Passing Variables

gazmania

The problem I have is that when I have entered the information i need into the fields below, the fields seem not to actually hold any information, and therefor all I get is empty entries.
I am no expert at MySQL and PHP, but I cannot see the problem as there are no errors and I think I have declared everything correctly.

Any help is greatly appreciated, it's going to help in my battle to rid our work of ASP '😃'

<html>
<body>
<form method="post" action="<?php echo $PHP_SELF?>">
Name:<input type="Text" name="namefield"><br>
age:<input type="Text" name="agefield"><br>
email:<input type="Text" name="emailfield"><br>
<input type="Submit" name="submit" value="Enter information">
</form>
<?
$db = mysql_connect("localhost", "gary", "*****");
mysql_select_db("test",$db)
or die("Unable to execute query:<br>$sql<br>Error: " . mysql_error());
$sql = "INSERT INTO wibble (name,age,email)
VALUES ('$namefield', '$agefield', '$emailfield')";
$result = mysql_query($sql)
or die("Unable to execute query:<br>$sql<br>Error: " . mysql_error());
mysql_close($db);
?>
</body>
</html>

jmcneese

two things:

first off, NEVER PUT YOUR PASSWORD IN AN EXAMPLE.

secondly,

$sql = "INSERT INTO wibble (name,age,email) 
VALUES ($namefield, $agefield, $emailfield)";

no need to quote variables.

piersk

I was gonna say that maybe you need the link identifier in the mysql_query bit:

Theoretically, you shouldnt need it, but you could try:

$result = mysql_query($sql,$db);

gazmania

Thanks for your suggestion.... no enough coffee made me forget about, what thankfully was a test password, but thanks 😃.

I have unquoted the variables and now it still does the same thing except with the error :

Unable to execute query:
INSERT INTO wibble (name,age,email) VALUES (, , )
Error: You have an error in your SQL syntax near ' , )' at line 2

Looking at it, it's as if that anything with '$' infront of it is ignored.
🙁

Thanks.

siwis

Your script as it stood was trying to input to the database before the form had been run.

With this modified one, once you've submitted, the entries should be echoed back as values to your form field which in turn means that you can check that your variables have been assigned on the second pass.

<html> 
<body>
<? 
if ($submit != "") {
$db = mysql_connect("localhost", "gary", "*****"); 
mysql_select_db("test",$db) 
or die("Unable to execute query:<br>$sql<br>Error: " . mysql_error()); 
$sql = "INSERT INTO wibble (name,age,email) 
VALUES ('$namefield', '$agefield', '$emailfield')"; 
$result = mysql_query("$sql", $db) 
or die("Unable to execute query:<br>$sql<br>Error: " . mysql_error()); 
mysql_close($db); 
};
?> 
<form method="post" action="<?php echo $PHP_SELF?>"> 
Name:<input type="Text" name="namefield" value="<?PHP echo $namefield ?>"><br> 
age:<input type="Text" name="agefield" value="<?PHP echo $agefield ?>"><br> 
email:<input type="Text" name="emailfield" value="<?PHP echo $emailfield ?>"><br> 
<input type="Submit" name="submit" value="Enter information"> 
</form> 
</body> 
</html>

gazmania

I have now entered the code you provided and now nothing gets entered into the database, I see the code at the top that makes sure that if there is nothing there then it won't submit it.

It seems the Fields are not passing anything to it, this is really strange, is there a parameter or something I should have in my php.ini or something, because as far as I see your code is perfect.

Many thanks.

siwis

Check "REGISTER GLOBALS" setting in php.ini

It sounds as if yours are turned off in which case you need to assign the values of $namefield etc:

$namefield = $HTTP_POST_VARS['namefield'];
etc

(I've never needed to do it so hope this is right!)

gazmania

Thank you all for your help, all it was that the Global Varibles had been disabled (duh!)

Thanks all. 😃