PHP Database only inputs the last value - php

Question

I have almost no experience with PHP and right now I'm stuck at the total beginning, which is really frustrating. I have a code, which seems to work. From my app I can input values and put them in my PHP database. Thing is that he only inputs the very last value from my PHP code. So the app aside: If I only use the PHP code to input something in my database he always only takes the last value. Here is my code:
<?php
$DB_HostName = "localhost";
$DB_Name = "xxx";
$DB_User = "xxx";
$DB_Pass = "xxx";
$DB_Table = "contacts";
if (isset ($_GET["name"]))
$name = $_GET["name"];
else
$name = "Blade";
if (isset ($_GET["lastname"]))
$lastname = $_GET["lastname"];
else
$lastname = "Xcoder";
if (isset ($_GET["number"]))
$number = $_GET["number"];
else
$number = "111";
$con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die(mysql_error());
mysql_select_db($DB_Name,$con) or die(mysql_error());
$sql = "insert into $DB_Table (Firtname) values('$name');";
$sql = "insert into $DB_Table (Lastname) values('$lastname');";
$sql = "insert into $DB_Table (Number) values('$number');";
$res = mysql_query($sql,$con) or die(mysql_error());
mysql_close($con);
if ($res) {
echo "success";
}else{
echo "faild";
}// end else
?>
To clarify: If I only have the firstname value, he inputs it in the right place (firstname). If I have a firstname and lastname value, he only inputs the lastname value, but not the firstname value (still at the right place lastname). And the same for number. If I have a firstname, lastname and a number, he only puts the number in the right place but not the other values. In addition I can only do it once. If I want to enter another contact he always says (Duplicate entry 'myentry' for key 'PRIMARY').

Answer 1

You overwrite your query, 2 ways to solve. Either 3 different variables, or 1 variable with 3 queries in it. I would prefer the 2nd option
$sql = "insert into $DB_Table (Firtname) values('$name');";
$sql .= "insert into $DB_Table (Lastname) values('$lastname');";
$sql .= "insert into $DB_Table (Number) values('$number');";
$res = mysql_query($sql,$con) or die(mysql_error());
Or if it can be 1 row in the table, as it is the same table anyways:
$sql = "insert into $DB_Table (Firtname,Lastname,Number) values('$name','$lastname','$number');";
$res = mysql_query($sql,$con) or die(mysql_error());

Answer 2

Because you are overwriting $sql on the next 2 lines, so the final sql line is what is inserted.
Your sql is wrong if you want them all in the same row.
$sql = "insert into $DB_Table (Firtname,lastname,number) values('$name','$lastname','$number');";

Answer 3

You are overwriting your $sql statements. You have to execute them. By doing:
$sql = "Insert INTO..."
you merely set a a variable. You need to ru each query using mysql_query().
I'm also guessing that you want to do this:
$sql = "INSERT INTO $DB_TABLE (Firstname, Lastname, Number) VALUES ('$name', '$lastname', '$number')
Finally, it is crucial that you sanitise your inputs:
$name = mysql_real_escape_string($_GET["name"]);
Thanks to this you avoid an SQL Injection attack.

Answer 4

You are overwriting $sql string each time.
Perhaps you meant to use the .= to append all three queries together. What is the structure for database tables?
$sql = "insert into $DB_Table (Firtname) values('$name');";
$sql .= "insert into $DB_Table (Lastname) values('$lastname');";
$sql .= "insert into $DB_Table (Number) values('$number');";

Answer 5

That is because you use the same variable name ($sql) for all 3 queries. Use $sql1, $sql2 and $sql3 instead. Also, call mysql_query for each one (but only if you've set it).
Like this:
$sql1 = "insert into $DB_Table (Firtname) values('$name');";
$sql2 = "insert into $DB_Table (Lastname) values('$lastname');";
$sql3 = "insert into $DB_Table (Number) values('$number');";
if (isset ($sql1))
{
$res1 = mysql_query($sql1,$con) or die(mysql_error());
}
if (isset ($sql2))
{
$res2 = mysql_query($sql2,$con) or die(mysql_error());
}
if (isset ($sql3))
{
$res3 = mysql_query($sql3,$con) or die(mysql_error());
}