[SOLVED] display table from database in browser

[Marcos01]

Hi all,

 

I would like to display everything from a table in the browser.

Do i need to use fetch_array or fetch_assoc? And what loop would work best?

 

Thanks.

 

The code i have so far:

 

$sql = "SELECT * FROM `trainingsdata` WHERE 1 LIMIT 0, 30 "; 
$result = mysql_query($sql) or die('Eror selecting data');
$row = mysql_fetch_array($result);
$num = mysql_num_rows($result);

[Marcos01]

I added a loop but get Resource id #5 multiple times

 

Here the code so far:

 

$sql = "SELECT * FROM `trainingsdata` WHERE 1 LIMIT 0, 30 "; 
$result = mysql_query($sql) or die('Error selecting data');
$row = mysql_fetch_array($result);
$num = mysql_num_rows($result);


for ($i=0; $i<$num; $i++)
{
echo $result;
}

[PFMaBiSmAd]

Directly out of the mysql_fetch_array section of the php manual -

 

$result = mysql_query("SELECT id, name FROM mytable");

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    // code to access your fields in the $row array
}

[Marcos01]

Thank you PFMaBiSmAd,

 

But can you explain on how to use this?

 

 

[JasonLewis]

$row becomes an array, and the keys are the name of the columns in your table. Inside the loop put this:

 

echo $row['columnName'];
echo "<br />";

 

Where columnName is the name of a column in the table you are selecting data from.

[Marcos01]

thanks

Thanks ProjectFear,

 

I still get Resource id #5 multiple times

 

This is the code i have now:

 

$result = mysql_query("SELECT id, datum, cursus, plaats FROM trainingsdata ");
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    echo $row['columnName'];
echo "<br />";
}

[PFMaBiSmAd]

So, what is the actual data in your table?

[Marcos01]

I have a date, course and place.

 

And want to display them all in the browser. Now I get resource Resource id #5 as many times

as I have rows in the database.

 

Thanks for helping me.

[PFMaBiSmAd]

With the code you just posted, that would mean your data in your table is not what you think it is.

[JasonLewis]

Hmm, try this:

 

$result = mysql_query("SELECT id, datum, cursus, plaats FROM trainingsdata ") or die(mysql_error());
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "<pre>", print_r($row), "</pre><br />";
}

 

And see if anything is outputted then.

[Marcos01]

I am getting this as output:

 

Array

(

    [id] => 1

    [datum] => 2008-09-09

    [cursus] => php

    [plaats] => Washington

)

1

 

 

Array

(

    [id] => 2

    [datum] => 2008-09-09

    [cursus] => php

    [plaats] => New York

)

1

 

 

Array

(

    [id] => 3

    [datum] => 2008-09-09

    [cursus] => php

    [plaats] => Chicago

)

etc etc.

 

The  code so far:

 

$result = mysql_query("SELECT id, datum, cursus, plaats FROM trainingsdata ") or die(mysql_error());

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

echo "<pre>", print_r($row), "</pre><br />";

}

 

I would like to have the output on one row.

 

[JasonLewis]

Strange, before it should've been echoing something. Try this now:

 

$result = mysql_query("SELECT id, datum, cursus, plaats FROM trainingsdata ") or die(mysql_error());

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo $row['cursus'];
echo "<br />";
}

 

See if that outputs anything.

[PFMaBiSmAd]

Strange, before it should've been echoing something.

 

The posted code was likely not the actual code.

[Marcos01]

Thanks  ProjectFear,

 

It now displays all from cursus.

 

So when I add date and place it works. But doesn't show nicely.

 

Here is the code now:

 

$result = mysql_query("SELECT id, datum, cursus, plaats FROM trainingsdata ") or die(mysql_error());

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo $row['datum']." ".$row['cursus']." ".$row['plaats'];
echo "<br />";
}

[JasonLewis]

Well you need to format it then. You can use normal markup here, just your basic HTML. Just make sure you place it in the right spots.

 

For example:

 

$result = mysql_query("SELECT id, datum, cursus, plaats FROM trainingsdata ") or die(mysql_error());

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "Date: ".$row['datum']."<br />Cursus: ".$row['cursus']."<br />Plaats: ".$row['plaats'];
echo "<br /><br />";
}

 

Of course you can do anything else, like place it in a table.

[Marcos01]

PFMaBiSmAd and ProjectFear, I made an error saving the changes to an older file.

 

 

[Marcos01]

ProjectFear and PFMaBiSmAd,

 

Thats what I need. Thank you both for your help!

[Adam]

$result = mysql_query("SELECT id, datum, cursus, plaats FROM trainingsdata ");

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

    echo $row['columnName'];

echo "<br />";

}

 

 

Perhaps the problem...    echo $row['columnName'];

 

...did you replace "columnName" with an actual field name from your table?

 

 

Adam

 

 

[Marcos01]

Hi Adam, yes I did. And works well now.

 

This is the final code:

 

$result = mysql_query("SELECT id, datum, cursus, plaats FROM trainingsdata ") 	or die(mysql_error());
echo '<table cellpadding="5" cellspacing="5"><tr>';
	while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "<td>".$row['datum']."</td><td>".$row['cursus']."</td><td> ".$row['plaats']."</td><tr>";
}
echo "</table>";