php mysql errors

devang23 · December 7, 2009

i have 2 scripts and one of them is to send info into mysql and the other one is to display it.

the problem with this script is that the page just refreshes, and i believe it doesnt store the information that im trying to store.

<html><head><title>Adding a User </title></head>
<body>
<?php
if( (!$firstname) or (!$lastname) or (!$username) or (!$password) )
{
$form ="Please enter all new user details";
$form.="<form action=\"$PHP_SELF\"";
$form.=" method=\"post\">First Name: ";
$form.="<input type=\"text\" name=\firstname\"";
$form.=" value=\"$firstname\"><br>Last Name: ";
$form.="<input type=\"text\" name=\"lastname\"";
$form.=" value=\"$lastname\"><br>UserName: ";
$form.="<input type=\"text\" name=\"username\"";
$form.=" value=\"$username\"><br>PassWord: ";
$form.="<input type=\"text\" name=\"password\"";
$form.=" value=\"$password\"><br>";
$form.="<input type=\"submit\" value=\"Submit\">";
$form.="</form>";
echo($form);
}
else
{ $conn = mysql_connect("mysql4.000webhost.com","a3213677_dolphin","******")or die("Try Again Punk");
  $db = mysql_select_db("a3213677_snorre",$conn)or die("NOOPE");
  $sql = "insert into User Table (first_name,last_name,username,password) values (\"$firstname\",\"$lastname\",\"$username\",password(\"$password\") )";
  $result = mysql_query($sql,$conn);
if($result) { echo("New User $username ADDED DUH"); }
}
?> </body></html>

and the 2nd one i get this error message "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''User Table' Where 1' at line 1"

<html><head><title>I'm a GENUIS</title></head>
<body>
<?php

$conn=mysql_connect("mysql4.000webhost.com","a3213677_dolphin","********") or die(mysql_error());
$rs = mysql_select_db("a3213677_snorre", $conn) or die("NOOOPE");
$sql="select * from 'User Table' Where 1";
$rs=mysql_query($sql,$conn) or die(mysql_error());;

$list = "<table border=\"1\" cellpadding=\"2\">";
$list.="<tr><th><First Name</th>";
$list.="<th>Last Name</th>";
$list.="<th>UserName</th>";
$list.="<th>PassWord</th></tr>";

while($row= mysql_fetch_array($rs) )
{
$list .= "<tr>";
$list .= "<td>".$row["first_name"]."</td>";
$list .= "<td>".$row["last_name"]."</td>";
$list .= "<td>".$row["username"]."</td>";
$list .= "<td>".$row["password"]."</td>";
$list .= "</tr>";
}
$list .= "<\table";

echo($list);

?>
</body><html>

Russia · December 7, 2009

Try this for the second code, it might work.

Also, whats the problem with the first code?

<html><head><title>I'm a GENUIS</title></head>
<body>
<?php

$conn=mysql_connect("mysql4.000webhost.com","a3213677_dolphin","********") or die(mysql_error());
$rs = mysql_select_db("a3213677_snorre", $conn) or die("NOOOPE");
$sql="select * from `User Table` Where 1";
$rs=mysql_query($sql,$conn) or die(mysql_error());;

$list = "<table border=\"1\" cellpadding=\"2\">";
$list.="<tr><th><First Name</th>";
$list.="<th>Last Name</th>";
$list.="<th>UserName</th>";
$list.="<th>PassWord</th></tr>";

while($row= mysql_fetch_array($rs) )
{
$list .= "<tr>";
$list .= "<td>".$row["first_name"]."</td>";
$list .= "<td>".$row["last_name"]."</td>";
$list .= "<td>".$row["username"]."</td>";
$list .= "<td>".$row["password"]."</td>";
$list .= "</tr>";
}
$list .= "<\table";

echo($list);

?>
</body><html>

devang23 · December 8, 2009

the page just refreshes itself...it doesnt reply that the data was submitted, and thanks for the last code and it work, its just that the info didnt show up, and im guessing its cause of the first code?

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