Cannot INSERT data into mysql using php - php

I have been trying for two days now to figure this one out. I copied verbatim from a tutorial and I still cant insert data into a table. here is my code with form
<font face="Verdana" size="2">
<form method="post" action="Manage_cust.php" >
Customer Name
<font face="Verdana">
<input type="text" name="Company" size="50"></font>
<br>
Customer Type
<font face="Verdana">
<select name="custType" size="1">
<option>Non-Contract</option>
<option>Contract</option>
</select></font>
<br>
Contract Hours
<font face="Verdana">
<input type="text" name="contractHours" value="0"></font>
<br>
<font face="Verdana">
<input type="submit" name="dothis" value="Add Customer"></font>
</form>
</font>
<font face="Verdana" size="2">
<?php
if (isset($_POST['dothis'])) {
$con = mysql_connect ("localhost","root","password");
if (!$con){
die ("Cannot Connect: " . mysql_error());
}
mysql_select_db("averyit_net",$con);
$sql = "INSERT INTO cust_profile (Customer_Name, Customer_Type, Contract_Hours) VALUES
('$_POST[Company]','$_POST[custType]','$_POST[contractHours]')";
mysql_query($sql, $con);
print_r($sql);
mysql_close($con);
}
?>
This is my PHPmyadmin server info:
Server: 127.0.0.1 via TCP/IP
Software: MySQL
Software version: 5.5.27 - MySQL Community Server (GPL)
Protocol version: 10
User: root#localhost
Server charset: UTF-8 Unicode (utf8)
PLEASE tell me why this wont work. when I run the site it puts the info in and it disappears when I push the submit button, but it does not go into the table. There are no error messages that show up. HELP

I have improved a little bit in your SQL statement, stored it in an array and this is to make sure your post data are really set, else it will throw a null value. Please always sanitize your input.
in your Manage_cust.php:
<?php
if (isset($_POST['dothis']))
{
$con = mysql_connect ("localhost","root","password");
if (!$con)
{
die ("Cannot Connect: " . mysql_error());
}
mysql_select_db("averyit_net",$con);
$company = isset($_POST['Company'])?$_POST['Company']:NULL;
$custype = isset($_POST['custType'])?$_POST['custType']:NULL;
$hours = isset($_POST['contractHours'])?$_POST['contractHours']:NULL;
$sql = "INSERT INTO cust_profile(Customer_Name,
Customer_Type,
Contract_Hours)
VALUES('$company',
'$custype',
'$hours')
";
mysql_query($sql, $con);
mysql_close($con);
}
?>

First of all, don't use font tags...ever
Secondly, because of this line:
if (isset($_POST['dothis'])) {
It looks like your HTML and PHP are combined into one script? In which case, you'll need to change the action on the form to something like this:
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
Plus, you can kill a bad connection in one line:
$con = mysql_connect("localhost","root","password") or die("I died, sorry." . mysql_error() );
Check your posts with isset() and then assign values to variables.
var $company;
if(isset($_POST['Company']) {
$company = $_POST['Company'];
} else {
$company = null;
}
//so on and so forth for the other fields
Or use ternary operators
Also, using the original mysql PHP API is usually a bad choice. It's even mentioned in the PHP manual for the API
Always better to go with mysqli or PDO so let's convert that:
//your connection
$conn = mysqli_connect("localhost","username","password","averyit_net");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$sql = "INSERT INTO cust_profile (Customer_Name, Customer_Type, Contract_Hours)
VALUES ($company,$custType,$contractHours)";
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// Assuming you set these
$stmt = mysqli_prepare($conn, $sql);
$stmt->execute();
$stmt->close();
Someone tell me if this is wrong, so I can correct it. I haven't used mysqli in a while.

Change the $sql to this:
$sql = "INSERT INTO cust_profile (Customer_Name, Customer_Type, Contract_Hours) VALUES ('".$_POST[Company]."','".$_POST[custType]."','".$_POST[contractHours]."')

