Delete row from sql table with php issue [duplicate] - php

This question already has answers here:
When to use single quotes, double quotes, and backticks in MySQL
(13 answers)
Closed 7 years ago.
Ask the user which row to delete based on the
value selected in the select list and the value from a textbox which contains
the matching data. For example, selected item = ‘city’ and the textbox =
‘Boston’. Include a button that removes the row from the table.
those are my instructions to follow. I am having two issues, the first issue is if I say.. select teamname and enter a team from the table I get an error saying:
DELETE FROM teams WHERE teamname = rockets
Column not found: 1054 Unknown column 'rockets' in 'where clause'
its trying to use rockets as the column name and not the row that's needs to be deleted. The second issue is if I select yearformed and enter in a correct year, it will delete the row/s with that year. If i enter a year that doesn't exist, my code says records deleted successfully, but nothing is actual deleted.
I think I have got somethings mixed up here, and not sure exactly how it halfway works for yearformed and doesn't for other columns....
html
<form method='post' action='phpfiles/deleteData.php'>
Select a column name, then enter which data to delete.
<br>
<br>
<label for='option1'>
<select name='selectColumn'>
<option value='teamname' id='team'>teamname</option>
<option value='city' id='city'>city</option>
<option value='bestplayer' id='best'>bestplayer</option>
<option value='yearformed' id='year'>year</option>
<option value='website' id='website'>website</option>
</select>
</label>
<label for='option2'>
Data to delete: <input type='text' name='dataDelete'>
</label>
<br><br>
<input type='submit' value='Submit Delete'>
</form>
php
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST"){
$servername = "localhost";
$username = "root";
$password = "enigma";
$dbname = "program09";
$columnSelect = $_POST['selectColumn'];
$deleteData = $_POST['dataDelete'];
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
// sql to delete a record
$sql = "DELETE FROM teams WHERE $columnSelect = $deleteData";
// use exec() because no results are returned
$conn->exec($sql);
echo "Record deleted successfully";
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
$conn = null;
}
?>

You need single quotes:
$sql = "DELETE FROM teams WHERE $columnSelect = '$deleteData'";
But it's better to use prepared statements to avoid SQL Injection. Also use a whitelist for your column names.
$sql = "DELETE FROM teams WHERE $columnSelect = :value1";
$stmt = $conn->prepare($sql);
$stmt->execute(array(':value1'=>$deleteData));

Related

PHP and Mysql trying to run code on DB [duplicate]

This question already has answers here:
mysqli_real_escape_string not working properly
(2 answers)
Closed 5 years ago.
Good Day
I am new to html, php I am trying to set up a page to run a sql query to move data from one table into another one. The trouble I am having is that I require a user to input a date that needs to pass to the query.
I have checked on the net and viewed the question that this is marked as a duplicate of. And am still unable to get this working.
My Html is as follows for the submit part.
<div class = "boxed">
<p> First Date to be used for comparasion</p>
<form method="post">
Date1: <input type="text" name="date1"/>
<input type="Submit" value="submit"/>
</form>
</div>
My Php is
<?php
session_start();
include_once 'include/dbconf.php';
if(isset($_POST['date1']))
$d1 = mysqli_real_escape_string($_POST['date1']);
$sql1 = "insert ignore into compare2 SELECT * FROM Compare where Date =
'$d1'";
$sqldata = mysqli_query($sql1,$conn) or die ('error updaing Table');
echo "Table updated";
mysql_close($con)
?>
My include/dbconfig.php works as I use it on another page and connect succesfully to the DB.
It looks like my query doesn't get the user defined input passed to it.
I don't get any errors either in the httpd error.log.
You advise and help would be most appreciated.
Updated PHP
<?php
session_start();
include_once 'include/dbconf.php';
if(isset($_POST['date1']))
{
$d1 = mysqli_real_escape_string($conn, $_POST['date1']);
$sql1 = "insert ignore into compare2 SELECT * FROM Compare where Date =
'$d1'";
$sqldata = mysqli_query($conn, $sql1) or die ('error updaing Table');
echo "Table updated";
}
mysqli_close($conn)
?>
Still not running the sql to update the table.
Change your code to:
session_start();
include_once 'include/dbconf.php';
if(isset($_POST['date1'])) {
$d1 = mysqli_real_escape_string($conn, $_POST['date1']);
$sql1 = "insert ignore into compare2 SELECT * FROM Compare where Date = '$d1'";
$sqldata = mysqli_query($conn, $sql1) or die ('error updaing Table');
echo "Table updated";
}
mysql_close($con);

