Hallo zusammen,
welcher Fehler steck in folgenen code:
Ich kriege leider die Fehlermeldung:
für eure Hilfe bedanke ich mich im Voraus.
welcher Fehler steck in folgenen code:
PHP-Code:
<?php
$dbuser="root";
$dbpass="root";
$dbname="dbtest";
$connect = mysql_connect("localhost", $dbuser, $dbpass)
or die("Connection Failure to Database");
echo "Connected to database server<br>";
mysql_select_db($dbname, $connect) or die ($dbname . " Database not found." . $dbuser);
echo "Database " . $dbname . " is selected";
$result = mysql_query("SELECT * FROM tbl-test");
while ($row = mysql_fetch_array($result)) {
$word = $row['word'];
$meanings =$row['meanings'];
echo $word." ".$meanings. "<br>";
}
mysql_close($connect);
?>
Code:
Connected to database server Database dbtest is selected Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\***** line 14
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