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PHP and MySQL: The Basics -problem

Discuss PHP and MySQL: The Basics -problem in the MySQL Development forum on Dev Articles.
PHP and MySQL: The Basics -problem MySQL Development forum to discuss administration, SQL syntax, and other MySQL-related topics. Find help with installing, configuring, and maintaining your MySQL databases.

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  #1  
Old April 17th, 2003, 11:02 AM
mjulson mjulson is offline
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PHP and MySQL: The Basics -problem
I was reading the PHP and MySQL: The Basics article here at DevArticles located at URL

And I got to the point of inserting data into the database. I get the message back that the data was entered, but it's not in the database. I don't get any errors. Any thoughts what might be going on?

Here's the code from the article.

Code:
<html> 

<body> 

<?PHP

if ($submit) { 

// process form 

$db = mysql_connect("localhost", "root"); 

mysql_select_db("mydb",$db);

$sql = "INSERT INTO address (name,,address,hnum,wnum) VALUES ('$name','$address','$hnum','$wnum')";

$result = mysql_query($sql); 

echo "Thank you! Information entered.\n"; 

} else{

// display form 

?>

<form method="post" action="<?PHP echo $PHP_SELF?>"> 

Name:<input type="Text" name="name"><br> 

Address:<input type="Text" name="address"><br> 

Home Number:<input type="Text" name="hnum"><br> 

Work Number:<input type="Text" name="wnum"><br> 

<input type="Submit" name="submit" value="Enter information">

</form> 

<?PHP 

} // end if

?>

</body> 

</html>
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  #2  
Old April 17th, 2003, 12:21 PM
ignitionMX ignitionMX is offline
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Did you leave out your mysql information purposely or is that the problem?
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  #3  
Old April 17th, 2003, 01:39 PM
mjulson mjulson is offline
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Quote:
Originally posted by ignitionMX
Did you leave out your mysql information purposely or is that the problem?


What information is that?

The only thing different in my code, is that I specified the password.
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  #4  
Old April 17th, 2003, 03:21 PM
Vasarab69 Vasarab69 is offline
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well, i'm not sure if this is the problem but...you have two comma's in your query, where there should be one

if it is, then change:

PHP Code:
 $sql = "INSERT INTO address (name,,address,hnum,wnum) VALUES ('$name','$address','$hnum','$wnum')"; 


to

PHP Code:
 $sql = "INSERT INTO address (name,address,hnum,wnum) VALUES ('$name','$address','$hnum','$wnum')"; 


Also...to make sure that there were no problems with the entering of data, you can change:

PHP Code:
 $result = mysql_query($sql); 

echo "Thank you! Information entered.\n"; 


to

PHP Code:
 $result = mysql_query($sql); 

if($result) {

echo "Thank you! Information entered.\n"; 

} else {

echo "We're sorry, there was a problem with the data you entered!";
echo "<br>";
echo "Please go <a href="javascript:history.go(-1)">back</a> and re-enter it, Thank you.";

} 
__________________
-Alexander
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  #5  
Old April 17th, 2003, 05:01 PM
torrent torrent is offline
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Change this:
PHP Code:
 $result = mysql_query($sql); 
to this...
PHP Code:
 $result = mysql_query($sql) or die(mysql_error()); 
This will show up any errors.
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  #6  
Old April 17th, 2003, 06:16 PM
mjulson mjulson is offline
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Ah, I didn't even notice that. It was cut and pasted from the article here. So if an admin is watching, you might want to change that.

I'll give it a try later tonight to see if that's the problem.

Thanks for catching it.

Mike
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  #7  
Old April 17th, 2003, 10:19 PM
Vasarab69 Vasarab69 is offline
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No problem! Glad to help...if you still are recieving errors, post back here and I'll do all i can
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