PHP Mysql Query (INSERT) Issue - php

Question

I am having an issue with a MySQL query as follows:
My script generates this as an example query:
INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('Test2', '123-456-7890', 'test#test.com', 'mesa', 'az', '04-14-2013')
Which if I drop directly into PHPMyA, works fine. However, the PHP script I am trying to use to send the query from my website is not working and I can't get it figured out. Here it is:
$sql = "INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('$name', '$phone', '$email', '$city', '$state', '$date')";
mysql_query($sql);
$result = mysql_query($sql);
if($result)
{
echo("<br>Data Input OK");
}
else
{
echo("<br>Data Input Failed");
}
Nothing makes it to the MySQL DB and no PHP errors are displayed, however, if I echo $sql I get the exact query I posted previously.

Answer 1

Just remove the single line mysql_query($sql); on your code and you will be fine.. But you should better start practicing PHP MySQLi which stands for PHP MySQL Improved, such:
$con = mysqli_connect($host, $user, $password, $password);
$sql = "INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('$name', '$phone', '$email', '$city', '$state', '$date')";
$result = mysqli_query($con, $sql);
if($result) {
echo("<br>Data Input OK");
} else {
echo("<br>Data Input Failed");
}

Answer 2

$sql = 'INSERT INTO Table_name (`id`, `name`) VALUES ("1", "php");

Answer 3

You are executing $sql twice in your script wich is causing the error, please remove
mysql_query($sql);
And it will be ready to go
I would also suggest to stop using mysql_query please switch to mysqli or PDO

Answer 4

Are you sure there is a valid connection (..mysql_connect())? Try using the full syntax like so..
$conn = mysql_connect(...);
$result = mysql_query($query, $conn);
Also try forcing a commit after you execute the statement -
$mysql_query("COMMIT", $conn);

Answer 5

You are running mysql_query twice. Reason of the error. Try running the following code.
$sql = "INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('$name', '$phone', '$email', '$city', '$state', '$date')";
$result = mysql_query($sql) or die(mysql_error());
if($result){
echo("<br>Data Input OK");
} else{
echo("<br>Data Input Failed");
}

Answer 6

use this
"INSERT INTO `contacts`(`name`, `phone`, `email`, `city`, `state`, `date`) VALUES ('$_POST[name]', '$_POST[phone]', '$_POST[email]', '$_POST[city]', '$_POST[state]', '$_POST[date]')";

Answer 7

Try to use mysql_query($sql,$con); instead of mysql_query($sql);.

Answer 8

if(isset($_POST['submit']))
{
$name=$_POST['name'];
$age=$_POST['age'];
$address=$_POST['address'];
$ins="insert into table_name(`name`,`age`,`address`)values('".$name."','".$age."','".$address."')";
mysql_query($ins);
echo 'data inserted successfully';
}