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PHP: passing php variables as arguments?

开发者 https://www.devze.com 2023-02-11 01:17 出处:网络
In my php code I need to invoke a script and passing it some arguments. How can I pass php variables (as arguments) ?

In my php code I need to invoke a script and passing it some arguments. How can I pass php variables (as arguments) ?

$string1="some value";
$string2=开发者_如何学运维"some other value";
header('Location: '...script.php?arg1=$string1&arg2=$string2');

thanks


header('Location: ...script.php?arg1=' . $string1 . '&arg2=' . $string2);


Either via string concatenation:

header('Location: script.php?arg1=' . urlencode($string1) . '&arg2=' . urlencode($string2));

Or string interpolation

$string1=urlencode("some value");
$string2=urlencode("some other value");
header("Location: script.php?arg1={$string1}&arg2={$string2}");

Personally, I prefer the second style. It's far easier on the eyes and less chance of a misplaced quote and/or ., and with any decent syntax highlighting editor, the variables will be colored differently than the rest of the string.

The urlencode() portion is required if your values have any kind of url-metacharacters in them (spaces, ampersands, etc...)


You could use the function http_build_query

$query = http_build_query(array('arg1' => $string, 'arg2' => $string2));

header("Location: http://www.example.com/script.php?" . $query);


Like this:

header("Location: http://www.example.com/script.php?arg1=$string1&arg2=$string2");


It should work, but wrap a urlencode() incase there is anything funny breaking the url:

header('Location: '...script.php?'.urlencode("arg1=".$string1."&arg2=".$string2).');
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