PHP post ID of dropdown statement not value

I have a PHP PDO statement that queries a table in my database to return a list of users.
The form is used to log maintenance events, so the username is used to log the person that has done the maintenance.
However when the form is submitted - its submitting the value of the dropdown box instead of the ID of the row it's returning. The table its submitting it into has a column "engineer" the is referencing the engineers table e.g. John Doe id = 1.
I think I need it to post the ID and not the name.. Here's my code:
<form action="" method="post">
<label>Engineer Name:</label>
<br/><br/>
<select id="teamlist" name="teamlist">
<?php
$pdo = new PDO('mysql:host=localhost;dbname=frontier_maintlog', 'user', 'pass');
#Set Error Mode to ERRMODE_EXCEPTION.
$pdo->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$stmt = $pdo->prepare('Select name from engineers');
$stmt->execute();
while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
echo '<option>'.$row['name'].'</option>';
}
?>
</select>
<br/><br/>
<input type="submit" value=" Submit " name="submit"/><br />
</form>
and here's the bit that does the inserting into database:
$sql = "INSERT INTO log (engineer, description, fa_description)
VALUES ('".$_POST["engineer"]."','".$_POST["description"]."','".$_POST["fa_description"]."')";
The "log" table looks a bit like this:logtableimage
Thank you! Sorry if it doesn't make perfect sense.
Here's the error I receive SQLSTATE[HY000]: General error: 1366 Incorrect integer value: '' for column 'engineer' at row 1
Error is that you are inserting string to int datatype.
You should also select id value from engineers table, to use for options:
...
$stmt = $pdo->prepare('Select id,name from engineers');
$stmt->execute();
// and assign value to options
while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
echo "<option value='$row[id]'>$row[name]</option>";
}
You have to assign a value to option. Every option needs to have it's own value. In you case you just echo the option.
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
echo '<option value="'.$row['id'].'">'.$row['name'].'</option>';
}

Mysqli and PHP not sending data [duplicate]