Related

PHP and MySQL query based on user input not working

I am creating a users database where there are 4 fields: ID, username, password, and occupation. This is a test database. I tried querying the db table and it worked but i have a lot of trouble having a user input and a MySQL query based off of it. I run an Apache server in Linux (Debian, Ubuntu).
I have 2 pages. The first one is a bare-bone test index page. this is where there are textboxes for people to input easy info to register their info in the db. Here is the code for it:
<html>
<form action="reg.php" method="POST">
Username:
<input type="text" name="u">Password:
<input type="password" name="p">Occupation:
<input type="text" name="o">
<input type="submit" value="register">
</form>
</html>
After the submit button is clicked. It goes to the reg.php file. This is where it gets complicated. The page goes blank!!! Nothing is displayed or inputted in the db. Normal queries work well, but when user interaction is added, something is wrong. Here is the code for reg.php:
<?php
$un = $_POST["u"]
$pk = $_POST["p"]
$ok = $_POST["o"]
$u = mysql_real_escape_string($un);
$p = mysql_real_escape_string($pk);
$o = mysql_real_escape_string($ok);
$link = mysql_connect('localhost', 'root', 'randompassword');
if (!$link){
die(' Oops. We Have A Problem Here: ' . mysql_error());
}
if ($link){
echo 'connected succesfully';
}
mysql_select_db("forum") or die(' Oops. We Have A Problem Here: ' . mysql_error());
$data = mysql_query("INSERT INTO users (username, password, occupation) VALUES ('{$u}', '{$p}', '{$o}')");
?>
Can anyone hep me to correct this code to make this work?
Thank you so much for your time. Much appreciated.
EDIT:
I noticed that i did not add semicolons in the first 3 lines. after doing so i got this error: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '{'', '', '')' at line 1." Can someone explain why?
EDIT: the website is just on my local machine...
on an apache server on linux
You are missing semi-colons in the first three lines.
$un = $_POST["u"];
$pk = $_POST["p"];
$ok = $_POST["o"];
mysql_real_escape_string() requires a db connection.
Try this ....
<?php
$un = $_POST["u"];
$pk = $_POST["p"];
$ok = $_POST["o"];
$link = mysql_connect('localhost', 'root', 'randompassword');
if (!$link){
die(' Oops. We Have A Problem Here: ' . mysql_error());
}
if ($link){
echo 'connected succesfully';
}
mysql_select_db("forum") or die(' Oops. We Have A Problem Here: ' . mysql_error());
$u = mysql_real_escape_string($un);
$p = mysql_real_escape_string($pk);
$o = mysql_real_escape_string($ok);
$sql = "INSERT INTO users (username, password, occupation) VALUES ('$u', '$p', '$o')";
$ins_sql = mysql_query($sql);
IF($ins_sql) {
echo 'Inserted new record.';
}ELSE{
echo 'Insert Failed.';
}
?>
Try adding this to the top of your script:
error_reporting(E_ALL);
ini_set("display_errors", 1);
This way you will see all errors that you made syntactically or even within your SQL.

How to insert long Strings into mySQL database using PHP?

