Why should use linebreak '\n' - php

I need some advice ...
I'd like to know if it is a good practice to use in a code line breaks "\n"? What is the purpose?
$my_string .= "\n" . "<p>Some values</p>" . "\n";
Till now I haven't use it and I'd really like to know ... your opinion.
Thanks

You only need to use "\n" to make the resulting html code neater/easier to read. It is not necessary.

The new line makes your generated code easier to read. You might think noone should read your code, however if you run into some problems the generated code is easier to read for debugging purposes as well.

\n can be used when streaming over sockets as well. Sometimes you need to use \r, depending on the operating system.

This is very common mistake of the newbie programmers.
They never have an idea that the result of the PHP script execution is plain HTML code.
And sometimes a programmer have to sort things out with that HTML.
While it's just impossible if there are no linegreaks in the code.
Anyway, a good practice would be
$my_string = "Some values";
?>
<p><?=$my_string</p>
so, you won't need no special linebreaks in the PHP code.
Also, there are some cases where you have to use linebreaks.
For example, if you are composing an HTML email message, you hve to add linebreaks, or they will be forcibly added in the unexpected places.

at time of writing to break line, means display text after line-break to next line
like enter in some text editor...

The only purpose is for people using the "View Source" feature of the browser - basically no one. While it is considered good practice, I almost never do (because it's pointless) :)

for pre-tag (pre formatted text):- http://webdesign.about.com/od/htmltags/f/blfaqpre.htm
so, you can do this in SO
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\n is just an escape for a new line. An escape is not really required in PHP, but it makes it seamless to output a new line mark.

Related

Replace // comments by /* comments */ Except in URLs [duplicate]

I need to remove the comment lines from my code.
preg_replace('!//(.*)!', '', $test);
It works fine. But it removes the website url also and left the url like http:
So to avoid this I put the same like preg_replace('![^:]//(.*)!', '', $test);
It's work fine. But the problem is if my code has the line like below
$code = 'something';// comment here
It will replace the comment line with the semicolon. that is after replace my above code would be
$code = 'something'
So it generates error.
I just need to delete the single line comments and the url should remain same.
Please help. Thanks in advance
try this
preg_replace('#(?<!http:)//.*#','',$test);
also read more about PCRE assertions http://cz.php.net/manual/en/regexp.reference.assertions.php
If you want to parse a PHP file, and manipulate the PHP code it contains, the best solution (even if a bit difficult) is to use the Tokenizer : it exists to allow manipulation of PHP code.
Working with regular expressions for such a thing is a bad idea...
For instance, you thought about http:// ; but what about strings that contain // ?
Like this one, for example :
$str = "this is // a test";
This can get complicated fast. There are more uses for // in strings. If you are parsing PHP code, I highly suggest you take a look at the PHP tokenizer. It's specifically designed to parse PHP code.
Question: Why are you trying to strip comments in the first place?
Edit: I see now you are trying to parse JavaScript, not PHP. So, why not use a javascript minifier instead? It will strip comments, whitespace and do a lot more to make your file as small as possible.

