Ik krijg de volgende foutmelding:
PHP Warning: sqlite_open() [function.sqlite-open]: malformed database schema - unable to open a temporary database file for storing temporary tables in C:\Inetpub\wwwroot\php5\eindoefeningen\test.php on line 3 PHP Warning: sqlite_query() expects parameter 1 to be resource, string given in C:\Inetpub\wwwroot\php5\eindoefeningen\test.php on line 5 PHP Warning: sqlite_fetch_array() expects parameter 1 to be resource, null given in C:\Inetpub\wwwroot\php5\eindoefeningen\test.php on line 6
Dit is mijn code:
Ik doe iets verkeerd, alleen wat. Kan iemand me misschien even helpen hiermee?
John83
PHP Warning: sqlite_open() [function.sqlite-open]: malformed database schema - unable to open a temporary database file for storing temporary tables in C:\Inetpub\wwwroot\php5\eindoefeningen\test.php on line 3 PHP Warning: sqlite_query() expects parameter 1 to be resource, string given in C:\Inetpub\wwwroot\php5\eindoefeningen\test.php on line 5 PHP Warning: sqlite_fetch_array() expects parameter 1 to be resource, null given in C:\Inetpub\wwwroot\php5\eindoefeningen\test.php on line 6
Dit is mijn code:
PHP:
<?php
function show_db(){
$db= sqlite_open("bc_php5.sdb");
$sql= "SELECT * FROM werknemers;";
$result = sqlite_query($db, $sql);
while ($rij = sqlite_fetch_array($result)){
echo ("<tr><td>". $rij['id'] . " </td> " .
"<td>" . $rij['voornaam'] . " " .
$rij['achternaam'] . " </td> " .
"<td>" . $rij['kamer'] . " </td> " .
"<td>" . $rij['toestel'] . " </td></tr>\n ");
}
}
?>
<html>
<head>
<title>Gegevens in HTML</title>
</head>
<body>
<table border="1" width="80%" align="center">
<tr>
<td colspan="4"><h2 align="center">Werknemers</h2></td>
</tr>
<tr>
<th>ID</th>
<th>Naam</th>
<th>Kamer</th>
<th>Toestel</th>
</tr>
<?php
show_db()
?>
<br>
</body>
</html>Ik doe iets verkeerd, alleen wat. Kan iemand me misschien even helpen hiermee?
John83
