ajax script (ajax.js) and a php-file (save.php).
I think i have some errors in my ajax.js, but after a long google search i couldn`t find a good solution. There must be a problem with the blob handling in JS....
HTML Form:
<div id="editEmployeeModal" class="modal fade">
<div class="modal-dialog">
<div class="modal-content">
<form id="update_form" enctype="multipart/form-data">
<div class="modal-header">
<h4 class="modal-title">table bearbeiten</h4>
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
</div>
<div class="modal-body">
<input type="hidden" id="id_u" name="id" class="form-control" required>
<div class="form-group">
<label>Image</label>
<input type="file" id="image_u" name="image" class="form-control" required>
</div>
<div class="form-group">
<label>Überschrift 1</label>
<input type="text" id="header1_u" name="header1" class="form-control" required>
</div>
<div class="form-group">
<label>Überschrift 2</label>
<input type="text" id="header2_u" name="header2" class="form-control" required>
</div>
<div class="form-group">
<label>Text</label>
<input type="text" id="text_u" name="text" class="form-control" required>
</div>
<div class="form-group">
<label>Link</label>
<input type="text" id="link_u" name="link" class="form-control" required>
</div>
<div class="form-group">
<label>Link-Text</label>
<input type="text" id="linkText_u" name="linkText" class="form-control" required>
</div>
<div class="form-group">
<label>Datum</label>
<input type="date" id="datum_u" name="datum" class="form-control" required>
</div>
</div>
<div class="modal-footer">
<input type="hidden" value="2" name="type">
<input type="button" class="btn btn-default" data-dismiss="modal" value="Cancel">
<button type="button" class="btn btn-info" id="update">Update</button>
</div>
</form>
</div>
</div>
</div>
In form.php i specified all my Inputs and for the image type=file
Ajax Code:
$(document).on('click','.update',function(e) {
var id=$(this).attr("data-id");
var image=$(this).attr("data-image");
var header1=$(this).attr("data-header1");
var header2=$(this).attr("data-header2");
var text=$(this).attr("data-text");
var link=$(this).attr("data-link");
var linkText=$(this).attr("data-linkText");
var datum=$(this).attr("data-datum");
$('#id_u').val(id);
$('#image_u').val(image);
$('#header1_u').val(header1);
$('#header2_u').val(header2);
$('#text_u').val(text);
$('#link_u').val(link);
$('#linkText_u').val(linkText);
$('#datum_u').val(datum);
});
$(document).on('click','#update',function(e) {
var data = $("#update_form").serialize();
$.ajax({
data: data,
enctype: 'multipart/form-data',
type: "post",
url: "save.php",
success: function(dataResult){
var dataResult = JSON.parse(dataResult);
if(dataResult.statusCode==200){
$('#editEmployeeModal').modal('hide');
alert('Data updated successfully !');
location.reload();
}
else if(dataResult.statusCode==201){
alert(dataResult);
}
}
});
});
My PHP-Code:
if($_POST['type']==2){
$id=$_POST['id'];
$image= base64_encode(file_get_contents($_FILES['image']['tmp_name']));
$header1=$_POST['header1'];
$header2=$_POST['header2'];
$text=$_POST['text'];
$link=$_POST['link'];
$linkText=$_POST['linkText'];
$datum=$_POST['datum'];
$sql = "UPDATE `tblTable` SET `image`='$image',`header1`='$header1',`header2`='$header2',`text`='$text',`link`='$link',`linkText`='$linkText',`datum`='$datum' WHERE id=$id";
if (mysqli_query($conn, $sql)) {
echo json_encode(array("statusCode"=>200));
}
else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
}
}
My Datbase-Connection works fine. I can write and read all other Data (Text and Date) to my local Database. I think my problem is in the ajax.js. How can i specify $('#image_u').val(image); that it must handle a blob file and not just normal text input. And do i have to modify these lines?
var data = $("#update_form").serialize();
$.ajax({
data: data,
enctype: 'multipart/form-data',
type: "post",
url: "save.php",
Related
I'm trying to create the Event's Calendar using PHP, Ajax and CKEditor 4. When I send the value to database via Ajax, it send the null value.
