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Author Topic: Newbie PHPer in Need  (Read 1954 times)

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Offline !!!!!Topic starter

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Newbie PHPer in Need
« on: December 10, 2006, 12:28:22 AM »
Okay, I am readin a BOOk on PHP and so I am just writing newbie scripts. Why isn't this working?

Code: [Select]
<?php
$outputString = "Hello world!"; 
$bus = "Thewwra";
$es = "Is a great Company"; 
$or = "50";
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">


<html>

<head>
<title>Hi</title>
</head>
<body>

<?php
if ($or > 58)
 print "58 greater than Variable.";
else
 print "58 is less than Variable.";
break;
echo "$bus $es";
break;
echo "Please join thewwra in our way to sell tons!";
?>

</body>
</html>
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Offline fert

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Re: Newbie PHPer in Need
« Reply #1 on: December 10, 2006, 12:35:07 AM »
change
Code: [Select]
<?php
if ($or > 58)
 print "58 greater than Variable.";
else
 print "58 is less than Variable.";
break;
echo "$bus $es";
break;
echo "Please join thewwra in our way to sell tons!";
?>
to
Code: [Select]
<?php
if ($or > 58)
{
 print "58 greater than Variable.";
}
else
{
 print "58 is less than Variable.";
}
echo $bus.$es;
echo "Please join thewwra in our way to sell tons!";
?>
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Offline !!!!!Topic starter

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Re: Newbie PHPer in Need
« Reply #2 on: December 10, 2006, 12:39:30 AM »
Quote from: fert on December 10, 2006, 12:35:07 AM
change
Code: [Select]
<?php
if ($or > 58)
 print "58 greater than Variable.";
else
 print "58 is less than Variable.";
break;
echo "$bus $es";
break;
echo "Please join thewwra in our way to sell tons!";
?>
to
Code: [Select]
<?php
if ($or > 58)
{
 print "58 greater than Variable.";
}
else
{
 print "58 is less than Variable.";
}
echo $bus.$es;
echo "Please join thewwra in our way to sell tons!";
?>
Umm, that didn't do anything, but thanks for trying. It says: Fatal error: Cannot break/continue 1 level in /home/thewwrac/public_html/mine.php on line 27

Code: [Select]
<?php
$outputString = "Hello world!"; 
$bus = "Thewwra";
$es = "Is a great Company"; 
$or = "50";
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">


<html>

<head>
<title>Hi</title>
</head>
<body>

<?php
if ($or > 58)
{
 print "58 greater than Variable.";
}
else
{
 print "58 is less than Variable.";
}
break;
echo "$bus $es";
break;
echo "Please join thewwra in our way to sell tons!";
?>

</body>
</html>still need some fixin', thanks!
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Offline fert

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Re: Newbie PHPer in Need
« Reply #3 on: December 10, 2006, 12:41:01 AM »
you need to remove th break; from your code, because you only use that when escaping a loop.
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Offline Mr_Pancakes

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Re: Newbie PHPer in Need
« Reply #4 on: December 10, 2006, 12:42:14 AM »
try this instead:

Code: [Select]
if (count($or) > 58)
print "58 greater than Variable.";
else
print "58 is less than Variable.";

also, you don't need those "break"s in your code. try it - it'll work fine without them. another way to create your if statements is with the { and } characters. you might find this method to be easier to organize and view your code when you start using more than 1 line of code in an if statement.

Code: [Select]
if (count($or) > 58) {
       print "58 greater than Variable.";
       // a second line of code is acceptable when you use { and }
       // and even more lines of code
} else {
       print "58 is less than Variable.";
       // even more code
}

hope this helps.
cheers,
.s
« Last Edit: December 10, 2006, 12:43:57 AM by Mr_Pancakes »
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Offline fert

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Re: Newbie PHPer in Need
« Reply #5 on: December 10, 2006, 12:43:55 AM »
count($or) won't work because $or isn't an array.
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Offline !!!!!Topic starter

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Re: Newbie PHPer in Need
« Reply #6 on: December 10, 2006, 12:45:53 AM »
Quote from: fert on December 10, 2006, 12:43:55 AM
count($or) won't work because $or isn't an array.
Okay, it works, but how do I break Echo and Print? Thanks a lot by the way
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Offline Mr_Pancakes

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Re: Newbie PHPer in Need
« Reply #7 on: December 10, 2006, 12:46:44 AM »
actually, upon a second review, count($or) woudn't work regardless because $or is equal to a string while the if statement is comparing to an integer.

try first setting $or equal to an integer as well.

Code: [Select]
$or = 50;
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Offline fert

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Re: Newbie PHPer in Need
« Reply #8 on: December 10, 2006, 12:47:07 AM »
echo "<br >";
« Last Edit: December 10, 2006, 12:49:51 AM by fert »
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Offline Mr_Pancakes

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Re: Newbie PHPer in Need
« Reply #9 on: December 10, 2006, 12:49:19 AM »
you can also insert an HTML break as well.

Code: [Select]
echo "blahhhhh <br />";

and if you wanna get picky, you could add a server line break to it all to line feed your code:


Code: [Select]
echo "blahhhhh <br />\n";
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