Proper Storing of DataURIs in a Variable

There is probably a much more elegant way to say the title, but that's the best I could come up with because frankly I am feeling silly that I don't know the answer and even have to ask but that's how you learn, so no shame necessary.
I have stored dataURIs such as :
data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAADAAAAAwCAYAAABXAvmHAAAE5klEQVR4nO1ZS0hjVxg+9xEJVFAnCzcuurVC3XfTWpqMMTGJbrSFQumAxhERFdE4YK4Bo5kY8zAvo0434nKghW46pXRhH5sKBXHXdikyDjGK0ETj7f/f5KY3N68bySSh9YPP/5x7Xv9/znfOubkS8oAH/L9B1XtAnufVYNTZbBqYpCgqVaKuCgySxiwwBXVvpHXYt+hrKajn5+fjNzc3hGEYolarT1ZXV0NLS0t74FwSK4DjLQ6H48nCwsJkKpXqSafTQl2v19sKxQ0PIAcM4vr6+r2rq6uQzWZ7ur+//8nd3R1ZXFz8Hp73YDnLsoSmaQwKm/DyPuj6u03+drvd7Z2dnf1dXV2fazSaX2DmSSKR6Dk+Pn55cnLy8vLysgeftbW1/Qz1Puvo6ND5fL53iGz2EXXfA3Kgzp1O59jZ2VkQZxmJzkNgT+12+w6kb8u1b3gACAwCJPRjPB7/APM48y6X68NKziMaIaEC4MnS2toaQr0jIR1U4jyioZtYCtiw57hZEbe3t+dK2zXFCiC6u7uP0eIeWFlZ+Ulpu6ZZgZGRkcTh4WHV7ZpmBUiRM14JGGnm4OCAHx4e5pBDaIeGuCGgBa3FwlmAZrRmM2c2WzgTWpOJM5nM3CDawUHBGsEajUbBGsAaDECjgRswAAfQDnD6AaAerZ7r1+u5346Onp2/fi348ceffz3T6rSc7nE/8DGn1ek4nS5jtVod9+rVdyuiz3kSQv2Njo4qnyYxQ4lpXpop3qbog8zDmZkZwQerdbzk+F6vLy9fEAACr/O8cSQD8lKvs/VzSfxDSQLgC5pncnx5vcAplOtaCpWKlYyfQd4euJcICSk62D16yU1guftVPhZbrFSs07//phae5fDtpx3KKpabEVlZUQmJuKvJzFaPcsPKJZQfgCyRfhsBlNK/uF8q6FGRhETUfgVq0GGzSogvMn6pelIUSCjD7HFarwCqOMbkQdZXQhXOf2V9lJWQmMqcw/VaAekwFTexLC+TEC+KUcjXOoDip091g1QlodrcsMqhZBMrkpBY5dcnj/IqU8J1T7LvO7lWxUx+eUnty6+lyqgsIQkoaQk6gY5nH2a+HuT3WvWC3WOF5atEy0qFXlmGFsjkmPmxzTJMljRRscy/eUgzbMYWMNtWVUAog7dLlYQiVLLn+eUV3kb3Xrwgu3t7And2d0lsBxkj27FtEt1GRkkkGiXhCDJCQuEwCYbCJBQKkWAwRLa2gsAtEghsEX8gQHwBP/H5/cTr9xGvz0c2vV7i8W6SjU2wHrAeD3FveIS0OMPP3Rtk3fWcrK27yNraOnECV51rQGfBvmyK70IIcJydmJgQvrxFIhGWoqi0knbN9Jv4XmiaAGZnZ98X09PT071K2zVNACCh9qxF0660XdPsgcnJyQD8Fp7CNMMwgXA4PN1onxRjeXlZY7Va42NjYzwS03Nzc4p+fzZ8BWw226NEIvFDOp3upbI3I34VoWn695aWlr5AIBAv177uewA0TjscDs3U1FQf8MuLi4sjdF5whqZnkRgIBNGbTCaPYDW+GB8f/6hUf434Nkqdnp6ei//MEP+hAbp3wGUofLWC+6ANjB3K3gV+hc9AWiQWixUopqGnEDoOrxnfAD+GTWuHPI+MRqMc2D7g1xX7qIejUqCEJOMKL1/odL39eMAD/iv4B7EvTT8yUT9VAAAAAElFTkSuQmCC
in a variable by simply doing: $some_var = "INSERT_ABOVE"; and has usually worked fine, but I am finding in a piece of code I am working on right now that the code breaks anytime one of the variable=>datauri scheme I showed above is used but deleting that line, fixes the code.