It send every values(title, start date, end date) except: c_details that I use CKEditor 4
Database's picture
My form's picture
Here's my form:
<div class="modal-body">
<form id="new_calendar">
<div class="form-group">
<label>Title</label>
<input type="text" class="form-control" name="title" placeholder="">
</div>
<div class="form-group">
<label>Description</label>
<textarea type="text" class="form-control" id="editor" name="c_details" placeholder=""></textarea>
</div>
<div class="form-group">
<label>Start Date</label>
<input type="text" class="form-control" name="start" placeholder="YYYY-mm-dd">
</div>
<div class="form-group">
<label>End Date</label>
<input type="text" class="form-control" name="end" placeholder="YYYY-mm-dd">
</div>
<input type="hidden" name="new_calendar_form">
</form>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-primary" onclick="return new_calendar();">Save</button>
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
</div>
Here's Ajax:
function new_calendar() {
for (instance in CKEDITOR.instances) {CKEDITOR.instances[instance].updateElement()}
$.ajax({
type: "POST",
url: "json-event.php",
data: $("#new_calendar").serialize(),
success: function (data) {
$(".close").trigger('click');
alert(data);
location.reload();
}
});
return false;}
Here's all my code on Code Sandbox, I just changed the name='description' to name='c_details'
: https://codesandbox.io/s/blissful-water-nop1y?file=/index.php
Save new data:
public function new_calendar($data)
{
$db = $this->connect();
$add_user = $db->prepare("INSERT INTO calendar (id,title,c_details,start,end) VALUES(NULL,?,?,?,?) ");
$add_user->bind_param("ssss", $data['title'], $data['c_details'], $data['start'], $data['end']);
if (!$add_user->execute()) {
echo $db->error;
} else {
echo "Saved !";
}
}
before send data to ajax request put this:
for (instance in CKEDITOR.instances) {CKEDITOR.instances[instance].updateElement()}
This will update the element textarea with information of ckeditor
I tried var $pic1 = document.getElementById("pic").value; give the image path: C:fakepath/pictures/pic.jpg
Solutions always include sending in form-data. However, I could not do it.
my images folder: /img
I want image upload to 'img' folder after INSERT process.
view : https://i.stack.imgur.com/t7Gxx.png
<div class="form-row">
<div class="form-group col-md-4"><input type="text" class="form-control" id="input1" name="name" placeholder="Name"></div>
<div class="form-group col-md-4">
<div class="custom-file">
<input type="file" class="custom-file-input" id="pic" aria-describedby="inputGroupFileAddon01" accept=".jpg,.gif,.png">
<label class="custom-file-label" for="inputGroupFile01">Choose Image</label>
</div>
</div>
<div class="form-group col-md-4">
<input type="hidden" name="projeid" value="<?php echo $_POST['proje_id']; ?>">
<button type="submit" name="add" id="add" class="btn btn-primary btn-md btn-block">Add</button>
</div>
</div>
$('#add').click(function(){
var name = document.getElementById('input1').value;
var action = 'insert';
$.ajax({
url:"data/action.php",
method:"POST",
data: {
name: name,
action: action
},
success:function(data) {
load_data(projid);
}
});
});
--data/action.php--
if($_POST["action"] == "insert") {
$query = "INSERT INTO grundriss (name) VALUES ('".$_POST["name"]."')";
$statement = $db->prepare($query);
$statement->execute();
echo '<p>Data Inserted...</p>';
}
Try to convert image to base64 and then post it with ajax.
Right now I am facing a problem : I am trying to insert records in the database with the help of jQuery & Ajax. Unfortunately, I tried to alert inserted values but It doesn't show. I also checked through serialize function and I am unable to do that.
