php中的字符串替换未按预期工作

我有一个字符串，如下所示：

$var = "Hello Hi World it is an awesome day. Test Hi guys Whats up";

function get_string_between($string, $start, $end){
        $string = " ".$string;
        $ini = strpos($string,$start);
        if ($ini == 0) return "";
        $ini += strlen($start);
        $len = strpos($string,$end,$ini) - $ini;
        return substr($string,$ini,$len);
        } 

$parsed0 = get_string_between($var, "Hello", "World");

So I get the String in `$parse0` 

$varnew = str_replace($parsed0,"NEWVALUE", $var);

echo $varnew;

Hello NEWVALUE World it is an awesome day. Test NEWVALUE guys Whats up

我正在尝试替换字符串中的值，如下所示：

$var = "Hello Hi World it is an awesome day. Test Hi guys Whats up";

function get_string_between($string, $start, $end){
        $string = " ".$string;
        $ini = strpos($string,$start);
        if ($ini == 0) return "";
        $ini += strlen($start);
        $len = strpos($string,$end,$ini) - $ini;
        return substr($string,$ini,$len);
        } 

$parsed0 = get_string_between($var, "Hello", "World");

So I get the String in `$parse0` 

$varnew = str_replace($parsed0,"NEWVALUE", $var);

echo $varnew;

Hello NEWVALUE World it is an awesome day. Test NEWVALUE guys Whats up

我将替换如下：

$var = "Hello Hi World it is an awesome day. Test Hi guys Whats up";

function get_string_between($string, $start, $end){
        $string = " ".$string;
        $ini = strpos($string,$start);
        if ($ini == 0) return "";
        $ini += strlen($start);
        $len = strpos($string,$end,$ini) - $ini;
        return substr($string,$ini,$len);
        } 

$parsed0 = get_string_between($var, "Hello", "World");

So I get the String in `$parse0` 

$varnew = str_replace($parsed0,"NEWVALUE", $var);

echo $varnew;

Hello NEWVALUE World it is an awesome day. Test NEWVALUE guys Whats up

我得到如下输出：

$var = "Hello Hi World it is an awesome day. Test Hi guys Whats up";

function get_string_between($string, $start, $end){
        $string = " ".$string;
        $ini = strpos($string,$start);
        if ($ini == 0) return "";
        $ini += strlen($start);
        $len = strpos($string,$end,$ini) - $ini;
        return substr($string,$ini,$len);
        } 

$parsed0 = get_string_between($var, "Hello", "World");

So I get the String in `$parse0` 

$varnew = str_replace($parsed0,"NEWVALUE", $var);

echo $varnew;

Hello NEWVALUE World it is an awesome day. Test NEWVALUE guys Whats up

我的预期结果是：

Hello NEWVALUE World it is an awesome day. Test Hi guys Whats up.

发生这种情况是因为在

$parse0

中，我得到的结果是

Hi

，它在该字符串中包含两次

有没有办法用一个函数获取起始和结束位置，然后用这些函数替换它？

我建议。检查下面的代码

function customReplace($string, $start, $end, $value = "NEWVALUE"){
    return preg_replace_callback('/'.$start.'(.*?)'.$end.'/', function($m) use($start, $end){
        return "$start NEWVALUE $end";
    }, $string);
}

$var = "Hello Hi World it is an awesome day. Test Hi guys Whats up";

echo customReplace($var, "Hello", "World");

输出：

Hello NEWVALUE World it is an awesome day. Test Hi guys Whats up

代码的问题是：

$varnew = str_replace($parsed0,"NEWVALUE", $var);

$parsed0

具有

Hi

（从函数中获取），并说

str\u replace

以将

Hi

替换为字符串

$var

（整个字符串）中的

NEWVALUE

。因此，

str_replace

按预期工作。希望现在一切都清楚了

更新

同时检查下面的@b263答案。您也可以使用，它不需要回调

你好，NEWVALUE World这是一个很棒的日子。嗨，伙计们，有什么事吗

你不能用str_replace做那件事

$var = "Hello Hi World it is an awesome day. Test Hi guys Whats up";
//$parsed0 = 'Hi';
$varnew = str_replace($parsed0,"NEWVALUE", $var);

因为$parsed0='Hi'，该字符串中出现的任何'Hi'都将被替换。如果只想替换第一个引用，可以执行以下操作

$varnew = preg_replace('/'.$parsed0.'/', ' NEWVALUE ', $var, 1);

---

@我需要的Justcode可能有非常大的字符串，可能包含更多的重复。我知道第一个和最后一个字符串，需要把它们放在中间。请引导我@Justcode它正在替换第一个或第二个引用，依此类推。在我的例子中，我需要替换两个字符串之间的值。希望你现在就得到它！！这有帮助吗？这个先前回答的链接可能会有帮助-