Insert Table Data into Array

[teknomage1]

The = operator is your friend

[Salchester]

Originally Posted by teknomage1

The = operator is your friend

For what?

Many Thanks,

[DJMaze]

Originally Posted by Salchester

how do you put information into the array using variables

Originally Posted by teknomage1

The = operator is your friend

there's nothing more to say about it.

P.S. Salchester, how old are you?

[Salchester]

DJMaze,

That's It? But how do I store the data in a variable in the array like following Visual Basic code does?

The name, species, & sex variables are linked with the array on the 6th line.

Private Type details
  name As String
  species As String
  sex As String
End Type
Dim record() As details

Private Sub Form_Load()
  ReDim record(0)
End Sub

Private Sub lblNext_Click()
  ReDim Preserve record(UBound(record) + 1)
  record(UBound(record)).name = txtName.Text
  record(UBound(record)).species = txtSpecies.Text
  record(UBound(record)).sex = txtSex.Text
  MsgBox record(1).name
End Sub

Many Thanks,
P.S I'm 20 Why?

[Salchester]

DJMaze,

The following code doesn't work, nothing gets display from the database.
Why is this?

<?php

$con = mysql_connect("localhost","peter","abc123");
if (!$con)
{
  die('Could not connect: ' . mysql_error());
}

mysql_select_db("WB", $con);

$result = mysql_query("select * from Birds");

// alternatively, you could use mysql_fetch_assoc if you wanted
// the array to be keyed by the field name
while ($array = mysql_fetch_assoc($result))
{
  echo $array[0]['Name'];
} 

?>

Many Thanks,

[DJMaze]

Well sometimes your question are like from a 12 year old in my feeling.
It's not to downgrade you, but some questions emply that you didn't learn a thing from answers due to the fact that basics like operator and associative arrays are missing in your knowledge.

PHP Code:

header('Content-type: text/plain');

$records = array();
while ($array = mysql_fetch_assoc($result))
{
    $records[] = $array;
} 

print_r($records); 


[redhead]

Just a quick out of thoughts rewrite of your VB thingy there into PHP, Absolutely no guarentie it will work of any sort

PHP Code:

<?

$con = mysql_connect("localhost","peter","abc123");
if (!$con)
{
  die('Could not connect: ' . mysql_error());
}
mysql_select_db("WB", $con);

class details {
  var $name;
  var $species;
  var $sex;
  function details($id){
    $result = mysql_query("select * from Birds where id = '$id'");
    if($array = mysql_fetch_array($result)){
      $this->name = $array['name'];
      $this->species = $array['species'];
      $this->sex = $array['sex'];
    }
  }
  function show(){
    echo "<table border=1 cellspacing=0 cellpadding=0>
    <tr><td>" . $this->name . "</td>
    <tr><td>" . $this->species . "</td>
    <tr><td>" . $this->sex. "</td></tr></table>"; 
  }
}

class record{
  var $items = array();
  var $count;
  function record(){
    $this->count = 0;
    $result = mysql_query("select id from Birds");
    while ($array = mysql_fetch_array($result)) {
      $this->items[$this->count++] = new details($array['id']);
    }
  }
  function show(){
    for($i=0; $i < $this->count; $i++)
      $this->items[$i]->show();
  }
}

$my_record = new record();
$my_record->show();
?>

For a refference, you can look at the Object Oriented PHP article.

And DjMs question regarding your age, is of some relevence, since some of your questions seems to come from one who dosn't grasp what is going on, and feel it is easier to ask for a solution than to piece together one from previus explanations.

[Salchester]

DJMaze,

How come the following doesn't work, when code is included from previous posts?

<?php

header('Content-type: text/plain');

$con = mysql_connect("localhost","peter","abc123");
if (!$con)
{
  die('Could not connect: ' . mysql_error());
}

mysql_select_db("WB", $con);

$result = mysql_query("select * from Birds");

$records = array();
while ($array = mysql_fetch_assoc($result))
{
  $records[0]['Name'];
}
print_r($records);

?>

[DJMaze]

Do you really want me to reply?
Did you ever try my code (and that from redhead and sde) ?
Did you see the output of that code?
Why did you modify my code?
Can you please read, copy and understand it all before screwing around with it?

[Salchester]

JMaze,

Do you really want me to reply?

Yes. Please

Did you ever try my code (and that from redhead and SDE)?

Yes, the code I posted previously was from help by SDE, only it don't work, nothing gets displayed. Somethings wrong, but what?

As far as Redhead code goes, that doesn't work either, I keep getting greeted with the following error message.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\Dale Piper\Desktop\xampplite\htdocs\Array.php on line 36

Did you see the output of that code?

Yes. See Above

Why did you modify my code?

The code you posted didn't include PHP opening/closing tags, nor did it declare where the data was coming from. e.g. Database, Table etc.

Can you please read, copy and understand it all before screwing around with it?

Understanding it, come with playing around. I am a hands on person, and mucking around with things at finding out how it works etc, is the way I learn. You can change the way people learn. We all learn in different ways, at within different time frames.

Many Thanks,

[Salchester]

DJMaze,

How come when i change the following code (In Bold)

from:

$records[] = $array;

to:

$records[1] = $array;

The the last record entered gets displayed as record 0 in stead of the second record being displayed?

<?php
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
{
  die('Could not connect: ' . mysql_error());
}

mysql_select_db("WB", $con);

$result = mysql_query("select * from Birds");
header('Content-type: text/plain');

$records = array();
while ($array = mysql_fetch_assoc($result))
{
    $records[0] = $array;
} 

print_r($records);
?>

[redhead]

$records[0] = $array;

You are overwriting index '0' of your array constantly, look at DjM code, it is not indexing a specific location when adding to the array.

[DJMaze]

Originally Posted by Salchester

As far as Redhead code goes, that doesn't work either, I keep getting greeted with the following error message.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\Dale Piper\Desktop\xampplite\htdocs\Array.php on line 36

That one is easy to debug.

PHP Code:

$result = mysql_query("select * from Birds");
if (!$result) { die(mysql_error()); } 


And i want you to tell me what this code does, step by step.

[Salchester]

Originally Posted by redhead

You are overwriting index '0' of your array constantly, look at DjM code, it is not indexing a specific location when adding to the array.

Redhead,

How do I change it so it does index a specific location when adding an entry to the array?

Resulting in each record (Name,Species, Sex) having its own location within the array, starting at 0.

This will be useful as I will need it for later.

Many Thanks,

[DJMaze]

Originally Posted by Salchester

JThe code you posted didn't include PHP opening/closing tags, nor did it declare where the data was coming from. e.g. Database, Table etc.Understanding it, come with playing around. I am a hands on person, and mucking around with things at finding out how it works etc, is the way I learn. You can change the way people learn. We all learn in different ways, at within different time frames.

It didn't have open/close tags because it shouldn't need that at all.
If you did compare codes, you would have noticed my code was just a portion of all the code.

Sorry but it looks like your way of learning stuff is bad. You haven't learn a thing so far.
If you did, tell me what you have learned, code line by code line. Else i don't see any use of replying to you again and again and again by any of us.