This question already has answers here:
When to use single quotes, double quotes, and backticks in MySQL
(13 answers)
Closed 7 years ago.
I'm building a recruiting website and need to save user data in my database but my form isn't sending anything to the database in phpmyadmin (using WAMP).
I checked the error logs for PHP, MySQL and Apache but don't see any errors. I also added "if/echo" blocks inside the $conn variables to test the connection, which returned true. Code below.
<!-- index.html-->
<form action="process.php" method="post">
<input type="text" name="first_name" placeholder="First Name" /><br/>
<input type="text" name="last_name" placeholder="Last Name" /><br/>
<button type="submit" name="submit"></button>
</form>
//database.php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "xxxx";
$dberror1 = "Could not connect to the database!";
$dberror2 = "Could not find selected table!";
// Connection to the database, Already tried this with echo statement and works
$conn = mysqli_connect($dbhost, $dbuser, $dbpass) or die ($dberror1);
// Selecting the database to connect to
$select_db = mysqli_select_db($conn, 'mainbase') or die ($dberror2);
//process.php
<?php include 'database.php';
if(isset($_POST['submit'])) {
// Creating variables to store form values
$first_name= $_POST['first_name'];
$last_name=$_POST['last_name'];
//Executing the query
mysqli_query($conn, " INSERT INTO 'candidates'('first_name', 'last_name') //Values in 'candidates' table on phpmyadmin
VALUES ('$first_name','$last_name')");/*variables from above*/
}
You're using myqli incorrectly. But on top of that, use PDO to connect to your database instead. It's safer and easy to expand in the future. Here is an example of how to connect to your database with PDO.
<?php
$myUser = "XXXXXX";
$myPass = "XXXXXX";
try{
$dbPDO = new PDO('mysql:host=localhost;dbname=xxxxxxxx', $myUser, $myPass);
$dbPDO->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
echo "Connection was successful";
} catch(PDOException $e){
print "Error!: " . $e->getMessage() . "<br />";
die();
}
?>
Simply change the Xs to your server's settings.
When you want to start a query simply you can do it like so:
$query = $dbPDO->prepare("SELECT * FROM Table_Name");
$query->execute();
Of course you'd want to pass variables to your queries so you can do that like this:
$query = $dbPDO->prepare("SELECT * FROM Table_Name WHERE ID = :id");
$query->bindParam(':id', $id);
$query->execute();
That keeps SQL injection off your worries. Just make sure to sanitize your variables before binding them to the query as well.
I figured I'd show how to insert your variables into your table with PDO.
$firstName = $_POST['first_name'];
$lastName = $_POST['last_name'];
$query = $dbPDO->prepare("INSERT INTO candidates first_name, last_name VALUES (:fname, :lname)");
$query->bindParam(':fname', $firstName);
$query->bindParam(':lname', $lastName);
$query->execute();
You could also make an array of both of your POST variables and pass that instead of binding each variable at a time.
$candidateName = array('$_POST['first_name']', '$_POST['last_name']');
$query = $dbPDO->prepare("INSERT INTO candidates first_name, last_name VALUES (?, ?)");
$query->execute($candidateName);
I hope that helps!
Happy coding!
The problem
Don't put table name and column names between apostrophes. That's what's causing your query to fail. Apostrophes are used to pass strings.
mysqli_query($conn, " INSERT INTO 'candidates'('first_name', 'last_name')
VALUES ('$first_name','$last_name')");
Should be
mysqli_query($conn, " INSERT INTO candidates(first_name, last_name)
VALUES ('$first_name','$last_name')");
Or
mysqli_query($conn, " INSERT INTO `candidates`(`first_name`, `last_name`)
VALUES ('$first_name','$last_name')");
if you like it better.
The error handling
In order to verify the problem you can echo the mysqli_error() function result whenever the query fails, it's a nice practice and would probably have helped you find a solution faster than asking it here.
$query= mysqli_query($conn, " INSERT INTO `candidates`(`first_name`, `last_name`)
VALUES ('$first_name','$last_name')");
if(!$query) //the query will return 0 if it fails
{
echo mysqli_error($conn);
}
The security issue
You're adding POST value directly into your query, which is dangerous.
On these lines:
$first_name= $_POST['first_name'];
$last_name=$_POST['last_name'];
You should be escaping user input.
This will escape any special characters that can cause issues in the mysql query.
$first_name = mysqli_real_escape_string($conn, $_POST['first_name']);
$last_name = mysqli_real_escape_string($conn, $_POST['last_name']);

