PHP str_replace 用数组中的随机替换来替换 need ？

Question

我进行了研究，需要找到用一系列可能性随机替换需求的最佳方法。

即：

$text = "Welcome to [city]. I want [city] to be a random version each time. [city] should not be the same [city] each time.";

$keyword = "[city]";
$values = array("Orlando", "Dallas", "Atlanta", "Detroit", "Tampa", "Miami");

$result = str_replace("[$keyword]", $values, $text);

结果是每个出现的城市都有“数组”。我需要用 $values 中的随机值替换所有出现的城市。我想以最干净的方式做到这一点。到目前为止我的解决方案很糟糕（递归）。对此最好的解决方案是什么？谢谢！

Answer 1

您可以使用 preg_replace_callback 为每个匹配执行一个函数并返回替换字符串：

$text = "Welcome to [city]. I want [city] to be a random version each time. [city] should not be the same [city] each time.";

$keyword = "[city]";
$values = array("Orlando", "Dallas", "Atlanta", "Detroit", "Tampa", "Miami");

$result = preg_replace_callback('/' . preg_quote($keyword) . '/', 
  function() use ($values){ return $values[array_rand($values)]; }, $text);

样品

$result

：

欢迎来到亚特兰大。我希望达拉斯每次都是随机版本。迈阿密不应该每次都是同一个亚特兰大。

Answer 2

您可以将

preg_replace_callback

与

array_rand

一起使用

<?php
$text = "Welcome to [city]. I want [city] to be a random version each time. [city] should not be the same [city] each time.";

$values = array("Orlando", "Dallas", "Atlanta", "Detroit", "Tampa", "Miami");

$result = preg_replace_callback("/\[city\]/", function($matches) use ($values) { return $values[array_rand($values)]; }, $text);

echo $result;

示例此处。

Answer 3

这是另一个想法

$text = "Welcome to [city]. I want [city] to be a random version each time. [city] should not be the same [city] each time.";

$pattern = "/\[city\]/";
$values = array("Orlando", "Dallas", "Atlanta", "Detroit", "Tampa", "Miami");

while(preg_match($pattern, $text)) {
        $text = preg_replace($pattern, $values[array_rand($values)], $text, 1);
}

echo $text;

还有一些输出：

Welcome to Orlando. I want Tampa to be a random version each time. Miami should not be the same Orlando each time.

Answer 4

您正在使用

$values

替换文本，它是一个数组，因此结果只是单词“Array”。替换应该是一个字符串。

您可以使用

array_rand()

从数组中随机选择条目。

$result = str_replace($keyword, $values[array_rand($values)], $text);

结果是这样的：

Welcome to Atlanta. I want Atlanta to be a random version each time. Atlanta should not be the same Atlanta each time.
Welcome to Orlando. I want Orlando to be a random version each time. Orlando should not be the same Orlando each time.

如果您希望城市每行是随机的，请检查@PaulP.R.O的答案。

Answer 5

如果：

1. 随机选择的四个城市占位符不应在 3 句字符串中重复并且
2. 你不介意改变输入数组并且
3. 您将始终有足够的值来容纳所有占位符，然后：

您可以将数组作为引用变量传递到自定义函数作用域中。 这意味着当您通过

array_pop()

访问和删除值时，不可能两次遇到同一个城市。 在执行替换之前，您只需

shuffle()

输入数组一次。

代码：（演示）

$text = "Welcome to [city]. I want [city] to be a random version each time. [city] should not be the same [city] each time.";

$keyword = "[city]";
$values = ["Orlando", "Dallas", "Atlanta", "Detroit", "Tampa", "Miami"];
shuffle($values);

echo preg_replace_callback(
         '#' . preg_quote($keyword) . '#',
         function() use(&$values) { return array_pop($values); },
         $text
     );

潜在产出：

Welcome to Detroit. I want Tampa to be a random version each time. Dallas should not be the same Orlando each time.

或

Welcome to Orlando. I want Miami to be a random version each time. Atlanta should not be the same Detroit each time.

或

Welcome to Atlanta. I want Tampa to be a random version each time. Orlando should not be the same Dallas each time.

等等

---

遵守本答案顶部最初描述的相同规则，将更容易阅读和维护不利用正则表达式的方法。只需对数组进行打乱，将方括号占位符替换为

printf()

系列 if 函数可以识别的占位符，然后将整个打乱后的数组提供给

vprintf()

。

这具有不通过

array_pop()

改变/消耗原始数组的额外好处。 (演示)

shuffle($values);
vprintf(
    str_replace(
         $keyword,
         '%s',
         $text
    ),
    $values
);

Answer 6

试试这个http://codepad.org/qp7XYHe4

<?
$text = "Welcome to [city]. I want [city] to be a random version each time. [city] should not be the same [city] each time.";

$keyword = "[city]";
$values = array("Orlando", "Dallas", "Atlanta", "Detroit", "Tampa", "Miami");

echo $result = str_replace($keyword, shuffle($values)?current($values):$values[0], $text);