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PHP数组关于数字键名的问题

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Release: 2016-06-21 08:48:07
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以下是对PHP数组数字键名的几点总结:

键名长度只能在 int 长度范围内,超过int 范围后将会出现覆盖等混乱情况

在键名长度为 int 范围内存取值时,PHP会强制将数字键名转换为 int 数值型

数字键名长度大于19位时,将变成 0


键名正常长度时,字符串或数值类型一样

$i = 126545165;
$arr['126545165'] = 'abc';
$arr[126545165] = 'uio';
var_dump($arr);
echo &#39;<br/>&#39;;
var_dump(isset($arr[$i]));
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长度超过整型时,键名混乱<喎?http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+PHByZSBjbGFzcz0="brush:java;">$i = 12312312312312; $arr['1000000000147483649'] = 'abc'; $arr[1000000000147483649] = 'uio'; var_dump($arr); echo '
'; var_dump(isset($arr[$i]));





长度超过20位时,键名将变成 0

$i = 123123123123123123123123123123;
var_dump($i);
echo &#39;<br/>&#39;;
$arr[123123123123123123123123123123] = &#39;abc&#39;;
$arr[strval(123123123123123123123123123123)] = &#39;abc&#39;;
var_dump($arr);
echo &#39;<br/>&#39;;
var_dump(isset($arr[$i]));
echo &#39;<br/>&#39;;
var_dump(isset($arr[strval($i)]));
echo &#39;<br/>&#39;;
var_dump(array_keys($arr));
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将变量直接作为键名存取,结果又有不同

$i = 123123123123123;
var_dump($i);
echo &#39;<br/>&#39;;
$arr[$i] = &#39;abc&#39;;
$arr[strval($i)] = &#39;abc&#39;;
var_dump($arr);
echo &#39;<br/>&#39;;
var_dump(isset($arr[$i]));
echo &#39;<br/>&#39;;
var_dump(isset($arr[strval($i)]));
echo &#39;<br/>&#39;;
var_dump(array_keys($arr));
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从上面的几种测试来看:

如果键名为数字,且范围在 int 以内,字符串或者 int 不会对存取有什么影响

如果长度大于 int 时会自动转化为 float ,再转换进行存取出现各种混乱情况,甚至直接变成 0,所以最好是统一转换为 string 类型


$i = 123123123123123123123123123123;
$j = &#39;123123123123123123123123123123&#39;;
$arr1[strval($i)] = &#39;abc&#39;;
$arr2[$j] = &#39;abc&#39;;
var_dump($arr1);
echo &#39;<br/>&#39;;
var_dump($arr2);
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所以在动态操作 PHP 数组时,如果不能确定键名是否会出现数字或者长度大于 int ,则统一将键名 strval 转换为 字符串来操作最为稳妥



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