PHP Can't enter data into database table

I am trying to enter the data that I get from the two variables stuname and book in the table's username and book columns !! I only want to enter data into those two columns since the id column is auto increment and the date is auto updated with time stamp!!! Each time I run my code I enter my data into the two text fields and when I press submit I get this message!!
Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:\xampp\htdocs\assignment.php on line 35
Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\assignment.php on line 36
Here is my Code:
<?php
$servername = "localhost";
$Username = "root";
$Password = "admin";
$Dbname = "nfc";
$conn = mysqli_connect($servername, $Username, $Password, $Dbname);
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
echo "Connected successfully";
if(isset($_POST["stuname"])&&($_POST["book"]))
{
$stuname = $_POST["stuname"];
$book =$_POST["bookname"];
$sql = "INSERT INTO library (id, username, book, date)
VALUES ('', '$stuname', '$book','')";
mysqli_select_db($conn, 'nfc') or die(mysqli_error($con));
$retval = mysqli_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
else
{
echo "Success";
}
echo " to stuname ". $stuname;
echo " to book ". $book;
}
?>
<form id="form1" name="form1" method="post" action="#">
<p>
<label for="1">student name</label>
<input type="text" name="stuname" id="1" />
</p>
<p>
<label for="12">book name</label>
<input type="text" name="bookname" id="12" />
</p>
<input name="submit" type="submit" value="Submit" />
</form>
In the mysqli_query you should put the conn first and then the query itself
$retval = mysqli_query( $conn, $sql );
The first problem was solved by #Ghost in the comments.
Now on to the rest of the problems:
1. Your database design is faulty
This should have failed immediately because you are inserting an empty value for id. id should be a primary key and therefore should be unique. An auto-increment doesn't work if you insert an empty value.
2. Your insert statement is faulty
You should exclude an auto-increment column in the INSERT statement and should not use an empty value for date. If date is a timestamp, you should either use NULL if the time is supposed to be empty or use NOW() to use the current timestamp.
3. You shouldn't be using insert on this page according to your comments.
You should be using UPDATE or REPLACE instead of INSERT if you are trying to update the existing row but you should be using the primary key to signify which row you are replacing. Right now, it looks like you don't have a primary key, so refer to my 1st point.
4. Security concerns: Your query is subject to SQL injections.
You use user input ($_POST) directly in a query. Any malicious user can take advantage of this and extract, delete, or manipulate data in your database. You should be using prepared statements, or at the very least escape functions.