I'm using a simple html-form and PHP to insert Strings into mySQL Database, which works fine for short strings, not for long ones indeed.
Using the phpmyadmin I'm able to insert Strings of all lengths, it's only doesn't work with the html file and PHP.
Will appreciate every kind of help, would love to learn more about this topic...
Thank you all a lot in advance and sorry if the question is to simple...
There are two very similar questions, I found so far... unfortunately they couldn't help:
INSERTing very long string in an SQL query - ERROR
How to insert long text in Mysql database ("Text" Datatype) using PHP
Here you can find my html-form:
<html>
<body>
<form name="input" action = "uploadDataANDGetID.php" method="post">
What is your Name? <input type="text" name="Name"><br>
Special about you? <input type="text" name="ThatsMe"><br>
<input type ="submit" value="Und ab die Post!">
</form>
</body>
</html>
and here is the PHP-Script named uploadDataANDGetID.php :
<?php
$name = $_POST["Name"];
$text = $_POST["ThatsMe"];
$con = mysql_connect("localhost", "username", "password") or die("No connection established.");
mysql_select_db("db_name") or die("Database wasn't found");
$q_post = mysql_query("INSERT INTO profiles VALUES (null, '{$name}' ,'{$text}')");
$q_getID =mysql_query("SELECT ID FROM profiles WHERE Name = '{$name}' AND ThatsMe = '{$text}'");
if(!$q_post) // if INSERT wasn't successful...
{
print('[{"ID": "-3"}]');
print("uploadDataAndGetID: Insert wasn't successful...");
print("about ME: ".$text);
}
else // insertion succeeded
{
while ($e=mysql_fetch_assoc($q_getID))
$output[]=$e;
//checking whether SELECTion succeeded too...
$num_results = mysql_num_rows($q_getID);
if($num_results < 1)
{
// no such profile available
print('[{"ID": "-1"}]');
}
else
{
print(json_encode($output));
}
}
mysql_close();
?>
Thank you guys!
Use the newer way to connect to MySQL and use prepared statements http://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php
you MUST escape your strings, with mysql_real_escape_string, like this:
$name = mysql_real_escape_string($_POST['Name']);
$text = mysql_real_escape_string($_POST["ThatsMe"]);
$q_post = mysql_query('INSERT INTO profiles VALUES (null, "' . $name . '" ,"' . $text . '")');
also read about SQL injection

How do I complete this login script?

I've almost completed my login script, but I don't know how to check if the username & password is correct.
Here's my script files.
index.php:
<html>
<body>
<form action="action1.php" method="post">
Username: <input type="text" name="uname">
Password: <input type="password" name="pword">
<input type="submit">
</form>
</body>
</html>
The index.php file is just the page that I use to collect the info from my users for registration.
action1.php:
<?php
$con = mysql_connect("localhost", "root", "");
if (!$con)
{
die ('Could not connect: ' . mysql_error());
}
mysql_select_db("user1", $con);
$sql="INSERT INTO useri1 (uname, pword)
VALUES
('$_POST[uname]','$_POST[pword]')";
if (!mysql_query($sql, $con))
{
die('Error: ' . mysql_error());
}
echo "1 record added";
mysql_close($con);
?>
The action1.php file is just the page that registers the users into the database.
login.php:
<html>
<body>
<form action="checklogin.php" method="post">
Username: <input type="text" name="uname1">
Password: <input type="password" name="pword1">
<input type="submit">
</form>
</body>
</html>
The login.php file is just the page I use for the users to type their login info in.
Now this is my problem, I have no idea of how to check the users login info so they can proceed to the members only area. I'm a newbie & any help is GREATLY appreciated.
Thanks,
--Devin
Firstly you want to use mysqli instead of mysql, because mysql is outdated and no longer actively developed. Secondly you want to start escaping your database queries to stop sql injection. In the code below, I used a session to keep track of the user. You can learn more about sessions here.
<?php
session_start();
$mysqli = new mysqli('localhost', 'root', DB_PASSWORD, 'user1');
/* check connection */
if ($mysqli->connect_error)
die('Connect Error (' . $mysqli->connect_errno . ') ' . $mysqli->connect_error);
/* escape string from sql injection */
$userName = $mysqli->real_escape_string($_POST['uname1']);
/* query database */
$result = $mysqli->query("SELECT `pword` FROM `user1` WHERE `uname` = '".$userName."'");
if ($result->num_rows == 1) {
while ($col = $result->fetch_array(MYSQLI_ASSOC)) {
// This presumes you're storing your passwords in plain text.
// If you hashed your passwords or anything, you would have to do the same to $_POST['pword']
if ($_POST['pword'] == $col['pword']) {
// You could do anything here, but sessions are a way of keeping track of a user.
$_SESSION['userName'] = $_POST['uname1'];
$_SESSION['loggedIn'] = true;
}
}
}
$result->close();
/* don't forget to close the connection */
$mysqli->close();
?>
You would run a query along the lines of this:
$user_check_query = "SELECT * FROM useri1 WHERE uname=" . $_POST['uname'] . " AND pword=" . $_POST['pword'] .";";
And if the query returns a value, then they can proceed. If the query returns nothing, they cannot pass. As for the logic, look at the code you have created and see how to create if and else statements to handle the logic.
You need to do a SELECT query - something along the lines of
SELECT STRCMP('{$_POST['pword']}', pword) FROM useri1 WHERE uname = '{$_POST['uname']}'
This should return a value in the range [-1, 1] - if it is not 0, the password is wrong. If there is no rows, the user does not exist.
You need to think about SQL injections as well - but that could very well be an exercise for another day.

