Nieuw lid |
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Volgens de foutmelding die ik krijg met dit script:
<?php
$db = mysql_connect("localhost","root","") or die ("Verbinding mislukt");
mysql_select_db("pokemon",$db);
$SQL_statement="SELECT Pokemon_tabel_nummer, Pokemon_naam, Pokemon_soort, Regio FROM begginner pokemons";
$resultset=mysql_query($SQL_statement);
while($data=mysql_fetch_array($resultset)){
echo $data['Pokemon_tabel_nummer'] . " " . $data['Pokemon_naam'] . " " . $data['Pokemon_soort'] . " " . $data['Regio'] . "<br />";
}
mysql_close();
?>
<?php $SQL_statement="SELECT Pokemon_tabel_nummer, Pokemon_naam, Pokemon_soort, Regio FROM begginner pokemons"; echo $data['Pokemon_tabel_nummer'] . " " . $data['Pokemon_naam'] . " " . $data['Pokemon_soort'] . " " . $data['Regio'] . "<br />"; } ?>
Om precies te zijn is dit de foutmelding:
Citaat: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in E:xampphtdocsmysql test.php on line 7
Kan iemand het oplossen alvast bedankt.
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