Displaying data that is in database in a webpage

max0005

So. I have setup two pages following the W3Schools tutorial.

One with this code:

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

$sql="INSERT INTO person (FirstName, LastName, Age)
VALUES
('$_POST[firstname]','$_POST[lastname]','$_POST[age]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

mysql_close($con)
?>

and the other with this

<html>
<body>

<form action="insert.php" method="post">
Firstname: <input type="text" name="firstname" />
Lastname: <input type="text" name="lastname" />
Age: <input type="text" name="age" />
<input type="submit" />
</form>

</body>

their purpous is to insert data in the mysql db, and it works fine. Now i'd like that data which is beeing stored in the db to be shown on another webpage. I tried with this code:

<html>
<head>
<title>Show</title>
<body>
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

// some code
echo $_POST["localhost . my_db"]
?>
</body>
</html>

it gives me no error, but not even any result... can you please help me?:queasy:

SFDonovan

Something like this.

$sql = sql_query("SELECT * FROM person");
while($row = sql_fetchrow($sql , $con)) {
$FirstName = $row[FirstName];
$LastName = $row[LastName ];
$Age = $row[Age];

echo" <table width=\"100%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\">"
  . "<tr> "
  . "  <th>First Name</th>"  

  . "  <th>Last Name </th>"
  . "  <th>Age</th>" 
  . "  </tr>"
  . "  <tr>"
  ."    <td align=\"center\">$FirstName</td>"  

  . "   <td align=\"center\">$LastName </td>"
  . "   <td align=\"center\">$Age</td>"
  . "</tr>"
  . "</table>";
}

max0005

I'm getting this error:

Fatal error: Call to undefined function sql_person() in C:\xampp\htdocs\demon.php on line 6

do I have to change something in line6? if yes to what?

SFDonovan

post your code.

max0005

<html>
<head>
<title>Show</title>
<body>
<?php
$sql = sql_query("SELECT * FROM person");
while($row = sql_fetchrow($sql , $con)) {
$FirstName = $row[FirstName];
$LastName = $row[LastName ];
$Age = $row[Age];

echo" <table width=\"100%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\">"
  . "<tr> "
  . "  <th>First Name</th>"  

  . "  <th>Last Name </th>"
  . "  <th>Age</th>"
  . "  </tr>"
  . "  <tr>"
  ."    <td align=\"center\">$FirstName</td>"  

  . "   <td align=\"center\">$LastName </td>"
  . "   <td align=\"center\">$Age</td>"
  . "</tr>"
  . "</table>";
} 
?>
</body>
</html>

SFDonovan

Maybe it needs to be mysql_query instead of sql_query.

max0005 wrote:

<html>
<head>
<title>Show</title>
<body>
<?php
$sql = mysql_query("SELECT * FROM person");
while($row = mysql_fetchrow($sql , $con)) {
$FirstName = $row[FirstName];
$LastName = $row[LastName ];
$Age = $row[Age];

echo" <table width=\"100%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\">"
  . "<tr> "
  . "  <th>First Name</th>"  

  . "  <th>Last Name </th>"
  . "  <th>Age</th>"
  . "  </tr>"
  . "  <tr>"
  ."    <td align=\"center\">$FirstName</td>"  

  . "   <td align=\"center\">$LastName </td>"
  . "   <td align=\"center\">$Age</td>"
  . "</tr>"
  . "</table>";
} 
?>
</body>
</html>

max0005

Using this code

<html>
<head>
<title>Show</title>
<body>
<?php
$sql = mysql_query("localhost" ,"root" ,"");
while($row = mysql_fetchrow($sql , $con)) {
$FirstName = $row[FirstName];
$LastName = $row[LastName ];
$Age = $row[Age];

echo" <table width=\"100%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\">"
  . "<tr> "
  . "  <th>First Name</th>"  

  . "  <th>Last Name </th>"
  . "  <th>Age</th>"
  . "  </tr>"
  . "  <tr>"
  ."    <td align=\"center\">$FirstName</td>"  

  . "   <td align=\"center\">$LastName </td>"
  . "   <td align=\"center\">$Age</td>"
  . "</tr>"
  . "</table>";
} 
?>
</body>
</html>

getting this

Warning: Wrong parameter count for mysql_query() in C:\xampp\htdocs\demon.php on line 6

Fatal error: Call to undefined function mysql_fetchrow() in C:\xampp\htdocs\demon.php on line 7

max0005

OKOK, got first error fixed, andpart of the second also.

