如何在不替换替换子字符串的情况下替换多个子字符串？

Question

帮助或协助更换我的案例中的此类变体：

$string = "This is simple string";

$search = array (
  "This is simple",
  "string",
  "simple",
  "apple"
);

$replace = array (
  "This is red",
  "apple",
  "false",
  "lemon"
);

$result = str_replace($search, $replace, $string);

Result must be: This is red apple

Not so: This is false apple OR This is red lemon OR This is false red lemon

如果在每次替换时，将更改的行切入某个变量然后稍后返回，则结果可能是正确的。但我不知道这是我的选择，但我无法实现它。

Answer 1

使用strtr():

$string = "This is simple string";

$search = array
(
  "This is simple",
  "string",
  "simple",
  "apple"
);

$replace = array
(
  "This is red",
  "apple",
  "false",
  "lemon"
);

echo strtr($string,array_combine($search, $replace));

输出：

This is red apple

重要

我必须告诉读者，这个漂亮的功能也是一个古怪的功能。如果您以前从未使用过此功能，我建议您 read the manual 以及它下面的评论。

对于这种情况很重要（而不是我的回答）：

If given two arguments, the second should be an array in the form array('from' => 'to', ...). The return value is a string where all the occurrences of the array keys have been replaced by the corresponding values. The longest keys will be tried first. Once a substring has been replaced, its new value will not be searched again.

在 OP 的编码尝试中，键 ($search) 按长度降序排列。这使函数行为与大多数人期望发生的事情保持一致。

但是，考虑一下这个演示，其中的键（及其值）被稍微打乱了：

代码：(Demo)

$string="This is simple string";
$search=[
   "string",  // listed first, translated second, changed to "apple" which becomes "untouchable"
   "apple",  // this never gets a chance
   "simple",  // this never gets a chance
   "This is simple"  // listed last, but translated first and becomes "untouchable"
];
$replace=[
   "apple",
   "lemon",
   "false",
   "This is red"
];
echo strtr($string,array_combine($search, $replace));

您可能会惊讶地发现这提供了相同的输出：This is red apple

如何在不替换替换子字符串的情况下替换多个子字符串？

How to replace multiple substrings without replacing replaced substrings?

php

string

strtr

preg-replace

str-replace

重要