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
Related
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($ins_sql) {
echo 'Inserted new record.';
}ELSE{
echo 'Insert Failed.';
}
?>
Try adding this to the top of your script:
error_reporting(E_ALL);
ini_set("display_errors", 1);
This way you will see all errors that you made syntactically or even within your SQL.
Okay, Here's my problem. I am trying to make a posting script for my website. However this script is not working; the script is below:
<?php
// Make sure the user is logged in before going any further.
if (!isset($_SESSION['user_id'])) {
echo '<p class="login">Please log in to access this page.</p>';
exit();
}
else {
echo('<p class="login">You are logged in as ' . $_SESSION['username'] . '. Log out.</p>');
}
// Connect to the database
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if (isset($_POST['submit'])) {
// Grab the profile data from the POST
$post1 = mysqli_real_escape_string($dbc, trim($_POST['post1']));
$query = "INSERT INTO ccp2_posts ('post') VALUES ('$post1')";
$error = false;
mysqli_close($dbc);
?>
<form enctype="multipart/form-data" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<legend>Posting</legend>
<label for="post">POST:</label>
<textarea rows="4" name="post1" id="post" cols="50">Write your post here...</textarea><br />
<input type="submit" value="submit" name="submit" />
</form>
</div>
<?php
include ("include/footer.html");
?>
</body>
</html>
Nothing shows up in the database when I submit the form. Help would be amazing. Thanks.
You haven't executed the query. All you've done is opened a connection, defined the query string and closed the connection.
Add:
if(msyqli_query($dbc, $query)) {
// Successful execution of insert query
} else {
// Log error: mysqli_error($dbc)
}
after this line:
$query = "INSERT INTO ccp2_posts ('post') VALUES ('$post1')";
Update:
Started editing but had to leave... As other answerers have pointed you need to either quote the post column with a backick or remove the single quote that you currently have altogether. The only case where you need to use backticks to escape identifiers that are one of the MySQL Reserved Words.
So the working version of your query would be:
$query = "INSERT INTO ccp2_posts (post) VALUES ('$post1')";
You may have other problems, but your SQL is bad. You can't use single quotes around 'post'. You want backticks or nothing:
INSERT INTO ccp2_posts(post) VALUES ('$post1')
You missed
mysqli_query($dbc,$query);
In your code,
$query = "INSERT INTO ccp2_posts ('post') VALUES ('$post1')";
mysqli_query($dbc,$query);
Your query is not quite right:
$query = "INSERT INTO `ccp2_posts` (`post`) VALUES ('$post1')";
Note that those are backticks `, not single-quotes. This is very important! Backticks are used to name databases, tables and column names, and in particular it means you don't have to remember the extensive list of every single reserved word. You could call your column `12345 once I caught a fish alive!` if you want to!
Anyway, more importantly, you aren't actually running your query!
mysqli_query($dbc,$query);
You are not submiting to the database using, for example, the mysql_query() function.
I'm using php and a database to add books to a database.
HTML
<form method="POST" action="addbook.php">
<p>Enter Book title :<input type="text" name="bookname"></p>
<p>Enter Book Author :<input type="text" name="bookauthor"></p>
<p><input type="submit" value="addbook"></p>
</form>
PHP
$bname = $_POST['bookname'];
$bauthor = $_POST['bookauthor'];
$dbcon = mysqli_connect('localhost','root','password','bookstore') or die('asd');
$dbquery = "INSERT INTO books (title,author) VALUES ($bname,$bauthor)";
mysqli_query($dbcon,$dbquery) or die('not queryed');
echo "Your book has been added to your online library";
I'm getting the reply ' not queryed'
try putting single quotes around the values
ie
$dbquery = "INSERT INTO books (title,author) VALUES ('$bname','$bauthor')";
You should be using PDO and prepared statements in order to prevent SQL injection. The resultant PHP would be something like this:
$bname = $_POST['bookname'];
$bauthor = $_POST['bookauthor'];
$dbh = new PDO("mysql:host=$host;dbname=$dbname", $user, $pass); //Fill in these variables with the correct values ('localhost' for host, for example)
$st = $dbh->prepare("INSERT INTO books (title,author) VALUES (?,?)");
$data = array($bname, $bauthor);
$st->execute($data);
You can then add logic to check if the statement executed successfully.
Also, I think you just gave us your root password?
For more information about PDO, see this tutorial.
Check the Column names in the table,whether they match with the one in the query.also check whether they are varchar itself.
I dont find any problem in the query, and also try putting
or die(mysqli_error());
and tell what exactly you can see.
If the type is varchar , you have to use single quotes around the values.
$dbquery = "INSERT INTO books (title,author) VALUES ('$bname','$bauthor')";
I have been trying for two days now to figure this one out. I copied verbatim from a tutorial and I still cant insert data into a table. here is my code with form
<font face="Verdana" size="2">
<form method="post" action="Manage_cust.php" >
Customer Name
<font face="Verdana">
<input type="text" name="Company" size="50"></font>
<br>
Customer Type
<font face="Verdana">
<select name="custType" size="1">
<option>Non-Contract</option>
<option>Contract</option>
</select></font>
<br>
Contract Hours
<font face="Verdana">
<input type="text" name="contractHours" value="0"></font>
<br>
<font face="Verdana">
<input type="submit" name="dothis" value="Add Customer"></font>
</form>
</font>
<font face="Verdana" size="2">
<?php
if (isset($_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]."')
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