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我已经将一个脚本与数组之类的东西放在一起。现在最后,它使用复选框检查分数并将其除以十。然后用数字格式将它四舍五入两位小数,它看起来像这样。

$number = $score / $total;
$number = $number * 10;
$number = number_format(round($number,2),2,',','.');
echo "The number is: $number <br/>";

然后稍后,我正在这样做。

if($number < 4 && $number > 0)
echo 'You're number is between zero and 4';
else if($number > 6 && $number < 4)
echo ' You're number is between 4and 6';
else if($number < 8 && $number > 6)
echo' You're number is between 6 and 8';
else if($number < 10 && $number > 8)
You're number is between 8 and 10';

现在,如果数字是 0.50 或 1.50 或 2 或 5.50,它会显示我提供的文本。但是,如果数字是 4,13 或 7,85 或 9,13,则不是。

现在搜索了很长时间,但无法弄清楚。你们看到解决方案了吗?

希望我足够清楚!

提前致谢。

4

2 回答 2

1

以下语句永远不会验证true

if($number > 6 && $number < 4) 

你知道任何大于六小于四的数字吗?

该代码还包含许多语法错误,例如:

echo 'You're number is between zero and 4';

撇号需要使用反斜杠进行转义:

echo 'You\'re number is between zero and 4';

或者对包含撇号的字符串使用双引号:

echo "You're number is between zero and 4";

最后,语法方面you're应该是your:)

于 2012-07-26T09:15:58.003 回答
1

这是你要找的吗?

<?php
$score = 7; //Example values
$total = 32;

$number = $score / $total;
$number = $number * 10;
$number = round($number,2);
echo 'The number is: ',number_format($number,2,'.',','),' <br/>';

if($number < 4 && $number > 0)
    echo 'Your number is between zero and 4';
else if($number < 6 && $number > 4)
    echo 'Your number is between 4 and 6';
else if($number < 8 && $number > 6)
    echo 'Your number is between 6 and 8';
else if($number < 10 && $number > 8)
    echo 'Your number is between 8 and 10';
?>
于 2012-07-26T09:17:05.240 回答