Here is my code of ajax
<script type="text/javascript">
$(document).ready(function(){
$("#add_new").click(function(){
$("#add").slideToggle();
});
$("#submit").click(function(){
var stud_no=$("#roll").val();
if(stud_no==''){
$("#msg").html("Input Roll Number");
} else {
var datastr = $("#sampleform").serialize();
alert(datastr);
$.ajax({
type:'POST',
url:'add_it.php',
data:datastr,
success:function(response){
$("#my_form")[0].reset();
$("#msg").html("Student Successfully Added");
},
error: function (xhr, ajaxOptions, thrownError) {
}
});
}
});
});
</script>
Here is body code :
<body>
<a id="add_new">+add new item</a><br /><br />
<div id="msg"></div><br />
<div id="add">
<form id="sampleform">
<fieldset>
<div class="form-group row">
<label for="roll" class="col-sm-2 col-form-label">Roll Number</label>
<div class="col-sm-6">
<input type="text" name="roll" class="form-control" id="roll">
</div>
</div>
<div class="form-group row">
<label for="name" class="col-sm-2 col-form-label">Name</label>
<div class="col-sm-6">
<input type="text" name="name" class="form-control" id="name">
</div>
</div>
<div class="form-group row">
<label for="clas" class="col-sm-2 col-form-label">Class</label>
<div class="col-sm-6">
<input type="text" name="standard" class="form-control" id="standard">
</div>
</div>
<div class="form-group row">
<label for="mail" class="col-sm-2 col-form-label">Email</label>
<div class="col-sm-6">
<input type="email" name="mail" class="form-control" id="mail">
</div>
</div>
<button type="submit" id="submit" class="btn btn-primary">Submit</button>
<button type="reset" class="btn btn-default">Reset</button>
</fieldset>
</fieldset>
</form>
</div>
Here is my add_it.php
<?php
include('connectdb.php');
$stud_no = trim($_POST['stud_no']);
$name = trim($_POST['name']);
$standard = trim($_POST['standard']);
$mail = trim($_POST['mail']);
$query = "insert into student (stud_no,name,standard,email) values ('$stud_no','$name','$standard','$mail')";
mysqli_query($con,$query) or die (mysqli_error());
?>
Your HTML form fields doesnt have any of these variables sent. You need to add name="email" etc to your form fields.
So for example the email field has to look like this:
<input type="email" name="email" class="form-control" id="mail">
id, class etc is not sent in POST - and therefor can not be recieved in the other end.
Jquery's serialize() function also only handles "name" fields.
https://api.jquery.com/serialize/
Snippet :
For a form element's value to be included in the serialized string, the element must have a name attribute
I'm trying to retrieve data from my database then display it into a modal, however, when I add the dataType:"json" in ajax, it seems like it no longer performs my php file.I am new to ajax and json so I don't really have an idea where I am having problem. BTW, I am trying to achieve a CRUD function that is why I am trying to load data into a modal for updating. I am using the same modal for create and update.
This is my html code.
<div class="modal fade" id="addmodal">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<h3>Add Product</h3>
</div>
<div class="modal-body">
<form class="" action="add.php" method="post" id="insert_form" enctype="multipart/form-data">
<div class="form-group input-width center-block">
<input class="form-control" type="file" name="img" id="img" value="" required/>
</div>
<div class="form-group input-width center-block">
<label>Product Name:</label>
<input class="form-control" type="text" name="pnametb" id="pnametb" placeholder="Product Name" required>
</div>
<div class="form-group input-width center-block">
<label>Price:</label>
<input class="form-control" type="text" name="pricetb" id="pricetb" placeholder="Price" required>
</div>
<div class="form-group input-width center-block">
<label>Category:</label>
<select style="" class="action form-control" name="category" id="category" required>
<option value="" disabled selected>Select a category:</option>
<?php
while($row = mysqli_fetch_assoc($result))
{
?>
<!-- Separated HTML and PHP -->
<option value="<?php echo $row['category']?>"><?php echo $row['category']?></option>
<?php
}
?>
</select>
</div>
<div class="form-group input-width center-block">
<label>Description:</label>
</div>
<div class="form-group input-width center-block">
<textarea style="color:black" name="destb" id="destb" class="message-box" placeholder="Description" rows="5" id="comment" required></textarea>
</div>
<input type="hidden" name="id" id="id">
<input class="btn btn-success" type="submit" name="add" value="Add" id="add">
</form>
</div>
<div class="modal-footer">
<button class="btn btn-default" type="button" name="button" data-dismiss="modal">Close</button>
</div>
</div>
</div>
</div>
My ajax code:
$(document).on('click', '.edit', function(){
var id = $(this).attr("id");
alert(id);
$.ajax({
url:"fetch.php",
method:"post",
data:{id:id},
dataType: "json",
success:function(data){
$('#img').val(data.img);
$('#pnametb').val(data.productname);
$('#pricetb').val(data.price);
$('#category').val(data.category);
$('#destb').val(data.description);
$('#id').val(data.id);
$('#add').val("Edit");
$('#addmodal').modal('show');
}
});
});
And here is my php file.
