i have code to insert values into mysql database. but the problem doesn't insert anything.
what i want is if user select from droplist table1. value inserting into table1 and son for table2.
my html code
<html>
<body>
<form method="post" action="update.php">
<input type="txt" name="name"/>
<select name="tables">
<option value="table1">table1</option>
<option value="table2">table2</option>
<option value="table3">table3</option>
<option value="table4">table4</option>
</select>
<input type="submit" value="Submit Pick" />
</body>
</html>
my php code
<?php
$table =$_POST['tables'];
$name =$_POST['name'];
$con=mysqli_connect("localhost","sqldata","sqldata","accounts");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="INSERT INTO $table (name)
VALUES
('$_POST[name]')";
echo ".$table.$name";
?>
You never actually run the query. After you define $sql, execute it.
mysqli_query($con, $sql);
Also of note, hopefully this is only test code. You must NEVER EVER take raw input from a request without passing it through something like mysqli_escape_string.
You code contains some errors. Once you never call mysqli_query to send the SQL to the server. In addition there's an error in your sql string. If you want to insert an array's value, you have to use brackets: {$_POST[name]}. Even better is the use of mysqli_escape_string. So your code should look like this:
$name = mysqli_escape_string($con, $_POST['name']);
$sql="INSERT INTO $table (name)
VALUES
('$name')"
mysqli_query($con, $sql);
Related
I have a problem with my code, I wanted to do that when I click on a checkbox and then on the acceptation button some information will be inserted into my sql database.
There is my code:
<form action="checkboxes.php" method="post">
<input type="checkbox" name="chk1"> 4K </input>
<input type="submit" name="Submit" value="Submit"></input>
</form>
<?php
/* Database connection */
$sDbHost = 'localhost';
$sDbName = 'testowanie';
$sDbUser = 'root';
$sDbPwd = '';
$Conn = mysql_connect ($sDbHost, $sDbUser, $sDbPwd);
mysql_select_db ($sDbName, $Conn)
$checkbox1 = $_POST['chk1'];
if ($_POST["Submit"]=="Submit") {
for ($i=0; $i<sizeof($checkbox1); $i++) {
$query="INSERT INTO cena (name) VALUES ('".$checkbox1[$i]."')";
mysql_query($query) or die (mysql_error() );
}
echo "Record is inserted";
}
?>
But when I click on the button this don't work and a text appears "Table 'testowanie.cena' doesn't exist" but the problem is that the table really exist.
So if someone can help me it will be great.
The name of the table is monitory, cena is a column in the table. There is no name column. So it should be:
$query="INSERT INTO monitory (cena) VALUES ('".mysql_real_escape_string($checkbox1[$i])."')";
You also need to use mysql_real_escape_string() to prevent SQL-injection. But it would be best if you converted to mysqli or PDO and used prepared queries, since the mysql extension is obsolete and has been removed from the current version of PHP.
first of all i am pretty new with mysql and php and for now i just want to insert some data in a mysql database form two text box using php.
here the database name is "info" and table name is "students" having three columns like id(primary key, auto increment activated), name and dept. There are two text boxes txtName and txtDept. I want that when i press the enter button the data form the text boxes will be inserted into the mysql database. I have tried the following code but data is not being inserted in the table....
<html>
<form mehtod="post" action="home.php">
<input type="text" name="txtName" />
<input type="text" name="txtDept" />
<input type="submit" value="Enter"/>
</form>
</html>
<?php
$con = mysqli_connect("localhost","root","","info");
if($_POST){
$name = $_POST['txtName'];
$dept = $_POST['txtDept'];
echo $name;
mysqli_query($con,"INSERT INTO students(name,dept) VALUES($name,$dept);");
}
?>
There are a few things wrong with your posted code.
mehtod="post" it should be method="post" - typo.
Plus, quote your VALUES
VALUES('$name','$dept')
DO use prepared statements, or PDO with prepared statements.
because your present code is open to SQL injection
and add error reporting
error_reporting(E_ALL);
ini_set('display_errors', 1);
You should also check for DB errors.
$con = mysqli_connect("localhost","root","","info")
or die("Error " . mysqli_error($con));
as well as or die(mysqli_error($con)) to mysqli_query()
Sidenote/suggestion:
If your entire code is inside the same file (which appears to be), consider wrapping your PHP/SQL inside a conditional statement using the submit button named attribute, otherwise, you may get an Undefined index... warning.
Naming your submit button <input type="submit" name="submit" value="Enter"/>
and doing
if(isset($_POST['submit'])){ code to execute }
Just doing if($_POST){ may give unexpected results when error reporting is set.
Rewrite: with some added security using mysqli_real_escape_string() and stripslashes()
<html>
<form method="post" action="home.php">
<input type="text" name="txtName" />
<input type="text" name="txtDept" />
<input type="submit" name="submit" value="Enter"/>
</form>
</html>
<?php
$con = mysqli_connect("localhost","root","","info")
or die("Error " . mysqli_error($con));
if(isset($_POST['submit'])){
$name = stripslashes($_POST['txtName']);
$name = mysqli_real_escape_string($con,$_POST['txtName']);
$dept = stripslashes($_POST['txtDept']);
$dept = mysqli_real_escape_string($con,$_POST['txtDept']);
echo $name;
mysqli_query($con,"INSERT INTO `students` (`name`, `dept`) VALUES ('$name','$dept')")
or die(mysqli_error($con));
}
?>
As per the manual: http://php.net/manual/en/mysqli.connect-error.php and if you wish to use the following method where a comment has been given to that effect:
<?php
$link = #mysqli_connect('localhost', 'fake_user', 'my_password', 'my_db');
if (!$link) {
die('Connect Error: ' . mysqli_connect_error());
}
?>
God save us all...
Use PDO class instead :). By using PDO you can additionally make prepared statement on client side and use named parameters. More over if you ever have to change your database driver PDO support around 12 different drivers (eighteen different databases!) where MySQLi supports only one driver (MySQL). :(
In term of performance MySQLi is around 2,5% faster however this is not a big difference at all. My choice is PDO anyway :).
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 . '" ,"' . $text . '")');
also read about SQL injection
Okay, Here's my problem. I am trying to make a posting script for my website. However this script is not working; the script is below:
<?php
// Make sure the user is logged in before going any further.
if (!isset($_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 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