[SOLVED] Listing / mysql

rugzo · September 25, 2008

Hi all,

I am new in the forum and also a beginner in php section. I am trying to do a simply listing in a php page. There is a simply table in mysql and the values in there should be listed in the page.

The code -->

<html>

<head>

</head>

<body>

<tr>

<td width="16%">Soyadı</td>

<td width="17%">Kullanıcı Adı</td>

<td width="13%">Şifre</td>

<td width="21%">email</td>

<td width="17%">Görüş</td>

</tr>

<?php

@mysql_connect ("localhost","root","123") or die ("nix sql") ;

@mysql_select_db ("uye") or die ("nix db") ;

$sql = mysql_query ("SELECT * FROM uyeler") ;

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

?>

<tr>

</tr>

<?

}

?>

</table>

</body>

</html>

But when i call the page i can see the table and the row numbers are also right but it look like ->

Adı Soyadı Kullanıcı Adı Şifre email Görüş

Array(0) Array(1) Array(2) Array(3) Array(4) Array(5)

the first line is ok. but the values are not listed instead of the names you just see "array(0)...

Can anyone help me. Where is my mistake. I am following the exact instructions of my php book.

Thanks very much...

jonsjava · September 25, 2008

try this:

<html>
<head>
<meta http-equiv="content-language" content="tr">
<meta http-equiv="content-type" type="text/html; charset=windows-1254">
</head>
<body>

<table border="1" bordercolor="#90022e">
<tr>
<td width="16%"><font face="Arial Black" size="2">Adı</font></td>
<td width="16%"><font face="Arial Black" size="2">Soyadı</font></td>
<td width="17%"><font face="Arial Black" size="2">Kullanıcı Adı</font></td>
<td width="13%"><font face="Arial Black" size="2">Şifre</font></td>
<td width="21%"><font face="Arial Black" size="2">email</font></td>
<td width="17%"><font face="Arial Black" size="2">Görüş</font></td>
</tr>

<?php

@mysql_connect ("localhost","root","123") or die ("nix sql") ;

@mysql_select_db ("uye") or die ("nix db") ;

$sql = mysql_query ("SELECT * FROM uyeler") ;
while ($liste = mysql_fetch_assoc($sql)) {
$liste0 = $liste[0];
$liste1 = $liste[1];
$liste2 = $liste[2];
$liste3 = $liste[3];
$liste4 = $liste[4];
$liste5 = $liste[5];
echo <<<END
<tr>
  <td width="%16"><font face="verdana" size="2">$liste0</font></td>
  <td width="%16"><font face="verdana" size="2">$liste1</font></td>
  <td width="%17"><font face="verdana" size="2">$liste2</font></td>
  <td width="%13"><font face="verdana" size="2">$liste3</font></td>
  <td width="%21"><font face="verdana" size="2">$liste4</font></td>
  <td width="%17"><font face="verdana" size="2">$liste5</font></td>
</tr>
END;
}
?>
</table>
</body>
</html>

Push Eject · September 25, 2008

Does this help?

<tr>
  <td width="%16"><font face="verdana" size="2"><?php echo $liste[0]; ?></font></td>
  <td width="%16"><font face="verdana" size="2"><?php echo $liste[1]; ?></font></td>
  <td width="%17"><font face="verdana" size="2"><?php echo $liste[2]; ?></font></td>
  <td width="%13"><font face="verdana" size="2"><?php echo $liste[3]; ?></font></td>
  <td width="%21"><font face="verdana" size="2"><?php echo $liste[4]; ?></font></td>
  <td width="%17"><font face="verdana" size="2"><?php echo $liste[5]; ?></font></td>
</tr>

rugzo · September 25, 2008

I tried that, but the result is that this time i just got the first line-->

Adı Soyadı Kullanıcı Adı Şifre email Görüş --->the headers of the table.

Thanks for the reply...

JsusSalv · September 25, 2008

Hell rugzo:

Give this a try:

<?php
// Create database connection.
@ mysql_connect ("localhost","root","123") or die ("nix sql") ;
@ mysql_select_db ("uye") or die ("nix db") ;

$sql = "SELECT * FROM uyeler";
$result = mysql_query($sql) or die(mysql_error());

?>
<table>
    <tr>
        <th>Adı</th>
        <th>Soyadı</th>
        <th>Kullanıcı Adı</th>
        <th>Şifre</th>
        <th>email</th>
        <th>Görüş</th>
    </tr>
   <?php
        while ($liste = mysql_fetch_array($result)) {
   ?>
   <tr>
      	<td><?php echo $liste[Adı]; ?></td>
        <td><?php echo $liste[soyadı]; ?></td>
        <td><?php echo $liste[KullanıcıAdı]; ?></td>
        <td><?php echo $liste[Şifre]; ?></td>
        <td><?php echo $liste[email]; ?></td>
        <td><?php echo $liste[Görüş]; ?></td>
    </tr>
    <?php }
        mysql_free_result($result);
    ?>
</table>

rugzo · September 26, 2008

Thanks for the reply but the result is -> Parse error: syntax error, unexpected ';' in C:\xampp\htdocs\t.php on line 23

I am also trying to understand why the example from my php book isn't working. It creates the table but list all values as array()...

Can the reason be the version of php i am using 5.2.5 or maybe the version of mysql...

Or can it be that i have to make a configuration in php.ini...

I am really lost ???

CroNiX · September 26, 2008

Its because you were using parenthesis on your array elements when they should be square brackets.

Wrong:

<? echo "$liste(0)"; ?>

Right:

<?php echo $liste[0]; ?>

You shouldnt need quotes around your array element.

Also, its bad practice to use php short tags (<?), use the full tag (<?php)

rugzo · September 26, 2008

Yesss, thats it...

Thanks very much Cronix

Push Eject · September 29, 2008

LOL - isn't that what I posted in my first reply?

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