现在我们已经有了允许用户输入一个笑话并将其加入到我们的数据库中的程序代码。现在剩下的就是将其加入到我们已做好的笑话显示页面。因为绝大多数的用户只会想要看看笑话,所以我们不想对我们的页面做大的更改,除非用户表示想要添加一个新的笑话。因为这个原因,我们的应用程序应该是一个多功能的页面。下面是程序的代码:
<HTML>
...
<BODY>
<?php
// If the user wants to add a joke
if (isset($addjoke)):
?>
<FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST>
<P>Type your joke here:<BR>
<TEXTAREA NAME="joketext" ROWS=10 COLS=40 WRAP></TEXTAREA><BR>
<INPUT TYPE=SUBMIT NAME="submitjoke" VALUE="SUBMIT">
</FORM>
<?php
else:
// Connect to the database server
$dbcnx = @mysql_connect("localhost",
"root", "mypasswd");
if (!$dbcnx) {
echo( "<P>Unable to connect to the " .
"database server at this time.</P>" );
exit();
}
// Select the jokes database
if (! @mysql_select_db("jokes") ) {
echo( "<P>Unable to locate the joke " .
"database at this time.</P>" );
exit();
}
// If a joke has been submitted,
// add it to the database.
if ("SUBMIT" == $submitjoke) {
$sql = "INSERT INTO Jokes SET " .
"JokeText='$joketext', " .
"JokeDate=CURDATE()";
if (mysql_query($sql)) {
echo("<P>Your joke has been added.</P>");
} else {
echo("<P>Error adding submitted joke: " .
mysql_error() . "</P>");
}
}
echo("<P> Here are all the jokes " .
"in our database: </P>");
// Request the text of all the jokes
$result = mysql_query(
"SELECT JokeText FROM Jokes");
if (!$result) {
echo("<P>Error performing query: " .
mysql_error() . "</P>");
exit();
}
// Display the text of each joke in a paragraph
while ( $row = mysql_fetch_array($result) ) {
echo("<P>" . $row["JokeText"] . "</P>");
}
// When clicked, this link will load this page
// with the joke submission form disp