I changed

mysql_fetchrow

mysql_fetch_row

and now the editor gives it as good. I'm now getting this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\demon.php on line 7

max0005

please someone!

steamPunk

you were getting your connection mixed up with your query

try this

<html>
<head>
<title>Show</title>
<body>
<?php
// open your connection
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

// select db
mysql_select_db("my_db", $con);

// query to run
$sql = sql_query("SELECT * FROM person");

// results
while($row = sql_fetch_row($sql , $con)) {
$FirstName = stripslashes($row[FirstName]);
$LastName = stripslashes($row[LastName ]);
$Age = $row[Age];

echo" <table width=\"100%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\">"
  . "<tr> "
  . "  <th>First Name</th>"  

  . "  <th>Last Name </th>"
  . "  <th>Age</th>"
  . "  </tr>"
  . "  <tr>"
  ."    <td align=\"center\">$FirstName</td>"  

  . "   <td align=\"center\">$LastName </td>"
  . "   <td align=\"center\">$Age</td>"
  . "</tr>"
  . "</table>";
} 
?>
</body>
</html>

steamPunk

while($row = sql_fetch_row($sql , $con)) {
$FirstName = stripslashes($row['FirstName']);
$LastName = stripslashes($row['LastName']);
$Age = $row['Age'];

max0005

Fatal error: Call to undefined function sql_query() in C:\xampp\htdocs\demon.php on line 17

🙁

steamPunk

something else i hadn't noticed - the sql_query should be mysql_query:

<html>
<head>
<title>Show</title>
<body>
<?php
// open your connection
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

// select db
mysql_select_db("my_db", $con);

// query to run
$sql = mysql_query("SELECT * FROM person");

// results
while($row = mysql_fetch_row($sql , $con)) {
$FirstName = stripslashes($row['FirstName']);
$LastName = stripslashes($row['LastName']);
$Age = $row['Age'];

echo" <table width=\"100%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\">"
  . "<tr> "
  . "  <th>First Name</th>"  

  . "  <th>Last Name </th>"
  . "  <th>Age</th>"
  . "  </tr>"
  . "  <tr>"
  ."    <td align=\"center\">$FirstName</td>"  

  . "   <td align=\"center\">$LastName </td>"
  . "   <td align=\"center\">$Age</td>"
  . "</tr>"
  . "</table>";
} 
?>
</body>
</html>

max0005

Now it's giving this

Warning: Wrong parameter count for mysql_fetch_row() in C:\xampp\htdocs\demon.php on line 20

steamPunk

try

mysql_fetch_row($sql)

max0005

That line of code is like this:

while($row = mysql_fetch_row($sql , $con)) {

steamPunk

if you use mysql_fetch_row() i think you need to do

$FirstName = stripslashes($row[0]);

etc

steamPunk

ok do it like this

while($row = mysql_fetch_array($sql , $con)) {
$FirstName = stripslashes($row['FirstName']);
$LastName = stripslashes($row['LastName']);
$Age = $row['Age'];

that way you can have more than 1 result

max0005

Getting this

Parse error: syntax error, unexpected ')' in C:\xampp\htdocs\demon.php on line 22

with this code

while($row = mysql_fetch_row($sql , $con)) {
$FirstName = stripslashes($row[0]);
$LastName = stripslashes($row[0]););
$Age = stripslashes($row[0]);

steamPunk

sorrryyyyyy

forgot to take the $con out - my bad :

while($row = mysql_fetch_array($sql)) {