Whats does pg_escape_string exactly do?

I'm using the following script which takes the data from a html form and stores in a Postgres DB. There is this pg_escape_string function which stores the value from the form to the php variable. Searching the web throughout, I found that pg_escape_string escapes a string for insertion into the database. I'm not much clear on this. What does it actually escaping? What actually happens when its said that a string is escaped?
<html>
<head></head>
<body>
<?php
if ($_POST['submit']) {
// attempt a connection
$dbh = pg_connect("host=localhost dbname=test user=postgres");
if (!$dbh) {
die("Error in connection: " . pg_last_error());
}
// escape strings in input data
$code = pg_escape_string($_POST['ccode']);
$name = pg_escape_string($_POST['cname']);
// execute query
$sql = "INSERT INTO Countries (CountryID, CountryName) VALUES('$code', '$name')";
$result = pg_query($dbh, $sql);
if (!$result) {
die("Error in SQL query: " . pg_last_error());
}
echo "Data successfully inserted!";
// free memory
pg_free_result($result);
// close connection
pg_close($dbh);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
Country code: <br> <input type="text" name="ccode" size="2">
<p>
Country name: <br> <input type="text" name="cname">
<p>
<input type="submit" name="submit">
</form>
</body>
</html>
Consider the following code:
$sql = "INSERT INTO airports (name) VALUES ('$name')";
Now suppose that $name is "Chicago O'Hare". When you do the string interpolation, you get this SQL code:
INSERT INTO airports (name) VALUES ('Chicago O'Hare')
which is ill-formed, because the apostrophe is interpreted as a SQL quote mark, and your query will error.
Worse things can happen, too. In fact, SQL injection was ranked #1 Most Dangerous Software Error of 2011 by MITRE.
But you should never be creating SQL queries using string interpolation anyway. Use queries with parameters instead.
$sql = 'INSERT INTO airports (name) VALUES ($1)';
$result = pg_query_params($db, $sql, array("Chicago O'Hare"));
pg_escape_string() prevent sql injection in your code

