为什么我的PHP计算器不能正常工作？ (字符串函数..)

该计算器应该在网页文本框中输入带空格的输入内容(例如“ 2 2”),然后回显上方的答案.当前,输出为“ 0”,或者看似会从输入中裁剪掉一个随机数.

我95％确信问题出在substr函数内-它们没有被分配正确的值.由于输出行为,这是我的猜测.展示,

1 1将= 0
2 2将= 0
…
9 9将= 0
10 10将= 1
20 20将= 2

注意：从第16行开始的$firstNumber,$secondNumber和$operator初始声明值是任意的.

以下是PHP手册中的substr(,)示例：

echo substr('abcdef', 1);     // bcdef   <br>
echo substr('abcdef', 1, 3);  // bcd<br>
echo substr('abcdef', 0, 4);  // abcd<br>
echo substr('abcdef', 0, 8);  // abcdef<br>
echo substr('abcdef', -1, 1); // f<br>

//'if' is executed when submit is pressed
if (isset($_POST['test']))
{
    $input = $_POST['test'];
    $answer = calculate($input);
    echo $answer;
}

//does string processing and number calculation
function calculate($input)
{
    //declarations (random)
    $firstNumber = 20;
    $secondNumber = 30;
    $operator = "+";
    $answer = 7;

    //string processing. 
    for($i = 0; $i <= strlen($input); $i++)  
    {
        //if current position of the string scan is an operator,
        if( trim(substr($input, $i, $i + 1), " ") == "+" ||trim(substr($input, $i, $i + 1), " ") == "-"
          ||trim(substr($input, $i, $i + 1), " ") == "*" ||trim(substr($input, $i, $i + 1), " ") == "/")
        {
            //then 

            //$operator = current position TO current position + 1
            $operator = substr($input, $i, $i + 1);
            //trim $operator
            $operator = trim($operator, " ");

            //$firstNumber = 0 TO current position - 1
            $firstNumber = substr($input, 0, $i - 1);
            //trim $operator
            $firstNumber = trim($firstNumber, " ");

            //$secondNumber = current position + 1 TO end of string
            $secondNumber = substr($input, $i + 1, strlen($input));
            //trim $operator
            $secondNumber = trim($secondNumber, " ");
        }

    }

    //if operator is ... then do that operation. 
    //example: if "+", then add $firstNumber and $secondNumber

    if($operator == "+")
    {
        $answer = $firstNumber + $secondNumber;
    }
    else if($operator == "-")
    {
        $answer = $firstNumber - $secondNumber;
    }
    else if($operator == "*")
    {
        $answer = $firstNumber * $secondNumber;
    }
    else if($operator == "/")
    {
        $answer = $firstNumber / $secondNumber;
    }

    //return the calculated answer for echo
    return $answer;
}

?>

<form action="test.PHP" method="POST">

    <textarea name="test" rows="3" cols="30"></textarea>
    <input type="submit" value="calculate!">

</form>

解决方法:

您的问题是您使用substr.

substr取一个字符串,一个起始位置和一个长度.这是您滥用的长度.在浏览输入字符串时,您正在增加要查找的子字符串的长度.由于运算符的长度为1,因此您应该使用

    //if current position of the string scan is an operator,
    if( trim(substr($input, $i, 1), " ") == "+" || trim(substr($input, $i, 1), " ") == "-"
      ||trim(substr($input, $i, 1), " ") == "*" || trim(substr($input, $i, 1), " ") == "/")
    {
         ....

您还需要在此处更新代码：

//$operator = current position TO current position + 1
$operator = substr($input, $i, 1);
//trim $operator
$operator = trim($operator, " ");

您可以通过做一些简化

  potentialOperator = trim(substr($input, $i, 1), " ")

并将其与您支持的操作符进行比较.这样可以为您节省多个不必要的修整和替换操作
这也意味着您只能一次识别一次运算符.

您可能还想查看带有PREG_SPLIT_DELIM_CAPTURE的preg_split来解析输入字符串,而不是逐字符扫描它.

为什么我的PHP计算器不能正常工作？ (字符串函数..)

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