Writing to MySQL Database please help

**lw2472** · 01-14-2010 12:25 AM

Can someone tell me why this isnt working? i am receiving a database not updated error, it should be showing the database updated, id, first name, last name, phone.

please thank you in advance..

// <?php
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$phone = $_POST['phone'];

mysql_connect ("localhost", "lddenter_xxx", "xxx") or die (' Error: ' .mysql_error());
mysql_select_db ("lddenter_test");

$query="INSERT INTO commentTable (ID, firstname, lastname, phone)
VALUES ('NULL','".$firstname."','".$lastname."','".phone. "')";

mysql_query($query) or die ('Error Updating Database');

$result = mysql_query ($query);

echo('Database Updated With: ' . $firstname . ' ' .$lastname . ' ' . $phone);

?> //

the html

// <body>
<form id="form1" name="form1" method="post" action="test.php">
<p>First Name:<br >
<input name="firstname" type="text" id="firstname" />
<br >
<br > Last Name:<br >
<label>
<input type="text" name="lastname" id="lastname" >
</label>
<br ><br >
Phone Number:<br >
<label>
<input type="text" name="phone" id="phone">
</label>
<br ><br >
<label>
<input type="submit" name="Submit" id="Submit" value="Submit Customer Information" >
</label>
<br >
</form>
</body> //

**misson** · 01-14-2010 04:42 AM

When asking for help, post the behavior you expect and the behavior you get, which includes error messages.
When posting any sort of code, use [CODE], [PHP] or [HTML] tags (as appropriate) to delineate the code.
Your code is vulnerable to SQL injection. See the next point.
The mysql driver is terribly out of date. Use PDO and prepared statements, which are invulnerable to SQL injection.

**guimar** · 01-14-2010 02:39 PM

Originally Posted by lw2472

Can someone tell me why this isnt working? i am receiving a database not updated error, it should be showing the database updated, id, first name, last name, phone.

please thank you in advance..

// <?php
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$phone = $_POST['phone'];

mysql_connect ("localhost", "lddenter_xxx", "xxx") or die (' Error: ' .mysql_error());
mysql_select_db ("lddenter_test");

$query="INSERT INTO commentTable (ID, firstname, lastname, phone)
VALUES ('NULL','".$firstname."','".$lastname."','".phone. "')";

mysql_query($query) or die ('Error Updating Database');

$result = mysql_query ($query);

echo('Database Updated With: ' . $firstname . ' ' .$lastname . ' ' . $phone);

?> //

the html

// <body>
<form id="form1" name="form1" method="post" action="test.php">
<p>First Name:<br >
<input name="firstname" type="text" id="firstname" />
<br >
<br > Last Name:<br >
<label>
<input type="text" name="lastname" id="lastname" >
</label>
<br ><br >
Phone Number:<br >
<label>
<input type="text" name="phone" id="phone">
</label>
<br ><br >
<label>
<input type="submit" name="Submit" id="Submit" value="Submit Customer Information" >
</label>
<br >
</form>
</body> //

you are running the sql iinsert twice, if any of the columns have a unique attribute the second insert fails.

mysql_query($query) or die ('Error Updating Database');

$result = mysql_query ($query);
should be
$result =mysql_query($query) or die ('Error Updating Database');
or even more helpful
$result =mysql_query($query) or die('Error Updating Database:'.mysql_error());

**descalzo** · 01-14-2010 03:14 PM

$query="INSERT INTO commentTable (ID, firstname, lastname, phone)
VALUES ('NULL','".$firstname."','".$lastname."','".phone. "')";

Don't you mean $phone

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