这个简单的php代码有什么错误 $number1=56; $number2=70; $number3=12.56; echo“添加$number1和$number2:”$编号1+$number2; 回声“”; echo“从$number1减去$number2,但使用绝对函数:”。abs($number-$number2)。“”;
当我刷新页面时,它会给出错误“遇到非数值”任何帮助,因为这个简单的php代码有什么错误 $number1=56; $number2=70; $number3=12.56; echo“添加$number1和$number2:”$编号1+$number2; 回声“”; echo“从$number1减去$number2,但使用绝对函数:”。abs($number-$number2)。“”;,php,Php,当我刷新页面时,它会给出错误“遇到非数值”任何帮助,因为$number未定义。定义如下 $number1 = 56; $number2 = 70; $number3 = 12.56; echo "Addition of $number1 and $number2 : " . $number1 + $number2 ; echo "<br>"; echo "Subtraction of $number2 from $number1 but with absolute functi
$number
未定义。定义如下
$number1 = 56;
$number2 = 70;
$number3 = 12.56;
echo "Addition of $number1 and $number2 : " . $number1 + $number2 ;
echo "<br>";
echo "Subtraction of $number2 from $number1 but with absolute function: " . abs ($number-$number2) . "<br>";
$number1=56;
$number2=70;
$number3=12.56;
$number=$number1+$number2//定义$number
echo“添加$number1和$number2:”$数量;
回声“
”;
echo“从$number减去$number2,但使用绝对函数:”。abs($number-$number2)。“
”;
输出:-
或按以下方式更改代码:
$number1 = 56;
$number2 = 70;
$number3 = 12.56;
$number = $number1 + $number2; //define $number
echo "Addition of $number1 and $number2 : " . $number ;
echo "<br>";
echo "Subtraction of $number2 from $number but with absolute function: " . abs ($number-$number2) . "<br>";
$number1=56;
$number2=70;
$number3=12.56;
echo“添加$number1和$number2:”。($number1+$number2)//加括号
回声“
”;
//使用预定义变量而不是$number
echo“从$number1减去$number2,但使用绝对函数:”。abs($number1-$number2)。“
”;
根据我的评论(您没有定义$number
。它应该是$number1
,而不是$number
)。同样根据@nick添加括号,如($number1+$number2)
$number1=56;
$number2=70;
$number3=12.56;
echo“添加$number1和$number2:”。($number1+$number2);
回声“
”;
echo“从$number1减去$number2,但使用绝对函数:”。abs($number1-$number2)。“
”;
在对abs
的调用中,您指的是未定义的$number
。abs($number-$number2)
您未定义$number
。它应该是$number1
而不是$number1
Thanx对于您的答案,我已经纠正了这个错误,但错误是由于上面的添加函数造成的。如果我删除字符串,它工作正常,但如果我添加字符串,它会error@Denshen显示您的错误您还需要更改“添加$number1和$number2:”$number1+$number2
到“$number1和$number2的添加:”。($number1+$number2)Thanx它也可以工作。我是新来的,所以我会犯一些错误,但这个网站非常有用。我不能投票,因为我是新来的,我没有那么多的答案,但我标记你的答案,因为它是有帮助的
$number1 = 56;
$number2 = 70;
$number3 = 12.56;
echo "Addition of $number1 and $number2 : " . ($number1 + $number2) ; //bracket added
echo "<br>";
//use predefined variable instead of $number
echo "Subtraction of $number2 from $number1 but with absolute function: " . abs ($number1-$number2) . "<br>";
$number1 = 56;
$number2 = 70;
$number3 = 12.56;
echo "Addition of $number1 and $number2 : " . ($number1 + $number2) ;
echo "<br>";
echo "Subtraction of $number2 from $number1 but with absolute function: " . abs ($number1-$number2) . "<br>";