过滤掉具有特定模式的字符串开头的数字答案

【问题标题】：Filtering out a number at the start of a string which has a certain pattern过滤掉具有特定模式的字符串开头的数字
【发布时间】：2010-08-14 21:17:53
【问题描述】：

如果字符串以@ 开头，我正在尝试从字符串中过滤出一个数字。这是我认为可以解决问题的方法，但它只返回一个空白页面。

（可能包含很多错误，因为我是 PHP 新手。）

<?php
$String = "@1234 Hello this is a message.";
$StringLength = 1;
Echo "Filtering the number after the @ out of " .$String;
If (substr($String , 0, 1)="@"){ //If string starts with @
    While (is_int(substr($String,1,$StringLength))){ //Check if the X length string after @ is a number.
            $StringLength=$StringLength+1; //If it was a number, up StringLength by one.
    }
    If ($StringLength >= 2){ //If the number is only 1 character long StringLength will be 2, loop completed once.
        $Number = substr($String,1,$StringLength-1);
        Echo $Number;
    }
    Else{ //The string started with @ but the While has never run because it was false.
        Echo "The @ isn't followed by a number.";
    }
Else{ //If string doesn't start with @
    Echo "String doesn't start with @.";
}
?>

我的脚本有什么问题？

提前致谢！

【问题讨论】：

标签： php string filter numbers

【解决方案1】：

if(substr($String , 0, 1)=="@")
//                       ^^ 2 equal signs for equality comparison.

顺便说一句，您的函数可以简单地用正则表达式 (example) 编写。要获取初始字符，请使用$string[0]。

if (preg_match('/^@(\\d+)/', $string, $results)) {
   echo $results[1];
} else {
   if ($string[0] != '@')
     echo "String doesn't start with @.";
   else
     echo "The @ isn't followed by a number.";
}

【讨论】：