计数strpos()发生

Question

如何计算已找到的关键字出现的次数？目前我有这个

$filesfound = false;
foreach ($arr as $file)
$filesfound = true;
...

echo $vouchers = ($filesfound === true) && (strpos(implode('/', $file), 'Voucher') !== false) ? '<div>+ 2 Files Available</div>' : '';

我试图获得"+2文件可用“，以实际帐户字符串发生的发现。我该怎么做？多个函数没有给出我期望的结果。在这种情况下，我期望2的正确阅读

谢谢！！

溶液

$filesfound = false;
foreach ($arr as $file) {
   // If occurrences were found. Returns 0 for each match, so we add + 1
   if (strpos(implode('/', $file), 'Voucher') === 0) { 
      $filesfound = true;
      $count = $count + 1; 
   } 
}
echo $count;

Answer 1

若要计数字符串中出现的文本，可以使用：

$text = 'This is a test';
// find how many "is" are within our $text variable:
echo substr_count($text, 'is'); // this will return 2, since there are 2 of "is" in our $text