在 PHP 中将对象转换为整数
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Convert object to integer in PHP
提问by ryonlife
- The value of $total_results = 10
- $total_results in an object, according to gettype()
- I cannot use mathematical operators on $total_results because it's not numeric
- Tried $total_results = intval($total_results) to convert to an integer, but no luck
- The notice I get is: Object of class Zend_Gdata_Extension_OpenSearchTotalResults could not be converted to int
- $total_results 的值 = 10
- $total_results 在一个对象中，根据 gettype()
- 我不能在 $total_results 上使用数学运算符，因为它不是数字
- 尝试 $total_results = intval($total_results) 转换为整数，但没有运气
- 我得到的通知是：类 Zend_Gdata_Extension_OpenSearchTotalResults 的对象无法转换为 int
How can I convert to an integer?
如何转换为整数？
回答by Sean Bright
Does this work?
这行得通吗？
$val = intval($total_results->getText());
回答by Karsten
$results_numeric = (int) $total_results;
or maybe this:
或者这个：
$results_numeric = $total_results->count();
回答by Pim Jager
perhaps the object has a build in method to get it as an integer?
也许该对象有一个内置方法可以将其作为整数获取？
Otherwise try this very hacky approach (relys on __toString() returning that 10)
否则尝试这种非常hacky的方法（依赖于 __toString() 返回 10）
$total_results = $total_results->__toString();
$total_results = intval($total_results);
However if the object has a build in non-magic method, you should use that!
但是，如果对象具有内置的非魔法方法，则应该使用它！
回答by Morten Christiansen
回答by CodeXP
try
尝试
class toValue {
function __toString()
{
return '3'; // you must return a string
}
}
$a = new toValue;
var_dump("$a" + 2);
result: int(5)
结果：int(5)
