Why is my form not working using POST method? - php

Question

I have written the below HTML code:
<form action="index.php" method="POST">
<input type="text" name="title" required>
<input type="text" name="brief_text" required>
<textarea name="text" required></textarea>
<input type="submit" name="add" value="Add">
</form>
My PHP code:
<?php
require_once('db.php');
if(isset($_POST['add'])){
$title = $_POST['title'];
$brief_text = $_POST['brief_text'];
$text = $_POST['text'];
$blog_cat_id = $_POST['blog_cat_id'];
if($title AND $brief_text AND $text AND $blog_cat_id){
$insert_blog = "insert into blog values ('','$title','$brief_text','$text','$blog_cat_id',NOW())";
$run_insertion = mysqli_query($con, $insert_blog);
if($run_insertion){
echo "Blog has been added!";
}
else{
echo "Error adding blog!!!";
}
}
else{
echo "All fields are required!";
}
}
else{
echo "GOODBYE";
}
?>
Every time I refresh the page, it only shows the form and "GOODBYE" and does not even insert the data into database table.
Help me out please.

Answer 1

Is it still showing 'GOODBYE' now you've changed
$_POST['add_blog']
to
$_POST['add']
when you click submit?

Answer 2

You have few mistakes,
1) should be: $_POST['add'] instead of $_POST['add_blog']
2) don't have $_POST['blog_cat_id'] as not in form
EDIT
Copied your code and made some changes:
code:
if(isset($_POST['add'])){
print_r($_POST);
$title = $_POST['title'];
$brief_text = $_POST['brief_text'];
$text = $_POST['text'];
$blog_cat_id = $_POST['blog_cat_id'];
if($title AND $brief_text AND $text AND $blog_cat_id){
echo "inside condition";
$insert_blog = "insert into blog values ('','$title','$brief_text','$text','$blog_cat_id',NOW())";
$run_insertion = mysqli_query($con, $insert_blog);
if($run_insertion){
echo "Blog has been added!";
}
else{
echo "Error adding blog!!!";
}
}
else{
echo "All fields are required!";
}
}
else{
echo "GOODBYE";
}
HTML:
<form action="index.php" method="POST">
<input type="text" name="title" required>
<input type="text" name="brief_text" required>
<input type="text" name="blog_cat_id" required>
<textarea name="text" required></textarea>
<input type="submit" name="add" value="Add">
</form>
output
Array ( [title] => test [brief_text] => test [blog_cat_id] => 1 [text] => testing [add] => Add )
inside condition
Now, check your query if doesn't work.
Hope this will help you.

Answer 3

Your query is wrong, columns are not specified and you are open to sql injection you should learn to use parameterized query. but for this time you can use the following.
Try this:
htmlcode
<form action="index.php" method="POST">
<input type="text" name="title" required>
<input type="text" name="brief_text" required>
<input type="text" name="blog_cat_id" required>
<textarea name="text" required></textarea>
<input type="submit" name="add" value="Add">
</form>
index.php
<?php
require_once('db.php');
if(isset($_POST['add'])){
$title = $_POST['title'];
$brief_text = $_POST['brief_text'];
$text = $_POST['text'];
$blog_cat_id = $_POST['blog_cat_id'];
if($title AND $brief_text AND $text AND $blog_cat_id){
$insert_blog = "insert into blog('col1','col2','col3','col4','col5','col6') values ('','$title','$brief_text','$text','$blog_cat_id',NOW())";
$run_insertion = mysqli_query($con, $insert_blog);
if($run_insertion){
echo "Blog has been added!";
}
else{
echo "Error adding blog!!!";
}
}
else{
echo "All fields are required!";
}
}
?>
Note : col1, col2, col3, col4, col5 and col6 will be your column name.