Question

输入：

古杰拉特（24）

预期输出：

24

字符串格式：

州名（州代码）

如何删除带有括号的州代码？

Answer 1

您也可以使用php explode：

$state = "GUJARAT (24)";
$output = explode( "(", $state );
echo trim( $output[0] ); // GUJARAT

Answer 2

您可以使用功能preg_match

$state = 'GUJARAT (24)'
$result = [];
preg_match('/[\(][\d]+[\)]/', $state, $result);
$stateCode = $result[0] // (24)
$stateCode = substr($stateCode, 1, strlen($stateCode) - 2); // will output 24

Answer 3

如果您知道stateCode的长度，并且它是固定的

$state = "GUJARAT (24)";
$state = substr($state, strpos($state, "(") + 1, 2);
// get string upto "(", and remove right white space    
echo $state;

您可以在php手册中查看有关substr和strpos的更多信息。

如果stateCode的长度不固定，则可以使用正则表达式：-

  $state = 'GUJARAT (124)';
  preg_match('/(\d+)/', $state, $res);
  $stateCode = $res[0]; 
  var_dump($stateCode);

Answer 4

string id = txt.Substring(0, txt.LastIndexOf("UID:") + 9);

Answer 5

let str = "LOREM Ipsum is simply DUMMY text of the printing and typesetting industry. Lorem IPSUM. Has been the industry's standard dummy text ever since the 1500s, WHEN an unknown printer took a galley."

let op = str.toLowerCase().replace(/^([a-z])|\.\s*([a-z])/g, (match)=>match.toUpperCase())
console.log(op)

PHP如何用空格替换字符串？

5 个答案: