i have searched this function on google a lot. However, i can't understand this function clearly.
i have a example:
<?php
//eval dangerous to use
$motto="lksdfasdkf";
$str= "<h1>Welcome</h1><?php echo $motto;?><br/>";
echo $str.'<br />'; //result: welcome
eval("?>"." $str"."<?php echo $motto;"); //error
echo $str;
?>
eval() takes a string and evaluates it as PHP code. Here are some important points to note:
eval() takes PHP Code as it's argument -- not mixed HTML markup. Currently, you're passing a string containing HTML markup.
You don't need to add <?php ... ?> tags in the string. eval() already knows the argument is going to be PHP code (it's supposed to be), so you don't need to tell it
Here's a very short example:
$motto = "lksdfasdkf";
$str = 'echo $motto;';
eval($str); // => lksdfasdkf
Here, the string $str contains the literal string echo $motto;, which is a valid statement in PHP. When you call eval($str); the string gets evaluated as PHP code. In this case, it will echo the contents of the variable.
Note that this wouldn't work if you use double-quotes instead:
$motto = "lksdfasdkf";
$str = "echo $motto;";
eval($str);
If you have error reporting enabled, then you'll get the following error:
Notice: Use of undefined constant lksdfasdkf - assumed 'lksdfasdkf' in
The reason is that variables are not parsed when they're wrapped in single-quotes. When you use double-quotes to define your variable, the variable value gets interpolated into the resulting string, meaning $str will contain the literal string echo lksdfasdkf; -- which is not valid PHP code. The solution is to escape the dollar character to avoid it being interpreted as a variable:
$motto = "lksdfasdkf";
$str = "echo \$motto;";
eval($str); // => lksdfasdkf
eval — Evaluate a string as PHP code - your code also working fine
try
$motto="lksdfasdkf";
$str= "<h1>Welcome</h1>$motto<br/>";
echo $str.'<br />'; //result: welcome
eval("\$str = \"$motto\";");
echo $str;
Related
I'm a designer trying to upgrade myself into a coder-designer. Lately I've been looking into some PHP codes and manuals, then I ran into an example code for the eval() function :
<?php
$string = 'cup';
$name = 'coffee';
$str = 'This is a $string with my $name in it.';
echo $str. "\n";
eval("\$str = \"$str\";");
echo $str. "\n";
?>
This is an example code of eval() function in official PHP website, and although it did help me understand the eval() function, I can't figure out how the example code works. to be more specific, I can't understand why
("\$str = \"$str\";")
results in a merged string.
I really can't figure out why this should work.
Ok, here is what we have:
eval("\$str = \"$str\";")
Look, the string is in double quotes: it means that the $ character will be interpreted as a variable start. So we screen this character with a backslash: \$, then it will "mean" just a normal dollar sign. Also, the double quotes inside the string had to be screened too.
In the end we are getting this string: (I changed the quotes to single so $ dont confuse you): '$str = "$str";'. Look, it looks more like a normal code now :)
Evaling it, PHP will do the following (I removed the outer quotes for convenience):
eval( $str = "$str" );
Notice the double quotes here, too. It means that the variable inside, again, will be parsed/interpreted.
As $str was originally == 'This is a $string with my $name in it.', it will be inserted into the expression, and now it will look like:
$str = "This is a $string with my $name in it.";
And, again, double quotes! It parses and substitutes variables $name and $string, giving us at the end:
$str = "This is a cup with my coffee in it."
Voila!
A mindbreaker, but a really good example to learn the mechanisms.
you should get two console.log like this
This is a $string with my $name in it.';
This is a $cup with my $coffe in it.';
Why? well first you print the value, of $str , without eval, and later, you eval them, basically this happens,
First.
Print $str, without eval.
This is a $string with my $name in it.';
Second
this piece of code, runs eval("\$str = \"$str\";");
Eval replace $string and $name. with the 2 new variables values, which are $cup and $coffe
Hope you get it
Weird situation and I'm not even sure what to call it (hence unable to find any previous posts).
I'm passing in a variable that is a string as function, and then attempting to concat it to another string as such:
function saveFunction($number){
$myStr = "drawer-$number[]";
return $myStr;
}
So the output should be :
saveFunction("2")
"drawer-2[]"
However, because it thinks I am accessing $number as an array because of the brackets, the output is:
"0"
I even tried this:
function saveFunction($number){
$myStr = "drawer-$number";
return $myStr + "[]";
}
And got the same result.
Suggestions?
Double quotes in PHP are tricky. They interpret the string. This is why "drawing-$number" works in the first place.
