PHP/mysql showing blank page

palisim

I think I'm missing something simple but I'm tired so I may need a second pair of eyes🙂

This is just a sample code right now. I want this php script to ead from a recipes database and just display the directions. My page is not displaying anything though when I load it in the browser though. I know that the php is working because when I do view/source no html code comes up. It's not doing anything, yet there are no errors being spit out. Here is the code. Thanks
<html>
<head>
<title>Untitled Document</title>
</head>

<?php
//connect to database server
$connect= @mysqlconnect ("localhost", "username", "password");
If (!connect) {
echo ("Unable to connect to server at this time. Please try again later.");
exit();
}

//select recipes database
If (!@mysql_select_db("Recipes")) {
echo ("Database unavailable at this time. Please try again later.");
exit();
}

//request text of all the recipes
$result=mysql_query ("select directions from Afghanistan");
If (!result) {
echo ("Error performing query: " . mysql_error() . " ");
exit();
}

// Display the text of each directions in a paragraph
while ( $row = mysql_fetch_array($result) ) {
echo(" $row[Directions] ");
}
?>

</body>
</html>

Kudose

Sounds like error reporting isnt setup.

Add this to the top of your script:

ini_set('display_errors', 'On');
error_reporting(E_ALL);

You can also take the error suppresion out of your code (the @) to check for errors.

Pikachu2000

palisim wrote:
I think I'm missing something simple but I'm tired so I may need a second pair of eyes🙂

This is just a sample code right now. I want this php script to ead from a recipes database and just display the directions. My page is not displaying anything though when I load it in the browser though. I know that the php is working because when I do view/source no html code comes up. It's not doing anything, yet there are no errors being spit out. Here is the code. Thanks
<html>
<head>
<title>Untitled Document</title>
</head>

<body text="#000000">
<?php
//connect to database server
$connect= @mysqlconnect ("localhost", "username", "password");
If (!connect) {
echo ("Unable to connect to server at this time. Please try again later.");
exit();
}

//select recipes database
If (!@mysql_select_db("Recipes")) {
echo ("Database unavailable at this time. Please try again later.");
exit();
}




//request text of all the recipes
$result=mysql_query ("select directions from Afghanistan");
If (!result) {
echo ("Error performing query: " . mysql_error() . " ");
exit();
}

// Display the text of each directions in a paragraph
while ( $row = mysql_fetch_array($result) ) {
echo(" $row[Directions] ");
}
?>

</body>
</html>

In line 3, you're missing an underscore in mysqlconnect, should be mysql_connect.

Is there a reason for not using this?

or die( mysql_error() );

Also, if you are on a Mac or Linux box (not sure/don't care about WinDoh's) you can validate the code via command line. You do need to cd to the directory containing the file, however.

php -l <[I]filename[/I]>

bradgrafelman

As Pikachu2000 points out, you have a parse error in your code.

The "blank screen" is likely due to the fact that you have display_errors turned Off (something you definitely want in a production environment). As such, you should be logging PHP errors to a log file somewhere.

Pikachu2000

palisim wrote:
I think I'm missing something simple but I'm tired so I may need a second pair of eyes🙂

This is just a sample code right now. I want this php script to ead from a recipes database and just display the directions. My page is not displaying anything though when I load it in the browser though. I know that the php is working because when I do view/source no html code comes up. It's not doing anything, yet there are no errors being spit out. Here is the code. Thanks
<html>
<head>
<title>Untitled Document</title>
</head>

<body text="#000000">

<?php
//connect to database server
$connect= @mysqlconnect ("localhost", "username", "password");
If (!connect) {
echo ("Unable to connect to server at this time. Please try again later.");
exit();
}

//select recipes database
If (!@mysql_select_db("Recipes")) {
echo ("Database unavailable at this time. Please try again later.");
exit();
}

//request text of all the recipes
$result=mysql_query ("select directions from Afghanistan");
If (!result) {
echo ("Error performing query: " . mysql_error() . " ");
exit();
}

// Display the text of each directions in a paragraph
while ( $row = mysql_fetch_array($result) ) {
echo(" $row[Directions] ");
}
?>

</body>
</html>

Just noticed something else:

echo("<P> $row[Directions]  </P>");

Should read:

echo("<P>" . $row[ 'directions' ] . "</P>");

bradgrafelman

Actually, neither will cause any parse errors/warnings, so it's really a matter of preference.

laserlight

Actually, neither will cause any parse errors/warnings, so it's really a matter of preference.

No, since Pikachu2000's "correction" is incorrect as the string index should be quoted.

bradgrafelman

Oops, didn't notice that. :o Anyway, all I meant was that string indexes can appear in double quoted strings so long as they aren't quoted.

adguru

Use php.net for specific function descriptions.

thanks
http://www.adguru.org

Pikachu2000

laserlight wrote:
No, since Pikachu2000's "correction" is incorrect as the string index should be quoted.

Missed that... Edited post.

palisim

ok looking better thanks. But is there any reason I get a parse error referring to an undeclared variable in this section? As you can see, the variable is declared.....
oh yeah and I had the error debugging turned on. I think that was on by default. Is that a firefox thing? That's why I was so confused about it coming up blank🙂

//request text of all the recipes
$result=mysql_query ("select directions from Afghanistan");

// Display the text of each directions in a paragraph
while ( $esult = mysql_fetch_array($result) ) {
echo("" $result[Directions] "");
}

Pikachu2000

palisim wrote:
ok looking better thanks. But is there any reason I get a parse error referring to an undeclared variable in this section? As you can see, the variable is declared.....
oh yeah and I had the error debugging turned on. I think that was on by default. Is that a firefox thing? That's why I was so confused about it coming up blank🙂

//request text of all the recipes
$result=mysql_query ("select directions from Afghanistan");

// Display the text of each directions in a paragraph
while ( $esult = mysql_fetch_array($result) ) {
echo("" $result[Directions] "");
}

I didn't get an undeclared variable warning when I pasted it and checked it. However, your variable in the while statement reads $esult, so the echo() statement needs to reflect that.

I know it's a matter of preference, but I usually write queries in this form

$q1 = "SELECT directions FROM Afghanistan";
$r1 = mysql_query( $q1 );

while( $a1 = mysql_fetch_array( $r1, MYSQL_ASSOC ) )
{
     // Note that this echo() statement now has the concatenation operators in it also!
     echo( "<P>" . $a1[ 'Directions' ] . "</P>" );
}

laserlight

Pikachu2000 wrote:
I know it's a matter of preference, but I usually write queries in this form

I hope you use better variable names than $q1, $r1 and $a1.

palisim, concerning your most recent code snippet, Pikachu2000 is right to say that you have a typo error ($esult versus $result). Not only that, but assigning the row result to the result set pointer variable is pretty bad as it messes up the loop. The correct variant of your code would then be:

$query = "SELECT directions FROM Afghanistan";
$result = mysql_query($query);

while ($row = mysql_fetch_assoc($result))
{
    echo "<p>$row[Directions]</p>";
}

It is arguably better to use concatenation instead of variable interpolation (embedding PHP variables in a double quoted PHP string) here since readability is not affected and concatenation is slightly faster than variable interpolation. I say "arguably", because the speed difference is usually negligible.