更新不在Php和MySql中工作的数据

Question

So, I'm new in php and using mysql. In my program, I tried making the simple edit, add and delete functions. Well, the add and delete functions are working but there is something wrong with the edit one... Whenever I go to phpMyAdmin to check if there are changes in the database, I found none…
I would be very thankful if someone could guide me in this part…

A part of adminpanel.php:

if(array_key_exists('editshirt', $_POST))
{
    include 'editshirt.php';
    exit();
}if(array_key_exists('changeshirt', $_POST))
{
    if(!mysqli_query($dbconnect, 'update shirts set 
        shirtName="'.mysqli_real_escape_string($dbconnect, $_POST['shirtName']).'",
        shirtDescription="'.mysqli_real_escape_string($dbconnect, $_POST['shirtDescription']).'",
        shirtGender="'.mysqli_real_escape_string($dbconnect, $_POST['shirtGender']).'",
        shirtColor="'.mysqli_real_escape_string($dbconnect, $_POST['shirtColor']).'",
        shirtPrice="'.mysqli_real_escape_string($dbconnect, $_POST['shirtPrice']).'",
        shirtPicture="'.mysqli_real_escape_string($dbconnect, $_POST['shirtPicture']).'",
        where shirts.shirtId="'.mysqli_real_escape_string($dbconnect, $_POST['shirtId']).''))
        echo ' ';
    exit();
} 

And here's my editShirt.php

<?php
$yas = mysqli_query($dbconnect, "select * from shirts where shirtId=".mysqli_real_escape_string($dbconnect, $_POST['shirtId']));
$roww = mysqli_fetch_array($yas);

echo "<p>Edit Shirt:</p>";
echo "<form action='?' method='post'>";

/*echo "<label for='shirtName'>Shirt Name: </label>
<input type='text' name='shirtName' value=".$roww['shirtName']."><br>";*/
echo "<label for='shirtName'>Shirt Name: </label>
<textarea name='shirtName' style='resize:none'>".$roww['shirtName']."</textarea><br>";

echo "<label for='shirtDescription'>Description: </label>
<textarea name='shirtDescription' rows='10' cols='30' style='resize:none'>".$roww['shirtDescription']."</textarea><br>";

echo "<label for='shirtGender'>Gender: </label>
<input type='radio' name='shirtGender' value='0'";
if($roww['shirtGender']=='0') echo"checked='checked'";
    echo ">Male&nbsp&nbsp<input type='radio' name='shirtGender' value='0'";
if($roww['shirtGender']=='1') echo "checked='checked'";
    echo ">Female&nbsp&nbsp<input type='radio' name='shirtGender' value='1'";
if($roww['shirtGender']=='2') echo "checked='checked'";
    echo ">Unisex<br><br>";

echo "<label for='shirtColor'>Color: </label>
<input type='radio' name='shirtColor' value='0'";
if($roww['shirtColor']=='0') echo "checked='checked'";
    echo ">Colored&nbsp&nbsp<input type='radio' name='shirtColor' value='1'";
if($roww['shirtColor']=='1') echo "checked='checked'";
    echo ">White&nbsp&nbsp<input type='radio' name='shirtColor' value='1'";
if($roww['shirtColor']=='2') echo "checked='checked'";
    echo ">Black<br><br>";

echo "<label for='shirtPrice'>Shirt Price: </label>
<input type='text' name='shirtPrice' value=".$roww['shirtPrice']."><br>";

echo "<label for='shirtPicture'>Shirt Picture: </label>
<input type='text' name='shirtPicture' value=".$roww['shirtPicture']."><br>";

echo "<input type='hidden' name='shirtId' value=".mysqli_real_escape_string($dbconnect, $_POST['shirtId']).">";

echo "<input type='submit' name='changeshirt' value='Update shirt'>";

echo "</form>";?>

Answer 1

This code is written in very bad way, although I have corrected the code.

mysqli_query($dbconnect, "update shirts set 
    shirtName='".mysqli_real_escape_string($dbconnect, $_POST['shirtName'])."',
    shirtDescription='".mysqli_real_escape_string($dbconnect, $_POST['shirtDescription'])."',
    shirtGender='".mysqli_real_escape_string($dbconnect, $_POST['shirtGender'])."',
    shirtColor='".mysqli_real_escape_string($dbconnect, $_POST['shirtColor'])."',
    shirtPrice='".mysqli_real_escape_string($dbconnect, $_POST['shirtPrice'])."',
    shirtPicture='".mysqli_real_escape_string($dbconnect, $_POST['shirtPicture'])."',
    where shirts.shirtId='".mysqli_real_escape_string($dbconnect, $_POST['shirtId'])."'");

Hope this will help.