PHP form not writing to mySQL database

I'm just learning PHP and am trying the most basic thing: capturing info from a form and sticking it into a table in a mySQL database. I'm embarrassed to ask such a stupid newbie question, but after reviewing two books, several Stack Overflow posts, and 7 different tutorials, I still can't get my pathetic code to write a few lousy metrics to my database.
Here's the latest version of the code. Could someone please tell me what I am doing wrong?
* Basic HTML Form *
<form method="post" action="post_metrics_stack.php" >
<p>Date<br />
<input name="date" type="text" /></p>
<p>Metric1<br />
<input name="metric1" type="text" /></p>
<p>Metric2<br />
<input name="metric2" type="text" /></p>
<input type="submit" name="submit" value="Submit" />
</form>
* Processor File *
<?php
$date=$_POST['date'];
$metric1=$_POST['metric1'];
$metric2=$_POST['metric2'];
$con = mysql_connect("localhost", "root", "mypassword");
if (!$con)
{die('Could not connect to mysql: ' . mysql_error());}
$mydb = mysql_select_db("mydatabasename");
if (!$mydb)
{die('Could not connect to database: ' . mysql_error());}
mysql_query("INSERT INTO my_metrics VALUES ('$date', '$metric1', '$metric2')");
Print "Your metrics have been successfully added to the database.";
mysql_close($con);
?>
Your mysql-syntax is wrong.
Try
INSERT INTO my_metrics
SET
date = '$date',
metric1 = '$metric1',
metric2 = '$metric2'
Depending on what the table looks like, your code may or may not work,
"INSERT INTO my_metrics VALUES ('$date', '$metric1', '$metric2')"
assumes that the fields are in that order, and that there are no fields before this one.
"INSERT INTO my_metrics (date, metric1, metric2) VALUES ('$date', '$metric1', '$metric2')"
would be more future proof, and may also solve your problem as they are going to insert into the correct fields.
It is also possible that you are getting some bad data for the field definitions, try doing the insert in phpmyadmin or at the command line instead of in php, then work backwards from there.
As far as the vulnerability to SQL injection, you should feed your input strings to mysql_real_escape_string();. This will escape any unwanted characters.
When connecting to the database, you write
$con = mysql_connect("localhost", "root", "mypassword");
if (!$con)
{die('Could not connect to mysql: ' . mysql_error());}
You can simplify this, and making this more readable by writing
mysql_connect('localhost','root','mypassword') or die('Could not connect to mysql:<hr>'.mysql_error());
For solving your problem, see if specifieng column names helps. If you don't, mysql will assume you enter values in the order of the columns, you might get some trouble with an ID field, or something like that. Your query could look like this:
"INSERT INTO my metrics (date,metric1,metric2) VALUES ('$data','$metric1','$metric2'))"
And finally, here's a speed concideration.
There are two ways to write strings: using single quotes ('string'), and using double quotes ("string"). in the case of 'string' and "string", they will work exactly the same, but there is a difference. Look at the following code
$age=3
echo 'the cat is $age years old.';
//prints out 'the cat is $age years old.'
echo "the cat is $age years old.";
//prints out 'the cat is 3 years old'
echo 'the cat is '.$age.' years old';
//prints out 'the cat is 3 years old'.
As you can see from this example, when you use single quotes, PHP doesn't check the string for variables and other things to parse inside the string. Doing that takes PHP longer than concatinating the variable to the string. so although
echo "the cat is $age years old"
is shorter to type than
echo 'the cat is '.$age.' years old';
it will boost your page loading when you write larger applications.
Hooray! Hooray! Hooray!
Thank you all for such helpful advice! It finally works! Here's the updated code in case any other newbies have the same issue. (Hope I didn't screw anything else up.)
Form
<form method="post" action="post_metrics_stack.php" >
<p>Date<br />
<input name="date" type="text" /></p>
<p>Metric1<br />
<input name="metric1" type="text" /></p>
<p>Metric2<br />
<input name="metric2" type="text" /></p>
<input type="submit" name="submit" value="Submit" />
</form>
Processor
<?php
ini_set('display_errors', 1); error_reporting(E_ALL);
// 1. Create connection to database
mysql_connect('localhost','root','mypassword') or die('Could not connect to mysql: <hr>'.mysql_error());
// 2. Select database
mysql_select_db("my_metrics") or die('Could not connect to database:<hr>'.mysql_error());
// 3. Assign variables (after connection as required by escape string)
$date=mysql_real_escape_string($_POST['date']);
$metric1=mysql_real_escape_string($_POST['metric1']);
$metric2=mysql_real_escape_string($_POST['metric2']);
// 4. Insert data into table
mysql_query("INSERT INTO my_metrics (date, metric1, metric2) VALUES ('$date', '$metric1', '$metric2')");
Echo 'Your information has been successfully added to the database.';
print_r($_POST);
mysql_close()
?>
Here you go love :) try W3c it a good place for new pepps
<?php
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("my_db", $con);
$sql="INSERT INTO my_metrics (date, metric1, metric2)
VALUES
('$_POST[date]','$_POST[mertric1]','$_POST[metric2]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "Your metrics have been successfully added to the database.";
mysql_close($con)
?>

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