php code parsing error - line 27

[syntax_error]

i try executing my php code where you can add a joke to the data base. My Name.php file reports a parsing error on line 27 (referenced) Please help. Thanks. This is what i have:

<?php

//If the user wants to add a joke to the database

if (isset ($addjoke) ) ;

?>

<form action=" <?php echo ($PHP_SELF); ?>" method="post">
<p>Type your Joke here:</P>
<textarea name="joketext" rows="10" cols="40" wrap></textarea><br>
<input type="submit" name="submitjoke" value="submit">
</form>

<?php
else: // THIS IS MY LINE 27

//Connect to database server
$dbcnx =@mysql_connect ("localhost",
"root", "mypassword");

if (!$dbcnx) {
echo (" <P> Unable to connect to the " .
"database server at this time </P>");
exit();
}

// Select 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 to database

if ( "submit" == $submitjoke) {
$sql = "INSERT INTO Jokes SET" .
"JokeText = '$joketext', " .
"JokeDate = CURDATE()";

if ( mysql_query($sql) ) {
echo ("<P> YOur joke has been submitted. </P?");
} else {
echo ("<P> Error submitting joke: " .
mysql_erro() . "</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 all text of each joke in 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 displyed.

echo ( "<P> <A HREF ='$PHP_SELF?addjoke=1'>" .
"Add a Joke!</A></P>");
endif;


?>

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In the example prior to this one in the tutorial where all the jokes are selected and displayed I get this error:

Call to undefined function: array() for my php code: Please look at code commented "this line..."


while ($row = mysql_fetch_array ($result) ) {
echo( "<P>" . $row ("JokeText") . "</P>" ); //This is line where I get an error

[This message has been edited by syntax_error (edited July 21, 2000).]

[This message has been edited by syntax_error (edited July 21, 2000).]

[edshuck]

hi

i would try $row["JokeText"] in place of the $row("JokeText")

best to you. ed

[ChrRmr]

I think this is the correct way (at least it works for me on several sites).

$result = mysql_query("SELECT name, age FROM db WHERE member_id=3");
while($data = mysql_fetch_arry($result)) {
echo $data['name'] ." - ". $data['age'];
}

OR, you can do:

$result = mysql_query("SELECT name, age FROM db WHERE member_id=3");
while($data = mysql_fetch_row($result)) {
echo $data[0] ." - ". $data[1];
}

the array() function is used to creact/populate an array.

Christian

[Kevin Yank]

syntax_error:

The parse error is being caused by your if(isset($addjoke)) line, which needs a colong, not a semicolon, on the end of it:

if (isset ($addjoke) ):



------------------
-Kevin Yank.
http://www.SitePoint.com/
Helping Small Business Grow Online!

[syntax_error]

The square brackets work [ ] also i missed the " : " after the isset function

Now I am getting an SQL syntax_error hmm my eyes are squinting. Let me know of you can catch it!

[Kevin Yank]

$sql = "INSERT INTO Jokes SET" .
"JokeText = '$joketext', " .
"JokeDate = CURDATE()";

Stick those three strings together, and you get this:

$sql = "INSERT INTO Jokes SETJokeText = '$joketext', JokeDate = CURDATE()";

See the problem? :-)


------------------
-Kevin Yank.
http://www.SitePoint.com/
Helping Small Business Grow Online!