Question

我正在尝试在PHP中执行以下操作，并非常感谢任何和所有帮助。

将字符的每个实例的位置存储在数组的字符串中
使用for循环来导航数组，并用另一个字符替换数组中每个元素的每个元素的字符。

这是我到目前为止所做的：

$character1="/";
$character2="%";
$string1="hello / how / are / you / doing";
$characterPositions = array();
/* store locations of $character1 in $string1 */
foreach($characterPositions as $position){
    /* replace what is at each $position in string $string1 */
}

我知道str_replace会做到这一点，但我想学习如何以上述方式做到这一点。

Answer 1

迭代每个角色并存储位置。然后迭代这些位置并设置角色。

for ($i = 0; $i < strlen($string1); $i++) {
    if ($string1[$i] == $character1) $characterPositions[] = $i;
}

foreach ($characterPositions as $position){
    $string1[$position] = $character2;
}

Answer 2

  <?php
  $character1="/";
  $character2="%";
  $string1="hello / how / are / you / doing";
  $characterPositions = array();
  /* store locations of $character1 in $string1 */
  $lastOffset = 0;

  while (($pos = strpos($string1, $character1, $lastOffset+1)) !== FALSE){
        echo $lastOffset;
        $characterPositions[] = $pos;
        $lastOffset = $pos;
  }
  print_r($characterPositions);
  foreach ($characterPositions as $v){
        $string1[$v] = $character2;
  }

在不同的字符串位置替换相同字符的不同实例

2 个答案: