How can I generate and store a completely unescaped string in PHP?
In JS I can do it this way:
// Generates unescaped string
const unescaped = String.raw`\/pass\nfoo=`;
// Interacts with unescaped string without changing it
const pass = unescaped.concat(`i15`);
console.log(pass);
// -> \/pass\nfoo=i15
while without String.raw it'd work differently:
console.log(`\/pass\nfoo=`);
// 1. /pass
// 2. foo=
How could I achieve the same in PHP?
Seems like the manual doesn't mention this topic.
In general, you could just use single quoted strings:
$unescaped = '\/pass\nfoo=';
$unescaped .= 'i15';
echo $unescaped;
Output:
\/pass\nfoo=i15
However there are two exceptions to this, when you have \\ or \' in your string, both of which are treated as an escaped character i.e. \ or '. e.g.
$unescaped = '\/pa\\s\'s\nfoo=';
$unescaped .= 'i15';
echo $unescaped;
Output:
\/pa\s's\nfoo=i15
You can work around that limitation by using nowdoc syntax:
$unescaped = <<<'EOT'
\/pa\\s\'s\nfoo=
EOT;
$unescaped .= <<<'EOT'
i15
EOT;
echo $unescaped;
Output:
\/pa\\s\'s\nfoo=i15
Demo on 3v4l.org
Related
I would like to replace array string values which contains multiple special characters to normal one.
Tried Code (array values):
$data['ENV_TEST'] = "rambo";
$data['ENV_DEV'] = "Project Bribara<"${ENV_TEST}"#gmail.com>"
echo str_replace("${ENV_TEST}", $data['ENV_DEV'], $data['ENV_DEV']);
also tried
echo str_replace("\"${ENV_TEST}\"", $data['ENV_DEV'], $data['ENV_DEV']);
Expected:
"Project Bribara<rambo#gmail.com>"
Actual:
"Project Bribara<"${ENV_TEST}"#gmail.com>"
How can I get the expected output?
You should on PHP strings sometime. The important part about double quoted strings for your question is that you need to put a backslash before every $ and every " inside your string. Your code will then look like this:
$data['ENV_TEST'] = "rambo";
$data['ENV_DEV'] = "Project Bribara<\"\${ENV_TEST}\"#gmail.com>";
echo str_replace("\${ENV_TEST}", $data['ENV_TEST'], $data['ENV_DEV']);
//also tried
echo "\n\n";
echo str_replace("\"\${ENV_TEST}\"", $data['ENV_TEST'], $data['ENV_DEV']);
If you use single quoted strings you don't need to escape $ (see the manual), and instead of \", you would need to escape single quotes (but there aren't any in your example).
$data['ENV_TEST'] = "rambo";
$data['ENV_DEV'] = 'Project Bribara<"${ENV_TEST}"#gmail.com>';
echo str_replace("\${ENV_TEST}", $data['ENV_TEST'], $data['ENV_DEV']);
//also tried
echo "\n\n";
echo str_replace('"${ENV_TEST}"', $data['ENV_TEST'], $data['ENV_DEV']);
I also fixed a missing semicolon and replaced DEV with TEST in one place.
String concatenation in PHP is done through the . operator. Your code would be:
$data['ENV_DEV'] = "Project Bribara<".$data['ENV_TEST']."#gmail.com>"
Let's say this is my script down here, as you can see I've used multiple " and '. These quotations conflict in ending the current php variable, so it basically sees this:
$message = "<?php echo '<div class="
As a string, whilst the quotation is only to define the class, not to end the variable. I've tried using ' but then it conflicts with the echo, so I'm kinda stuck at the moment.
<?php
$message = "
<?php
echo '<div class="gebruiker">';
$fh = fopen('_gebruiker.txt','r');
while ($line = fgets($fh)) {
echo($line);
}
fclose($fh);
echo '</div>';
?>
";
**MORE PHP CODE HERE**
?>
How can I use multiple quotations in one PHP script without them having conflicts.
If you use single quotes outside, you need to escape all single quotes inside, but can use double quotes and the dollar char without escaping.
If you use double quotes outside, you need to escape all double quotes and dollar chars inside, but can use single quotes without escaping.
If you use a heredoc string, you need to escape dollar chars but can use both single and double quotes without escaping.
If you use a nowdoc string, you do not need to escape anything unless you have FOO; in the string at the beginning of a new line.
So the solution is to use a nowdoc string:
$message = <<<'EOF'
your stuff with " or ' or $ here!
EOF;
Adding a backslash so PHP can recognize just only double quotes or quotes when escaped.
Example:
echo "<div class=\"gebruiker\">";
I suggest using either Heredoc or Nowdoc
Example
$foo = 'test';
// Heredoc
$here = <<<HERE
I'm here, $foo!
HERE;
// Nowdoc
$now = <<<'NOW'
I'm now, $foo!
NOW;
Heredoc will print the contents of $foo when echoed while Nowdoc will simply echo $foo.
In the references I added below you can do more reading up on this subject.
