$ajaxForm reply back from processing page using jquery and php

Question

Hello,

I have a page called guestbook.php in which contains

 $('#guest_form').ajaxForm({});

When the form is triggered it goes to a save.php page which contains and values inserted

if($_POST['x']){

$xx = $_POST['x'];
$yy = $_POST['y'];
$zz = $_POST['z'];


$query_one = "INSERT INTO xxx (x1,yl,z1,z2) values ('$xx','$yy','$zz','00000')";
mysql_select_db($database_1, $1);
$Result = mysql_query($query_guest_one, $1) or die(mysql_error());

So far so good.

Now I run a select query based on the insert and display it in a div on the guestbook.php page. That is where I cannot do it.

All help appreciated.

Thanks Jean

Answer 1

Assuming you have a primary key after the INSERT query you can do something like:

$id = mysql_insert_id();
$query = mysql_query("SELECT * FROM guestbook WHERE id = $id");
$row = mysql_fetch_array($query);

echo $row['x1'];
echo $row['y1'];

// etc

Answer 2

You can select all records in the table using

$result=mysql_query("SELECT * FROM xxx");
while ($record=mysql_fetch_array($result)) {
    // using your example, access the data you need
    print_r($record["x1"]);
    print_r($record["y1"]);
    // and so on..
}

You can filter the records requesting a column with a particular value

$result=mysql_query(sprintf(
    "SELECT * FROM xxx WHERE x1 = '%s' ",
    mysql_real_escape_string("some value")
));

Please remember to sanitize your values with mysql_real_escape_string before inserting or searching if they come from user input.

These are the most basic queries. I suggest you to find a tutorial about SQL, google is your friend and you'll learn many non obvious things