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We've all encountered it before, needing to print a variable in an input field but not knowing for sure whether the var is set, like this. Basically this is to avoid an e_warning.
<input value='<?php if(isset($var)){print($var);}; ?>'> How can I write this shorter? I'm okay introducing a new function like this:
<input value='<?php printvar('myvar'); ?>'> But I don't succeed in writing the printvar() function.
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PHP $_GET is a PHP super global variable which is used to collect form data after submitting an HTML form with method="get". $_GET can also collect data sent in the URL. Assume we have an HTML page that contains a hyperlink with parameters: <html> <body>
The Scope Resolution Operator (also called Paamayim Nekudotayim) or in simpler terms, the double colon, is a token that allows access to static, constant, and overridden properties or methods of a class.
PHP isset() FunctionThe isset() function checks whether a variable is set, which means that it has to be declared and is not NULL. This function returns true if the variable exists and is not NULL, otherwise it returns false.
$$var: $$var stores the value of $variable inside it. Syntax: $variable = "value"; $$variable = "new_value"; $variable is the initial variable with the value. $$variable is used to hold another value.
My recommendation would be to create a issetor function:
function issetor(&$var, $default = false) { return isset($var) ? $var : $default; } This takes a variable as argument and returns it, if it exists, or a default value, if it doesn't. Now you can do:
echo issetor($myVar); But also use it in other cases:
$user = issetor($_GET['user'], 'guest'); As of PHP 7 you can use the null-coalesce operator:
$user = $_GET['user'] ?? 'guest'; Or in your usage:
<?= $myVar ?? '' ?> Another option:
<input value="<?php echo isset($var) ? $var : '' ?>"> If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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