Update A Row Mysql In Php
Solution 1:
There is a lot going on here.
Always us a integer as a primary key, example: id MEDIUMINT NOT NULL AUTO_INCREMENT. Then make it the primary key.
you need to sanitize your input to the database using
mysql_real_escape_string()
you need to concatenate your query, so it should look like this:
mysql_query("UPDATE user SET username = '".mysql_real_escape_string($_POST[nusername])."' SET password = '".mysql_real_escape_string($_POST[npassword])."' SET name = '".mysql_real_escape_string($_POST[nname])."' SET surname = '".mysql_real_escape_string($_POST[nsurname])."' SET role = '".mysql_real_escape_string($_POST[nrole])."' WHERE username='".mysql_real_escape_string($_POST[us])."'");
Here is corrected code:
<?php$hostname = "localhost";
$database = "mydb";
$username = "myuser";
$password = "mypsw";
$link = mysql_connect( $hostname , $username , $password ) ordie("Prosoxi!Provlima stin sundesi me ton server : " . mysql_error());
mysql_select_db($database,$link);
mysql_query("UPDATE user
SET username = '".mysql_real_escape_string($_POST['nusername'])."',
SET password = '".mysql_real_escape_string($_POST['npassword'])."',
SET name = '".mysql_real_escape_string($_POST['nname'])."',
SET surname = '".mysql_real_escape_string($_POST['nsurname'])."',
SET role = '".mysql_real_escape_string($_POST['nrole'])."'
WHERE username='".mysql_real_escape_string($_POST['us'])."'");
mysql_close($link);
header("Location: users.php");
?>
notice the single quote surrounding the _POST var, $_POST['nusername'] you had $_POST[nusername].
Try it now, and see if it updates.
Solution 2:
$query = mysql_query("UPDATE user
SET username = '" .mysql_escape_string($_POST[nusername]) . "'
password = ' " .mysql_escape_string($_POST[npassword]) . "'
name = '" . mysql_escape_string($_POST[nname]) . " '
surname = '" . mysql_escape_string($_POST[nsurname])."'
SET role = '".mysql_escape_string($_POST[nrole]) . "'
WHERE username='" .mysql_escape_string( $_POST[us]) . "'");
If everything else fails echo the SQL statement then paste on SQL Browser/PHPMyAdmin then debug it there. Then you just replace the code with the error-free one.
You need to make sure that the data you are sending are sql-inject free as well. Someone might just bypass it..
Solution 3:
Your query need to be modified as bellow.
mysql_query("UPDATE user
SET username = '" .$_POST[nusername] . "' ,
password = ' " .$_POST[npassword] . "',
name = '" . $_POST[nname] . " ',
surname = '" . $_POST[nsurname]',
role = '$_POST[nrole] . "'
WHERE username='" . $_POST[us] . "'");
Also you are open for the SQL Injection attacks. So you better use mysql_real_escape_string()
function as well.
http://php.net/manual/en/function.mysql-real-escape-string.php
Also I would like to suggest you few steps to over come this kind of issue. This is just an example.
Step 1
When you need a SQL statement in your PHP code. You better write it in your MySQL tool first and test it with sample values.
UPDATE subscriber
SET
Subscriber_Name ='Test' ,
Email ='test@test.com'WHERE
Subscriber_ID ='2' ;
Step 2:
If the query works fine then copy it to php. And replace values with mysql_real_escape_string()
support.
$sql = "UPDATE subscriber
SET
Subscriber_Name = '" . mysql_real_escape_string($_POST['name']) . "' ,
Email = '" . mysql_real_escape_string($_POST['email']) . "'
WHERE
Subscriber_ID = '" . mysql_real_escape_string($_POST['id']) . "' ;"
Step 3:
Execute your query.
$result = mysql_query($sql);
Step 4 :
You can see any if there any errors available.
echo mysql_error();
EDIT:
Answer for you Question 2 "How can I achieve already filled boxes in the html file, with the right values, if I choose a certain username?" could be like this.
First you have to write a select statement and get whatever data you want. Ex.
$sql = "SELECT user.username, user.name, user.surname , user.role FROM USER WHERE user.username = '" . mysql_real_escape_string($_POST[us]) . "'";
$result = mysql_query($sql, $link) ordie(mysql_error());
$row = mysql_fetch_assoc($result);
Then put your HTML code. Ex:
<formaction="edit_user.php"method="post"><p>Username<inputtype="text"name="nusername"size="40"value="<?phpecho$row['username'];?>"></p><p>Password<inputtype="password"name="npassword"size="40"></p><p>Name<inputtype="text"name="nname"size="40"value="<?phpecho$row['name'];?>"></p><p>Surname<inputtype="text"name="nsurname"size="40"value="<?phpecho$row['surname'];?>"></p><p>Role<inputtype="text"name="nrole"size="40"value="<?phpecho$row['role'];?>"></p><p><inputtype="submit></p>
</form>
Solution 4:
Apparently there is a syntax error in the latest code you posted, when you obtain the data from post you have $_POST[nusername]
and it should be $_POST['nusername']
since it is an index of the array, and I also recommend echoing the query and commenting the header call so you can see what is the query that is being sent to MySQL
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