PHP Mysql Query (INSERT) Issue

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', '[email protected]', '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.

Did you print your query to see whether all the expected values are there? Also just a note you are executing your query two times, so once you get to fix it, it will be doing 2 inserts per run — Hanky Panky
– Hanky Panky, Commented Apr 14, 2013 at 13:31
you're running the query twice, your second attempt might be the one that's failing — ahmad
– ahmad, Commented Apr 14, 2013 at 13:31
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — ulentini
– ulentini, Commented Apr 14, 2013 at 13:41

5ervant - techintel.github.io · Accepted Answer · 2013-04-14 14:37:10Z

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");
}

answered Apr 14, 2013 at 14:37

5ervant - techintel.github.io

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1 Comment

Majid Over a year ago

problem cannot be twice usage of mysql_query($sql).

sudhakar · Accepted Answer · 2014-02-15 13:00:06Z

1

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

edited Feb 15, 2014 at 13:00

sudhakar

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answered Feb 15, 2014 at 12:26

rajesh

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Comments

Fabio · Accepted Answer · 2013-04-14 13:39:58Z

0

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

answered Apr 14, 2013 at 13:39

Fabio

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1 Comment

Jon Over a year ago

Once cleaned up to only run the query once, and $result = mysql_query($sql) or die(mysql_error()); is added I am greeted with the message "No database selected" which I find perplexing since I am staring at my mysql_select_db line... Not sure where the error is, but its time to backtrack my way through the connection string

manojtc · Accepted Answer · 2013-04-14 13:44:17Z

0

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);

answered Apr 14, 2013 at 13:44

manojtc

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Comments

Harpreet · Accepted Answer · 2013-04-14 14:19:52Z

0

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");
}

edited Apr 14, 2013 at 14:19

answered Apr 14, 2013 at 13:31

Harpreet

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3 Comments

TheEwook Over a year ago

Agree but it's not the main issue here

Hanky Panky Over a year ago

Query being run twice is not the error OP is asking about, they don't see data in there for even one run of the query

Harpreet Over a year ago

Try debugging the code by using $result = mysql_query($sql) or die(mysql_error()); Check and see what causes the problem.

Manish Nagar · Accepted Answer · 2013-04-15 09:12:15Z

0

use this

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

answered Apr 15, 2013 at 9:12

Manish Nagar

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Comments

Majid · Accepted Answer · 2013-04-20 16:55:25Z

0

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

edited Apr 20, 2013 at 16:55

Majid

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answered Apr 14, 2013 at 14:42

Ali Mohammadi

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Comments

johny kumar · Accepted Answer · 2016-02-11 18:54:30Z

0

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';
}

edited Feb 11, 2016 at 18:54

johny kumar

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answered Feb 4, 2014 at 12:11

dev4092

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