update multiple columns in mysql database using $_GET - php

Question

I want to update 3 fields in a row in 3 columns but I don't know how to do it. I already searched google and searcedh here but couldn't find any solution for it. I want to change title, paragraph and category of a blog post using $_GET using this way:
<?php
$id = $_GET['id'];
?>
<div class="middle">
<div class="content" style="width:100%;">
<div class="context" style="width:100%">
<?php
if(isset($_POST['submit'])){
$title = $_POST['title'];
$txt = $_POST['txt'];
$query = ("UPDATE tbl_post SET title='$title' WHERE id=$id");
$query = ("UPDATE tbl_post SET txt='$txt' WHERE id=$id");
when I use only one of $_title or $_txt, it works. But I couldn't find a way to update both fields together and couldnt update category selection.
full code of update.php page :
<?php require_once("config.php"); ?>
<?php require_once("header.php"); ?>
<?php
$id = $_GET['id'];
?>
<div class="middle">
<div class="content" style="width:100%;">
<div class="context" style="width:100%">
<?php
if(isset($_POST['submit'])){
$title = $_POST['title'];
$txt = $_POST['txt'];
$query = ("UPDATE tbl_post SET title='$title' WHERE id=$id");
$query = ("UPDATE tbl_post SET txt='$txt' WHERE id=$id");
$query = ("UPDATE tbl_post SET cat='$cat' WHERE id=$id");
mysql_query($query,$con);
header("location:insert.php");
exit();
}
?>
<form action="" method="post">
<?php
$id = $_GET['id'];
$query = "SELECT * FROM `tbl_post` WHERE(id=$id)";
$res = mysql_query($query,$con);
while($rows = mysql_fetch_array($res,MYSQL_ASSOC)){
?>
<p>عنوان مطلب</p>
<input type="text" name="title" style="width:200px; border:1px solid #8C8C8C" value="<?php echo $rows['title'] ?>">
<p>محتوای پست</p>
<textarea name="txt" style="width:300px"><?php echo $rows['txt'] ?></textarea>
<div class="clear"></div>
<?php } ?>
<p>دسته بندی</p>
<select name="cat" style="width:200px">
<?php
$query = "SELECT * FROM `tbl_cat` ORDER BY `id` ASC";
$res = mysql_query($query,$con);
while($rows = mysql_fetch_array($res,MYSQL_ASSOC)){
?>
<option value="<?php echo $rows ['id'] ?>"><?php echo $rows ['name'] ?></option>
</li>
<?php } ?>
</select>
<input type="submit" name="submit" class="" value="ثبت در دیتابیس" style="width:200px; margin-top:15px;">
</form>
</div>
</div>
</div>
<?php require_once("footer.php"); ?>
and insert.php :
<?php require_once("config.php"); ?>
<?php require_once("header.php"); ?>
<div class="middle">
<div class="content" style="width:100%;">
<div class="context" style="width:100%">
<?php
if(isset($_POST['submit'])){
$title = $_POST['title'];
$cat = $_POST['cat'];
$txt = $_POST['txt'];
echo 'title = '.$title.'<br>'.'category ='.$cat.'<br>'.'txt = '.$txt;
$query = "INSERT INTO tbl_post(`title`,`txt`,`cat_id`) VALUES ('$title','$txt','$cat')";
mysql_query($query,$con);
header("location:insert.php");
exit();
}
?>
<form action="" method="post">
<p>عنوان مطلب</p>
<input type="text" name="title" style="width:200px; border:1px solid #8C8C8C;">
<p>دسته بندی</p>
<select name="cat" style="width:200px">
<?php
$query = "SELECT * FROM `tbl_cat` ORDER BY `id` ASC";
$res = mysql_query($query,$con);
while($rows = mysql_fetch_array($res,MYSQL_ASSOC)){
?>
<option value="<?php echo $rows ['id'] ?>"><?php echo $rows ['name'] ?></option>
</li>
<?php } ?>
</select>
<p>محتوای پست</p>
<textarea name="txt" style="width:300px"></textarea>
<div class="clear"></div>
<input type="submit" name="submit" class="" value="ثبت در دیتابیس" style="width:200px; margin-top:15px;">
</form>
</div>
</div>
</div>
<?php require_once("footer.php"); ?>

Answer 1

Combine all the fields into a single query:
$title = $_POST['title'];
$txt = $_POST['txt'];
$cat = $_POST['cat'];
$query = "UPDATE tbl_post SET title='$title', txt = '$txt', cat = '$cat' WHERE id = $id";
Also, you should switch to parametrized queries instead of substituting into the SQL; this means using PDO or mysqli. Otherwise you need to escape the input data. See
How can I prevent SQL injection in PHP?