PHP Echo char变量错误

Question

我有这个代码

<?php
for($i="a";$i<="z";$i++)    {
echo $i;
}
?>

我有一个问题,如果

$i <="y"

它仍然是正确的和回声

a->y

但如果 $i<="z"像上面那样结果将是:

abcdefghijklmnopqrstuvwxyzaaabacadaeafagahaiajakalamanaoapaqarasatauavawaxayazbabbbcbdbebfbgbhbibjbkblbmbnbobpbqbrbsbtbubvbwbxbybzcacbcccdcecfcgchcicjckclcmcncocpcqcrcsctcucvcwcxcyczdadbdcdddedfdgdhdidjdkdldmdndodpdqdrdsdtdudvdwdxdydzeaebecedeeefegeheiejekelemeneoepeqereseteuevewexeyezfafbfcfdfefffgfhfifjfkflfmfnfofpfqfrfsftfufvfwfxfyfzgagbgcgdgegfggghgigjgkglgmgngogpgqgrgsgtgugvgwgxgygzhahbhchdhehfhghhhihjhkhlhmhnhohphqhrhshthuhvhwhxhyhziaibicidieifigihiiijikiliminioipiqirisitiuiviwixiyizjajbjcjdjejfjgjhjijjjkjljmjnjojpjqjrjsjtjujvjwjxjyjzkakbkckdkekfkgkhkikjkkklkmknkokpkqkrksktkukvkwkxkykzlalblcldlelflglhliljlklllmlnlolplqlrlsltlulvlwlxlylzmambmcmdmemfmgmhmimjmkmlmmmnmompmqmrmsmtmumvmwmxmymznanbncndnenfngnhninjnknlnmnnnonpnqnrnsntnunvnwnxnynzoaobocodoeofogohoiojokolomonooopoqorosotouovowoxoyozpapbpcpdpepfpgphpipjpkplpmpnpopppqprpsptpupvpwpxpypzqaqbqcqdqeqfqgqhqiqjqkqlqmqnqoqpqqqrqsqtquqvqwqxqyqzrarbrcrdrerfrgrhrirjrkrlrmrnrorprqrrrsrtrurvrwrxryrzsasbscsdsesfsgshsisjskslsmsnsospsqsrssstsusvswsxsysztatbtctdtetftgthtitjtktltmtntotptqtrtstttutvtwtxtytzuaubucudueufuguhuiujukulumunuoupuqurusutuuuvuwuxuyuzvavbvcvdvevfvgvhvivjvkvlvmvnvovpvqvrvsvtvuvvvwvxvyvzwawbwcwdwewfwgwhwiwjwkwlwmwnwowpwqwrwswtwuwvwwwxwywzxaxbxcxdxexfxgxhxixjxkxlxmxnxoxpxqxrxsxtxuxvxwxxxyxzyaybycydyeyfygyhyiyjykylymynyoypyqyrysytyuyvywyxyyyz

我不知道为什么,有人可以帮帮我吗？

Answer 1

看看PHP文档中有关增加字符的内容:

当处理字符变量而不是C的算术运算时,PHP遵循Perl的约定.例如,在PHP和Perl $ a ='Z'中; $ A ++; 将$ a变为'AA',而在C a ='Z'; 一个++; 将a转换为'['('Z'的ASCII值为90,'['的ASCII值为91).请注意,字符变量可以递增但不递减,即使只支持纯ASCII字符(az和AZ).递增/递减其他字符变量无效,原始字符串不变.