如何替换＆rdquo; A＆QUOT;与＆QUOT; D＆QUOT;，＆QUOT; B＆QUOT;与＆QUOT; E＆QUOT;，＆QUOT; C＆QUOT;与＆QUOT;餐饮QUOT ;,。 。 。英寸×＆QUOT;与＆QUOT; A＆QUOT;，＆QUOT; Y＆QUOT;与＆QUOT; B＆QUOT;和＆QUOT; Z＆QUOT;与＆QUOT; C＆QUOT;？ [英] How to replace "a" with "d", "b" with "e", "c" with "f", . . . "x" with "a", "y" with "b", and "z" with "c"?
问题描述
我想执行一个字符串此搜索和替换，比如密码。
I would like to perform this search and replacement on a string, say password.
你可以从这个问题明白了，更换后的字符串应该成为 sdvvzrug 。
As you can understand from the question, after replacement the string should become sdvvzrug.
但遗憾的是以下code输出 bbbbcaab ：
But unfortunately the following code outputs bbbbcaab:
$search = range("a", "z");
$replace = array_merge(range("d", "z"), range("a", "c"));
echo str_replace($search, $replace, "password");
什么可能导致此问题？起初我怀疑是使用数组的 str_replace转换，所以我用了一个循环，并尝试一次更换一个字符：
What may be causing the issue here? At first I suspected the use of arrays in str_replace, therefore I used a loop and tried replacing one character at a time:
$search = range("a", "z");
$replace = array_merge(range("d", "z"), range("a", "c"));
$str = "password";
for($i = 0; $i < 26; $i++)
{
$str = str_replace($search[$i], $replace[$i], $str);
}
echo $str;
要我彻底失望，结果仍然是相同的。我在做什么错了，我应该怎么实现正确的结果？
To my utter dismay, the result remained the same. What am I doing wrong and how should I achieve the correct results?
推荐答案
其他答案已经解释你如何能做到这一点，所以我将重点放在了我在做什么了？一部分。
Other answers have explained how you can do this, so I will focus on the "What am I doing wrong?" part.
所以，你的为 -loop相当于这样的：
So, your for-loop is equivalent to this:
$str = str_replace('a', 'd', $str);
$str = str_replace('b', 'e', $str);
$str = str_replace('c', 'f', $str);
$str = str_replace('d', 'g', $str);
$str = str_replace('e', 'h', $str);
$str = str_replace('f', 'i', $str);
$str = str_replace('g', 'j', $str);
// ...
$str = str_replace('z', 'c', $str);
正如你所看到的，这种字母通过扫荡的： A 获取与替换ð，后来 D 被替换先按g ，后来先按g 变与Ĵ替换。问题是，当你与先按g 替换 D ，你这样做，无论是否<$的C $ C> D 重新presents什么原是 D 或什么最初是一个 A 。每个 str_replace转换被丢弃的信息;第一行后，你不能告诉字符串 DDD 是否原是添加，或者原来父亲，当您更改 D 到先按g ，你得到 GGG 而不是 DGG 或 GDG 或诸如此类的东西。
As you can see, this sort of "sweeps through" the alphabet: a gets replaced with d, and later d gets replaced g, and later g gets replaced with j. The problem is that when you're replacing d with g, you're doing it regardless of whether the d represents what was originally a d or what was originally an a. Each str_replace is discarding information; after the first line, you can't tell whether the string ddd was originally add, or originally dad, and when you change d to g, you get ggg instead of dgg or gdg or whatnot.
原因的版本只是一个单一的 str_replace转换有同样的结果是 str_replace转换做同样的事情，你的循环：它只是搜索阵列上迭代，进行替换，并不能保证它永远不会重新替换了一子。这就是为什么你需要使用的方法，如 strtr函数的效率就是专门为此而设计的。
The reason the version with just a single str_replace had the same result is that str_replace does the same thing as your loop: it just iterates over the search-array, performing the replacements, and doesn't ensure that it never "re-replaces" a substring. That's why you need to use a method such as strtr that's specifically designed for this purpose.
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