So I am not sure ?WHY? this particular datauri is breaking the code and shouldn't anything inside "" QUOTATIONS be treated as a literal string? Is there a best practice or should I say recommended way of storing datauris as variables to be used through the code, so it doesn't break?
Any thoughts and/or suggestions would be greatly appreciated, TIA.
PS. This wasn't just due to the backslash at the end, it was also that the string by itself wasn't being respected, now it is storing into variable like this, thanks. Since the title most accurately describes what was being achieved, I think its more appropriate.
Use Heredoc string quoting. It's another way to represent strings in PHP. It can eliminate your problem of ending quote getting escaped by string data.
$str = <<<EOD
data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAADAAAAAwCAYAAABXAvmHAAAE5klEQVR4nO1ZS0hjVxg+9xEJVFAnCzcuurVC3XfTWpqMMTGJbrSFQumAxhERFdE4YK4Bo5kY8zAvo0434nKghW46pXRhH5sKBXHXdikyDjGK0ETj7f/f5KY3N68bySSh9YPP/5x7Xv9/znfOubkS8oAH/L9B1XtAnufVYNTZbBqYpCgqVaKuCgySxiwwBXVvpHXYt+hrKajn5+fjNzc3hGEYolarT1ZXV0NLS0t74FwSK4DjLQ6H48nCwsJkKpXqSafTQl2v19sKxQ0PIAcM4vr6+r2rq6uQzWZ7ur+//8nd3R1ZXFz8Hp73YDnLsoSmaQwKm/DyPuj6u03+drvd7Z2dnf1dXV2fazSaX2DmSSKR6Dk+Pn55cnLy8vLysgeftbW1/Qz1Puvo6ND5fL53iGz2EXXfA3Kgzp1O59jZ2VkQZxmJzkNgT+12+w6kb8u1b3gACAwCJPRjPB7/APM48y6X68NKziMaIaEC4MnS2toaQr0jIR1U4jyioZtYCtiw57hZEbe3t+dK2zXFCiC6u7uP0eIeWFlZ+Ulpu6ZZgZGRkcTh4WHV7ZpmBUiRM14JGGnm4OCAHx4e5pBDaIeGuCGgBa3FwlmAZrRmM2c2WzgTWpOJM5nM3CDawUHBGsEajUbBGsAaDECjgRswAAfQDnD6AaAerZ7r1+u5346Onp2/fi348ceffz3T6rSc7nE/8DGn1ek4nS5jtVod9+rVdyuiz3kSQv2Njo4qnyYxQ4lpXpop3qbog8zDmZkZwQerdbzk+F6vLy9fEAACr/O8cSQD8lKvs/VzSfxDSQLgC5pncnx5vcAplOtaCpWKlYyfQd4euJcICSk62D16yU1guftVPhZbrFSs07//phae5fDtpx3KKpabEVlZUQmJuKvJzFaPcsPKJZQfgCyRfhsBlNK/uF8q6FGRhETUfgVq0GGzSogvMn6pelIUSCjD7HFarwCqOMbkQdZXQhXOf2V9lJWQmMqcw/VaAekwFTexLC+TEC+KUcjXOoDip091g1QlodrcsMqhZBMrkpBY5dcnj/IqU8J1T7LvO7lWxUx+eUnty6+lyqgsIQkoaQk6gY5nH2a+HuT3WvWC3WOF5atEy0qFXlmGFsjkmPmxzTJMljRRscy/eUgzbMYWMNtWVUAog7dLlYQiVLLn+eUV3kb3Xrwgu3t7And2d0lsBxkj27FtEt1GRkkkGiXhCDJCQuEwCYbCJBQKkWAwRLa2gsAtEghsEX8gQHwBP/H5/cTr9xGvz0c2vV7i8W6SjU2wHrAeD3FveIS0OMPP3Rtk3fWcrK27yNraOnECV51rQGfBvmyK70IIcJydmJgQvrxFIhGWoqi0knbN9Jv4XmiaAGZnZ98X09PT071K2zVNACCh9qxF0660XdPsgcnJyQD8Fp7CNMMwgXA4PN1onxRjeXlZY7Va42NjYzwS03Nzc4p+fzZ8BWw226NEIvFDOp3upbI3I34VoWn695aWlr5AIBAv177uewA0TjscDs3U1FQf8MuLi4sjdF5whqZnkRgIBNGbTCaPYDW+GB8f/6hUf434Nkqdnp6ei//MEP+hAbp3wGUofLWC+6ANjB3K3gV+hc9AWiQWixUopqGnEDoOrxnfAD+GTWuHPI+MRqMc2D7g1xX7qIejUqCEJOMKL1/odL39eMAD/iv4B7EvTT8yUT9VAAAAAElFTkSuQmCC\
EOD;
echo $str;

