[SOLVED] PHP database code issue

acrikey · #1 Mar 22nd, 2008, 15:17

I am struggling to get a bit of code to work, have i got it right ? I want to access the database once logged in to display customer details.

PHP: Select all


// issue the query
$sql = "SELECT f_name, 1_name FROM users WHERE
        username = '".$_POST["username"]."' AND
        password = PASSWORD('".$_POST["password"]."')";
$result = mysqli_query($mysqli, $sql) or die(mysqli_error($mysqli));

    
        //display string
    $display_block = "
    <p>".$f_name." <p>
    <p>".$l_name." <p>
    <p>".$h_name." <p>
    <p>".$2l_address." <p>
    <p>".$vill_address." <p>
    <p>".$ar_address." <p>
    <p>".$cout_address." <p>
    <p>".$pc_address." <p>
    <p>".$hc_phone." <p>
    <p>".$wc_phone." <p>
    <p>".$mc_phone." <p>
    <p>".$email." <p>

CloudedVision · #2 Mar 22nd, 2008, 16:00

Use this:

PHP: Select all


m// issue the query
$sql = "SELECT * FROM users WHERE
        username = '".$_POST["username"]."' AND
        password = PASSWORD('".$_POST["password"]."')";
$result = mysqli_query($mysqli, $sql) or die(mysqli_error($mysqli));
$info = mysqli_fetch_array($result);
    
        //display string
    $display_block = "
    <p>".$info['f_name']." <p>
    <p>".$info['l_name']." <p>";

etc.

acrikey · #3 Mar 22nd, 2008, 16:26

Ok just getting a blank screen on test.

RohanShenoy · #4 Mar 22nd, 2008, 16:32

^Did you echo $display_block ?

PHP: Select all


echo $display_block;

add this at the end of the code.

CloudedVision · #5 Mar 22nd, 2008, 16:32

You have to print it! add this to the end.

PHP: Select all


echo $display_block;

acrikey · #6 Mar 22nd, 2008, 16:35

no, echo it below?

CloudedVision · #7 Mar 22nd, 2008, 16:38

This should be the entire code

PHP: Select all


$sql = "SELECT * FROM users WHERE
        username = '".$_POST["username"]."' AND
        password = PASSWORD('".$_POST["password"]."')";
$result = mysqli_query($mysqli, $sql) or die(mysqli_error($mysqli));
$info = mysqli_fetch_array($result);
    
        //display string
    $display_block = "
    <p>".$info['f_name']." <p>
    <p>".$info['l_name']." <p>";

echo $display_block;

acrikey · #8 Mar 22nd, 2008, 16:45

ok still nothing

CloudedVision · #9 Mar 22nd, 2008, 16:49

PHP: Select all


$sql = "SELECT * FROM users WHERE
        username = '".$_POST["username"]."' AND
        password = PASSWORD('".$_POST["password"]."')";
$result = mysqli_query($mysqli, $sql) or die(mysqli_error($mysqli));
if($info = mysqli_fetch_array($result)) { 
        //display string
    $display_block = "
    <p>".$info['f_name']." <p>
    <p>".$info['l_name']." <p>";

echo $display_block; 
} else {
   echo '<p>Wrong username and password!</p>';
}

acrikey · #10 Mar 22nd, 2008, 16:49

As the page has got to look up the db details again, I am right in saying I need to remind the page of the db data again or just remind it to look at the db page again

acrikey · #11 Mar 22nd, 2008, 16:53

Ok this is what I have got on the new page:

Code: Select all

<?php
ob_start();
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name=""; // Table name

m// issue the query
$sql = "SELECT * FROM users WHERE
        username = '".$_POST["username"]."' AND
        password = PASSWORD('".$_POST["password"]."')";
$result = mysqli_query($mysqli, $sql) or die(mysqli_error($mysqli));
if($info = mysqli_fetch_array($result)) { 
        //display string
    $display_block = "
    <p>".$info['f_name']." <p>
    <p>".$info['l_name']." <p>";

echo $display_block; 
} else {
   echo '<p>Wrong username and password!</p>';
} 
?>

CloudedVision · #12 Mar 22nd, 2008, 16:58

you gotta connect to the database!

acrikey · #13 Mar 22nd, 2008, 17:03

Ok, just to make sure we all know what I am trying to do.
Member logs in to the login screen and gets to the members area, from there they have a new menu, if they select account details, the new page is created with their personal details on it from the database. The new page must cross reference the username and password so as to display the correct details.

acrikey · #14 Mar 22nd, 2008, 17:05

Code: Select all

<?php
ob_start();
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name=""; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

m// issue the query
$sql = "SELECT * FROM members WHERE
        username = '".$_POST["username"]."' AND
        password = PASSWORD('".$_POST["password"]."')";
$result = mysqli_query($mysqli, $sql) or die(mysqli_error($mysqli));
if($info = mysqli_fetch_array($result)) { 
        //display string
    $display_block = "
    <p>".$info['f_name']." <p>
    <p>".$info['l_name']." <p>";

echo $display_block; 
} else {
   echo '<p>Wrong username and password!</p>';
} 
?>

CloudedVision · #15 Mar 22nd, 2008, 17:05

i understand

you still gotta connect to the database.

Jack Franklin · #16 Mar 22nd, 2008, 17:05

In your code you posted a couple of posts up, you havent actually connected to the database.

(Okay you have now)

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[SOLVED] PHP database code issue