cannot get the simplest of forms to work

D-ToX

I have tried several tutorials, and cannot seem to get any of them to work. just trying to insert user input into table.

Any one thats got small scripts for me to test? am using php 4.2.2 and mysql

they are working together that much i can get working.

cheers

D-ToX

chinni_77

D-Tox,

that should be really a simple one.Anyways, try this example:


<HTML> 
<?php 
if($submit) 
{ 
$db = mysql_connect("localhost", "root",""); 
mysql_select_db("learndb",$db); 
$sql = "INSERT INTO personnel (firstname, lastname, nick, email, salary) VALUES ('$first', '$ last', '$nickname','$email','$salary')";
$result = mysql_query($sql); 
echo "Thank you! Information entered.\n"; 
} 
else 
{ 
?> 
<form method="post" action="input.php"> 
First name:<input type="Text" name="first"><br> 
Last name:<input type="Text" name="last"><br> 
Nick Name:<input type="Text" name="nickname"><br> 
E-mail:<input type="Text" name="email"><br> 
Salary:<input type="Text" name="salary"><br> 
<input type="Submit"name="submit" value=" Enter information"></form> 
<? 
} 
?> 
</HTML>

Note: Rename the database, table, and the fields with yours.

Hope this script gives you some basic idea.

cheers

votum76

I'm having a problem too
I cannot connect to mysql I tried your code
I set up a test personnel database with the fields you have in your sample code
but when I run it I get
Notice: Undefined variable: submit in c:\program files\apache group\apache\htdocs\index.php on line 5
I also get the form but if I fill it out and click submit it goes nowhere and no data is entered into the form

stolzyboy

if register_globals is off in your php.ini, you have to access the variables with

$variable = $POST['variable'];
$variable1 = $POST['variable1'];

and so forth, then you can use just the $variable in your insert statement, register_globals is off by default in php 4.2+

D-ToX

This is the table i am trying to insert to

CREATE TABLE reklame(
id INT( 7 ) NOT NULL PRIMARY KEY AUTO_INCREMENT,
Kunde CHAR( 30 ) NOT NULL ,
Title CHAR( 30 ) NOT NULL
);

here's the php:

<?php

// SQL database Variables

$hostname="localhost";
$user="root";
$pass="pass";
$dbase="database";

// Connection to database

MySQL_connect("$hostname","$user","$pass")
or die ("Could not connect to MYSQL");

MySQL_select_db("$dbase") or die ("Could not connect select database");

//Assign contents to variables
$Kunde = $Post['Kunde'];
$Title = $Post['Title'];

$result = mysql_query($sql);

$sql = "INSERT INTO reklame(id, Kunde, Title) VALUES ('','$Kunde','$Title')
)";

//Code that checks if the statment executed properly and display the message

if ($result)
{
echo ("Kunde and Title have been added");
}
else
{
echo ("an error has occured");
}

//Close connection with MySQL
MySQL_close()
?>

The out put to the screen is "an error has occured"

what am i dong wrong?

D-ToX

Replaced this
$sql = "INSERT INTO reklame(id, Kunde, Title) VALUES ('','$Kunde','$Title')
)";

with this

$sql = "INSERT INTO reklame(id, Kunde, Title) VALUES ('','$Kunde','$Title')";

but the error persists

xblue

I'd suggest to put your query together before executing it 😉

put the line $sql=...; before $result = mysql_query($sql);

also, while you are developing instead of
$result = mysql_query($sql);
it is better to write
$result = mysql_query($sql) or die(mysql_error());
this will display a more helpful error than just "an error has occured".

and remember using the php tags when posting code in the forum, it is so much easier to read.

D-ToX

That seems to have done the trick. but when lookin at the database it has not inserted the user input. but it has placed an id number in that field. if i do this several time the id number increments but the other two cols in the DB remain blank.

xblue

write $_POST all uppercase as in stolzyboy's example.

if that doesn't help, insert echo $sql; after you define $sql for debugging.

D-ToX

That did it.. thanks very much to the both of.. 😃