Connecting html form to php page according to primary key

Ok so essentially what I'm trying to do is add a q&a component to my website (first website, so my current php knowledge is minimal). I have the html page where the user's input is recorded, and added to the database, but then I'm having trouble pulling that specific info from the database.
My current php page is pulling info where the questiondetail = the question detail (detail='$detail') in the database, but that could potentially present a problem if two users enter the same information as their question details (unlikely, but still possible, especially if the same person accidentally submits the question twice). What I want to do is have the page load according to the database's question_id (primary key) which is the only thing that will always be unique.
HTML CODE:
<form id="question_outline" action="process.php" method="get">
<p><textarea name="title" id="title_layout" type="text" placeholder="Question Title" ></textarea> </p>
<textarea name="detail" id= "detail_layout" type="text" placeholder="Question Details" ></textarea>
<div id="break"> </div>
<input id="submit_form" name="submit_question" value="Submit Question" type="submit" />
</form>
PROCESS.PHP CODE:
$name2 = $_GET['name2'];
$title = $_GET['title'];
$detail = $_GET['detail'];
$query= "INSERT INTO questions (title, detail) VALUES ('$title', '$detail')";
$result = mysql_query("SELECT * FROM questions where detail='$detail' ")
or die(mysql_error());
The info is being stored correctly in the database, and is being pulled out successfully when detail=$detail, but what I'm looking to do is have it pulled out according to the question_id because that is the only value that will always be unique. Any response will be greatly appreciated!
Updated Version
QUESTION_EXAMPLE.PHP CODE
<?php
$server_name = "my_servername";
$db_user_name ="my_username";
$db_password = "my_password";
$database = "my_database";
$submit = $_GET['submit'];
$title = $_GET['title'];
$detail = $_GET['detail'];
$conn = mysql_connect($server_name, $db_user_name, $db_password);
mysql_select_db($database) or die( "Unable to select database");
$result = mysql_query("SELECT title, detail FROM questions WHERE id =" .
mysql_real_escape_string($_GET["id"]), $conn);
$row = mysql_fetch_assoc($result);
mysql_close($conn);
?>
<h1><?php echo htmlspecialchars($row["title"]);?></h1>
<p><?php echo htmlspecialchars($row["detail"]);?></p>
Firstly, if that is code to be used in production, please make sure you are escaping your SQL parameters before plugging them in to your statement. Nobody enjoys a SQL injection attack. I would recommend using PDO instead as it supports prepared statements and parameter binding which is much much safer.
How can I prevent SQL injection in PHP?
So you have a form...
[title]
[details]
[submit]
And that gets inserted into your database...
INSERT INTO questions (title, details) VALUES (?, ?)
You can get the last insert id using mysql_insert_id, http://php.net/manual/en/function.mysql-insert-id.php.
$id = mysql_insert_id();
Then you can get the record...
SELECT title, details FROM questions WHERE id = ?
And output it in a preview page.
I have written an example using PDO instead of the basic mysql functions.
form.php:
<form action="process.php" method="post">
<label for="question_title">Title</label>
<input id="question_title" name="title"/>
<label for="question_detail">Detail</label>
<input id="question_detail" name="detail"/>
<button type="submit">Submit</button>
</form>
process.php:
<?php
// Create a database connection
$pdo = new PDO("mysql:dbname=test");
// Prepare the insert statement and bind parameters
$stmt = $pdo->prepare("INSERT INTO questions (title, detail) VALUES (?, ?)");
$stmt->bindValue(1, $_POST["title"], PDO::PARAM_STR);
$stmt->bindValue(2, $_POST["detail"], PDO::PARAM_STR);
// Execute the insert statement
$stmt->execute();
// Retrieve the id
$id = $stmt->lastInsertId();
// Prepare a select statement and bind the id parameter
$stmt = $pdo->prepare("SELECT title, detail FROM questions WHERE id = ?");
$stmt->bindValue(1, $id, PDO::PARAM_INT);
// Execute the select statement
$stmt->execute();
// Retrieve the record as an associative array
$row = $stmt->fetch(PDO::FETCH_ASSOC);
?>
<h1><?php echo htmlspecialchars($row["title"]);?></h1>
<p><?php echo htmlspecialchars($row["detail"]);?></p>
Without PDO...
form.php:
<form action="process.php" method="post">
<label for="question_title">Title</label>
<input id="question_title" name="title"/>
<label for="question_detail">Detail</label>
<input id="question_detail" name="detail"/>
<button type="submit">Submit</button>
</form>
process.php:
<?php
// Create a database connection
$conn = mysql_connect();
// Execute the insert statement safely
mysql_query("INSERT INTO questions (title, detail) VALUES ('" .
mysql_real_escape_string($_POST["title"]) . "','" .
mysql_real_escape_string($_POST["detail"]) . "')", $conn);
// Retrieve the id
$id = mysql_insert_id($conn);
// Close the connection
mysql_close($conn);
header("Location: question_preview.php?id=$id");
question_preview.php:
<?php
// Create a database connection
$conn = mysql_connect();
// Execute a select statement safely
$result = mysql_query("SELECT title, detail FROM questions WHERE id = " .
mysql_real_escape_string($_GET["id"]), $conn);
// Retrieve the record as an associative array
$row = mysql_fetch_assoc($result);
// Close the connection
mysql_close($conn);
?>
<h1><?php echo htmlspecialchars($row["title"]);?></h1>
<p><?php echo htmlspecialchars($row["detail"]);?></p>
I assume you want to sort the questions according to the question_id. You could try using the ORDER BY command
example -
$result = mysql_query("SELECT * FROM questions where detail='$detail' ORDER BY question_id")
For these type of examples, you need to run Transaction within database
below are the
http://dev.mysql.com/doc/refman/5.0/en/commit.html
Or else
Create an random variable stored in session and also insert into database and you call it from database and you can preview it easily.
id | question_code | q_title
question_code is the random value generated before insertion into database,
and save the question_code in a session and again call it for preview.

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