php 检查变量是否存在且 === true
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Check if Variable exists and === true
提问by Sebastian Sebald
I want to check if:
我想检查是否：
- a field in the array isset
- the field === true
- 数组 isset 中的一个字段
- 该字段 === 真
Is it possible to check this with one ifstatement?
是否可以用一个if语句来检查这一点？
Checking if ===would do the trick but a PHP notice is thrown. Do I really have to check if the field is set and then if it is true?
检查是否===可以解决问题，但会抛出 PHP 通知。我是否真的必须检查该字段是否已设置，然后是否为真？
回答by Madara's Ghost
If you want it in a single statement:
如果你想在一个单一的声明：
if (isset($var) && ($var === true)) { ... }
If you want it in a single condition:
如果你想在单一条件下使用它：
Well, you could ignore the notice (aka remove it from display using the error_reporting()function).
好吧，您可以忽略该通知（也就是使用该error_reporting()功能将其从显示中删除）。
Or you could suppress it with the evil @character:
或者你可以用邪恶的@角色压制它：
if (@$var === true) { ... }
This solution is NOT RECOMMENDED
不推荐使用此解决方案
回答by askome
I think this should do the trick ...
我认为这应该可以解决问题......
if( !empty( $arr['field'] ) && $arr['field'] === true ){
do_something();
}
回答by Phill Pafford
Alternative, just for fun
另类，仅供娱乐
echo isItSetAndTrue('foo', array('foo' => true))."<br />\n";
echo isItSetAndTrue('foo', array('foo' => 'hello'))."<br />\n";
echo isItSetAndTrue('foo', array('bar' => true))."<br />\n";
function isItSetAndTrue($field = '', $a = array()) {
return isset($a[$field]) ? $a[$field] === true ? 'it is set and has a true value':'it is set but not true':'does not exist';
}
results:
结果：
it is set and has a true value
it is set but not true
does not exist
Alternative Syntax as well:
替代语法也是：
$field = 'foo';
$array = array(
'foo' => true,
'bar' => true,
'hello' => 'world',
);
if(isItSetAndTrue($field, $array)) {
echo "Array index: ".$field." is set and has a true value <br />\n";
}
function isItSetAndTrue($field = '', $a = array()) {
return isset($a[$field]) ? $a[$field] === true ? true:false:false;
}
Results:
结果：
Array index: foo is set and has a true value