Do I really need to add "\n" to every PHP statement that I write in a script if I dont want my output to be like totally messed up?

I'm learning PHP and writing it and executing in the browser is cumbersome.
So I write it as a script and execute it on the terminal, such as
me#machine $ php script.php
However, it seems to me, all statements are printed to the same line, if not explicitly a newline character is also printed.
<?php
echo "Hello World.\n";
?>
If I omit \n, I end up with
me#machine $ php hello_world.php
Hello World. > me#machine $
which kind of is lame.
Do I really, like really really (as in "totally really"), need to type \n for every statement I like to test?
You've got a choice:
Include a \n on the end, and have a line feed.
Leave out the \n and don't have a line feed.
It's up to you. No, you don't need to have it there, but if you want to output text to the command line, you probably do want it.
I guess there's one other alternative. Since PHP outputs content that is outside of the <?php .. ?> tags as plain text, you could just put a blank line at the end of your code after the final ?>. That will cause PHP to output a new line at the end without you needing to write \n.
But to be honest, putting the \n in your string is better coding practice. (And frankly, \n isn't exactly the worst thing in the world to having in your code. if you can't cope with the horrors of seeing \n in your code, then you're going to have a hard time reading most program code anyway... just wait to you learn Regex!!!)
No. You might write your own writeLn() function and maybe use PHP_EOL instead of \n, depending on what the script is for. (New-line string differ across systems, and PHP_EOL is your server's version of new-line, so it makes your script portable at least in regard of the running environment.)
Yes, you need to use \n to print a new line.
PHP will only print that which you tell it to print. So, if you want a new line then you need to print a new line.
Depending on your requirements.
\n is representing a new line break.
So if you are testing a single command with a single echo, from command line you might beable to go without the newline break.
If you have multiple echos, all the output will be jumbled into one long text.
function NewLine ($Text)
{
$Text = $Text."\n";
return $Text;
}
echo NewLine('This Text Will Have A New Line Appended To The End');

Removing Break Lines

I've asked this question before but I didn't seem to get the right answer. I've got a problem with new lines in text. Javascript and jQuery don't like things like this:
alert('text
text);
When I pull information from a database table that has a break line in it, JS and jQuery can't parse it correctly. I've been told to use n2lbr(), but that doesn't work when someone uses 'shift+enter' or 'enter' when typing text into a message (which is where I get this problem). I still end up with separate lines when using it. It seems to correctly apply the BR tag after the line break, but it still leaves the break there.
Can anyone provide some help here? I get the message data with jQuery and send it off to PHP file to storage, so I'd like to fix the problem there.
This wouldn't be a problem normally, but I want to pull all of a users messages when they first load up their inbox and then display it to them via jQuery when they select a certain message.
You could use a regexp to replace newlines with spaces:
alert('<?php preg_replace("/[\n\r\f]+/m","<br />", $text); ?>');
The m modifier will match across newlines, which in this case I think is important.
edit: sorry, didn't realise you actually wanted <br /> elements, not spaces. updated answer accordingly.
edit2: like #LainIwakura, I made a mistake in my regexp, partly due to the previous edit. my new regexp only replaces CR/NL/LF characters, not any whitespace character (\s). note there are a bunch of unicode linebreak characters that i haven't acknowledged... if you need to deal with these, you might want to read up on the regexp syntax for unicode
Edit: Okay after much tripping over myself I believe you want this:
$str = preg_replace('/\n+/', '<br />', $str);
And with that I'm going to bed...too late to be answering questions.
I usually use json_encode() to format string for use in JavaScript, as it does everything that's necessary for making JS-valid value.

Debugging PHP Output

I have a php website that on certain pages is adding a dot or space before the first html tag. I can't figure out where it is coming from - is there a way to debug the code so i can see where it is coming from?
Thanks,
Josh
To help prevents this happening it is considered a good practice to don't end your PHP file with a ?>.
You possibly have some file that are this way (notice the extra space after the ?>):
<?php
// Some code //
?>
If you would remove the ?> at the end, the extra space at the end of the file won't be interpreted as something to output.
For files that contain only PHP code,
the closing tag ("?>") is never
permitted. It is not required by PHP,
and omitting it´ prevents the
accidental injection of trailing white
space into the response.
Source: http://framework.zend.com/manual/en/coding-standard.php-file-formatting.html
Maybe it is a BOM character?
Maybe you should check your templates if you are using them... the problem could be there and not in your main code.
and yes is a GOOD PRACTICE in PHP not to close the ending tag.
There really is no good way to go about debugging this. You need to go through every file the page is hitting and figure out where the output is coming from. If you really wanted to be lazy about it you could do some output buffering, but this isn't the right way to do things.
Problems like this can be difficult to track down. If you're in some kind of framework or system that includes a lot of files, you might try a var_dump(get_included_files()) on the line before your error occurs, and that will give you a place to start. If that isn't sufficient, xdebug might get you further. Things to look out for are space before and after the PHP tags, and functions that might send output.

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