<?php
//fetch.php
$connect = mysqli_connect("localhost", "root", "", "testing");
if(isset($_POST["id"]))
{
$query = "SELECT * FROM productsa WHERE id = '".$_POST["id"]."'";
$result = mysqli_query($connect, $query);
$row = mysqli_fetch_array($result);
echo json_encode($row);
}
?>
Tell me if I need to clarify more things, I'm not really good in expressing what I want to happen. Any help will be highly appreciated.
First add a JSON header to your PHP script like so:
header("Content-Type: application/json", true);
Next instead of echo $row you do:
echo json_encode($row);
And after that you probably have to change your succes: function of your Ajax call to fix printing out the data, but i will leave that up to you :)
Further reading and a more detailed explanation here:
jQuery $.ajax request of dataType json will not retrieve data from PHP script
I have a page called page2.php. It has a form that allows you to search via jquery ajax call. The code is below. My question is, I have a different form on page1.php. How can I submit the form on page1.php to go to page2.php and have it execute the page2.php code below using the data from the form on page1.php? I know this is most likely simple, having a brain fart right now.
$(function() {
$.validate({
form: '#form_search',
validateOnBlur: false,
errorMessagePosition: 'top',
onSuccess: function(form) {
var formval = $(form).serialize();
var formurl = '/page2.php?a=search';
$('#form_results').html('<div class="form_wait_search">Searching, please wait...<br><img src="/images/search-loader.gif"></div>');
$.ajax({
type: 'POST',
url: formurl,
data: formval,
success: function(data){
var json = $.parseJSON(data);
$('#form_errors').html(json.message);
$('#form_results').html(json.results);
}
});
return false;
}
});
});
UPDATE
Here is the forms Im referring to.
On page1.php this is like a module on the right side bar. Just a form that I want to post to page2.php
<div class="scontent_box1">
<strong class="box_title"><i class="fa fa-search fa-flip-horizontal"></i> Find Locations</strong>
<form method="post" action="/page2.php">
<div class="form-group">
<label>Street Address</label>
<input type="text" class="form-control" name="ad" placeholder="Enter Street Address...">
</div>
<div class="form-group">
<label>City, State or Zip Code</label>
<input type="text" class="form-control" name="ct" placeholder="Enter City, State or Zip Code...">
</div>
<button type="submit" class="btn btn-default blue_btn">Search</button>
</form>
</div>
Now here is the form on page2.php that executes the ajax code above. I want page1.php to submit to page2.php and envoke the same jquery code above.
<div class="row no_gutters">
<div class="search_page">
<div id="form_errors"></div>
<form method="post" id="form_search">
<div class="form-group">
<label for="ad">Street Address<span class="reqfld">*</span></label>
<input type="text" class="form-control" data-validation="required" id="ad" name="ad" placeholder="Enter Street Address..." value="">
</div>
<div class="row">
<div class="col-md-12">
<div class="form-group">
<label for="ct">City, State or Zip Code<span class="reqfld">*</span></label>
<input type="text" class="form-control input-sm" data-validation="required" id="ct" name="ct" placeholder="Enter City Name..." value="">
</div>
</div>
</div>
<div class="row">
<div class="col-md-4 col-xs-12">
<button type="submit" class="btn btn-default blue_btn btn-block">Search Locations</button>
</div>
<div class="col-md-8"></div>
</div>
</form>
</div>
<div id="form_results"></div>
</div>