Sometimes your string doesn't lend itself for automatic interpretation. You can use the always safe concatenation:
return "drawer-" . $number . "[]";
Or use {} to help the automatic detection:
return "drawer-{$number}[]"; // as opposed to: "drawer-{$number[]}"
Or use sprintf:
return sprintf("drawer-%s[]", $number); // %s because $number is actually a string, not an int
You can just concat the variable with the string like this:
<?php
function saveFunction($number){
return $myStr = "drawer-" . $number . "[]";
}
echo saveFunction("2");
?>
Output:
drawer-2[]
For further information see also: http://php.net/manual/en/language.operators.string.php
I have a string (a c code) in a vairable. I want to print it php. But i don't why everything after double quotes in not printing.. pls help. Below is the code.
$answer_something='printf("\\n")';
echo $answer_something;
//OUTPUT: printf(
//WHAT I WANT TO PRINT IS printf("\\n");
The PHP syntax for strings is explained in the Strings chapter of the manual. To produce static strings with code samples into variables I'd go for nowdoc:
<?php
$str = <<<'EOD'
printf("\\n");
You can write almost anything you want. No 'escaping' "needed" \r \n \
EOD;
var_dump($str);
Of course, this does not apply if you read information from the $_POST superglobal array: the array will automatically contain whatever the user submitted.
How about using:
$x = 'printf("\\\\n");';
echo $x;
I suppose this is for some kind of trivia / questionnaire. You have to escape each backslash with 2 backslashes.
You can check the output here: http://ideone.com/fFvb28
try this:
$answer_something = $_POST['option'];
if($answer_something == '\\n'){
printf("\\n");
}
I'm a bit confuse with
$hello = "hello";
echo "Say $hello";
echo "Say {$hello}";
echo "Say ${hello}";
and the output is same Say hello. When should I use {$hello} and ${hello}? and why it cannot be used in single quote?
$animal = 'cat';
echo "I have 14 $animals";
This may lead to problems, thus you will "escape" it
echo "I have 14 ${animal}s";
or
echo "I have 14 {$animal}s";
In single caused variables/expression were never substituted.
Single quoted string will never expand variables in PHP. See:
http://php.net/manual/en/language.types.string.php
for more detail of the string formats in PHP. There are 4 in total (including nowdoc introduced in PHP 5.3). Only double quoted and heredoc string formats cause variables to be expanded.
According to http://www.php.net/manual/en/language.types.string.php#language.types.string.parsing ,
this is a simple syntax:
echo "Say ${hello}";
and this is a curly syntax:
echo "Say {$hello}";
Why does them both output the same? Becaus in PHP you can use variable variables in every place you want. For example:
$var = 'somevar';
$bar = 'var';
echo $$bar; // "somevar", simple variable variable
echo ${$bar}; // "somevar", complex syntax
echo ${bar}; // "var", because {bar} treated as a string constant:
// Notice: Use of undefined constant bar - assumed 'bar'
So, using variable variables syntax ${hello} simply translated to $hello.
i have a string with double quote like this
$count = 5;
$str = "result: $count";
echo $str; //result: 5
variable parsing work well, and my problem is $count var must be define later than $str
$str = "result: $count";
$count = 5;
echo $str; //result:
So i will use single quote and ask a question here to finding a way to parse var whenever i want
$str = 'result: $count';
$count = 5;
//TODO: parse var process
echo $str; //result: 5
I'm will not using regex replace.
For this type of thing, I'd probably use string formatting. In PHP, that'd be printf.
?php
$str="result: %d"
....dostuff.....define $count.....
printf($str,$count)
?
edit:
although, the best way to do this probably depends partly on why you have to define $string before $count.
If it's a string that's repeated a lot, and you wanted to put it in a global variable or something, printf would be my choice, or putting it in a function as other answers have suggested.
If the string is only used once or twice, are you sure you can't refactor the code to make $count be defined before $string?
finally, a bit of bland theory:
when you write '$string = "result: $count"',
PHP immediately takes the value of $count and puts it into the string. after that, it isn't worried about $count anymore, for purposes of $string, and even if $count changes, $string won't, because it contains a literal copy of the value.
There isn't, as far as I'm aware, a way of binding a position in a string to a yet-to-be-defined variable. The 'printf' method leaves placeholders in the string, which the function printf replaces with values when you decide what should go in the placeholders.
So, if you wanted to only write
$string = "result: $count"
$count=5
$echo string
(and not have to call another function)
and get
"result: 5",
there's no way to do that. The closest method would be using placeholders and printf, but that requires an explicit call to a function, not an implicit substitution.
Why don't you use a function?
function result_str($count) { return "result: $count"; }
preg_replace is the simplest method. Something like this:
$str = preg_replace("/\\$([a-z0-9_]+)/ie", "$\\1", $str);
But if you really don't want to use a regex, then you'll have to parse the string manually, extract the variable name, and replace it.