MySQL : Can't connect to table

Question

I'm trying to connect my website to the MySQL table.

Here is my code;

<?php
$v1 = mysql_query('SELECT * FROM ('.$nompage.') ORDER BY id');
if($v1 === FALSE) {
    die('Erreur lors de la selection de la table.'); // TODO: meilleur erreur
}
while($liste = mysql_fetch_array($v1)){
?>

<tr>
<th><?php if(isset($liste['nom'])){
  echo $liste["nom"];} ?></th>
<th><a href="<?php echo $liste['lien1']; ?>"><i class="fa fa-external-link"></i>
<?php echo $liste['lien1']; ?></a></th>
<th><?php echo $liste['autre']?></th>
</tr>
<?php
}
?>

The $nompage fonction is ->

$nompage = htmlspecialchars($_GET["nom"]);
$nompage = mysql_real_escape_string($nompage);

I can see 4 row, because I have 4 items in my tables.

This is the error I get on the second and third column.

Notice: Undefined index: lien1 in C:\xampp\htdocs\show.php on line 102
Notice: Undefined index: autre in C:\xampp\htdocs\show.php on line 103

Do not add the table name between quotes. You can escape the name of the table with ` characters. — vaso123, Dec 08 '14 at 12:46
@RakeshSharma Error : `Erreur lors de la selection de la table.` Unable to select the table. — Félix Desjardins, Dec 08 '14 at 12:47
Are you not first connecting? http://php.net/manual/en/function.mysql-connect.php - Plus, where is `$nompage` being first defined? You've got code that's marked as defined "below". Show us the **actual way** you're using it. — Funk Forty Niner, Dec 08 '14 at 12:48
do you have columns called `lien1` and `autre` in the table set in `$nompage` in your database? — RichardBernards, Dec 08 '14 at 12:49
@lolka_bolka `mysql_query('SELECT * FROM '$nompage' ORDER BY id'); if($v1 === FALSE) {` This won't work, I get an error : `Parse error: syntax error, unexpected '$nompage' (T_VARIABLE) in C:\xampp\htdocs\show.php on line 91` — Félix Desjardins, Dec 08 '14 at 12:49
@RichardBernards Yes I have these and `$nompage` is set as;`$nompage = htmlspecialchars($_GET["nom"]); $nompage = mysql_real_escape_string($nompage);` — Félix Desjardins, Dec 08 '14 at 12:51
@InxDev change you sql query `"SELECT * FROM ".$nompage." ORDER BY id"` — Altmish-E-Azam, Dec 08 '14 at 12:53
possible duplicate of [PHP: "Notice: Undefined variable" and "Notice: Undefined index"](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index) — andrewsi, Dec 09 '14 at 05:03

score 0 · Accepted Answer · answered Dec 08 '14 at 12:52

0

The query should be -

'SELECT * FROM '.$nompage.' ORDER BY id' // if $nompage contains the table name

And avoid using mysql. It is deprecated. Try to use mysqli or PDO.

answered Dec 08 '14 at 12:52

Sougata Bose

31,517
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score 0 · Answer 2 · answered Dec 08 '14 at 12:55

This might be annoying for you, but you should use PDO instead of MySQL. MySQL syntax is deprecated. Anyhow, you may try this:

<?php
    //Pretty sure you need to give it either ASC or DESC.
    $v1 = mysql_query('SELECT * FROM "'.$nonpage.'" ORDER BY id DESC');
    if($v1 === FALSE) {
        die('Erreur lors de la selection de la table.'); // TODO: meilleur erreur
    }
        while($liste = mysql_fetch_array($v1)){
    ?>
<tr>
    <th><?php if(isset($liste['nom'])){
    echo $liste["nom"];} ?></th>
    <th><a href="<?php echo $liste['lien1']; ?>"><i class="fa fa-external-link"></i>
    <?php echo $liste['lien1']; ?></a></th>
    <th><?php echo $liste['autre']?></th>
</tr>
<?php
}
?>

score 0 · Answer 3 · answered Dec 08 '14 at 13:32

<?php
$query= 'SELECT * FROM "'.$nonpage.'" ORDER BY id ASC'
$v1 = mysql_query($query);
if($v1 === FALSE) {
    die('Erreur lors de la selection de la table.'); 
}
while($liste = mysql_fetch_assoc($v1)){
?>

<tr>
<th><?php if(isset($liste['nom'])){
  echo $liste["nom"];} ?></th>
<th><a href="<?php echo $liste['lien1']; ?>"><i class="fa fa-external-link"></i>
<?php echo $liste['lien1']; ?></a></th>
<th><?php echo $liste['autre']?></th>
</tr>
<?php
}
?>

use mysql_fetch_assoc insted of mysql_fetch_array

MySQL : Can't connect to table

3 Answers3