References:
php.net - strings
stackoverflow - advantages of heredoc vs nowdoc
You can escape the quotes with a slash character like so: \"
This should work, you can't open Php tag without closing before.
<?php
$message = '<div class="gebruiker">';
$fh = fopen('_gebruiker.txt','r');
while ($line = fgets($fh)) {
$message .=$line.'<br/>';
}
fclose($fh);
$message.='</div>';
echo $message;
**MORE PHP CODE HERE**
?>
i'm using php on_start and ob_get_contents to echo html and store in a variable. However when I json encode and check the output it doesn't output the entire string. Could anyone help point out what I'm doing wrong
ob_start();
echo'<img src=\"images/editphotohover.png\"/>\"';
$photo = ob_get_contents();
ob_end_clean();
I get only get the ending anchor tag
in the json encode output
There is no need to escape double quotes here
echo'<a href=\"javascri...
just write this:
echo'<a href="javascri...
Double quotes are kept while in single quotes!
Additionally, note that escaping within single quotes has no effect:
"\t" renders as a TABULATOR character
'\t' renders as \t
The PHP documentation states this:
To specify a literal single quote, escape it with a backslash (\).
To specify a literal backslash before a single quote, or at the end of the string, double it (\\).
Note that attempting to escape any other character will print the backslash too.
Therefore, how about this code:
echo'<a href="javascript:pixlr.edit(
{ image: \'http://mywebite.com/uploads/$photo\',
title: \'' . $photoFileNameProper . '\',
service: \'express\',
exit:\'http://mywebsite.com/home\',
method: \'get\',
locktarget: \'true\',
target: \'http://mywebsite.com/plixr.php\',
locktitle: \'true\'
});"
id = "uploadedPhoto"
title = "click to enhance photo">
<img src="images/editphotohover.png"/>
</a>'
;
$city = 'London Paris Lisabona';
And i need print this string in textarea.
How print city in new line?
I need in textarea get this:
London
Paris
Lisabona
Code:
$city = 'London\nParis\nLisabona';
echo '<textarea>'.$city.'</textarea>';
result:
London\nParis\nLisabona
In general: Use \n for line breaks.
In your case (only works of cites don't consist of two words, i.e. each word must be a city):
$city = str_replace(' ',"\n", $str); // generates 'London\nParis\nLisabona'
Or if possible build the string with \n instead of spaces from the beginning.
Update:
Escaped character sequences like \n are only processed in double quoted strings. They are taken literally in single quoted strings (with two exceptions). Read more in the documentation:
To specify a literal single quote, escape it with a backslash (\). To specify a literal backslash, double it (\\). All other instances of backslash will be treated as a literal backslash: this means that the other escape sequences you might be used to, such as \r or \n, will be output literally as specified rather than having any special meaning.
Thus, you have to declare your strings as
$cities = "London\nParis\nLisabona";
Further note:
Whenever possible avoid echoing HTML with PHP. It makes it more difficult to debug the HTML. Instead, embed the PHP into HTML like so:
<?php
$cities = "London\nParis\nLisabona";
?>
<textarea><?php echo $cities; ?></textarea>
<?php
$city = "London\nParis\nLisabona";
?>
<textarea rows="3" cols="20">
<?php echo $city; ?>
</textarea>
$city = str_replace(' ', "<br />", $city);
If you echo it in HTML.
<textarea><?= str_replace(" ", "<br />", $city); ?></textarea>
If you want "\n" to be converted to line breaks, you need to use double-quotes instead of single quotes.
i.e.
$foo = 'a\nb\nc\n';
echo $foo;
> a\nb\nc\n
$foo = "a\nb\nc\n";
echo $foo;
> a
> b
> c
works in <textarea> form me, lines might get soft-wrapped though (but that is expected)
simple problem baffling me...
i have a function:
function spitHTML() {
$html = '
<div>This is my title</div>\n
<div>This is a second div</div>';
return $html
}
echo $spitHTML();
Why is this actually spitting out the \n's?
Backslashes used in single quote strings do not work as escape characters (besides for the single quote itself).
$string1 = "\n"; // this is a newline
$string2 = '\n'; // this is a backslash followed by the letter n
$string3 = '\''; // this is a single quote
$string3 = "\""; // this is a double quote
So why use single quotes at all? The answer is simple: If you want to print, for example, HTML code, in which naturally there are a lot of double quotes, wrapping the string in single quotes is much more readable:
$html = '<div class="heading" style="align: center" id="content">';
This is far better than
$html = "<div class=\"heading\" style=\"align: center\" id=\"content\">";
Besides that, since PHP doesn't have to parse the single quote strings for variables and/or escaped characters, it processes these strings a bit faster.
Personally, I always use single quotes and attach newline characters from double quotes. This then looks like
$text = 'This is a standard text with non-processed $vars followed by a newline' . "\n";
But that's just a matter of taste :o)
Because you're using single quotes - change to double quotes and it will behave as you expect.
See the documentation for Single quoted strings.
Change ' to " :) (After that, all special chars and variable be noticed)
$html = "
<div>This is my title</div>\n
<div>This is a second div</div>";