douyou1937 2014-04-11 23:06
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尝试回显数组数据时出错

<?php
    $greeting = "Hello";
    $place = "World";
    $num = 3;
    $data = array(
        0 => "zero", 
        1 => "one"
    );
    echo "<pre>";
    echo $greeting;
    echo "

";
    echo '$greeting' . "$place";
    echo "

";
    echo $num * 8;
    echo "

";
    echo "Greeting
World";
    echo "

";
    echo $data['$num'].'is a 
 number';
    echo "

";
    echo "It's" . "$data[1]" . "small" . "$place";
    echo "

";
    echo $data;
    echo "

";
    echo substr($hello . " and good luck! ", 3, 15);
    echo "

";
    echo "<pre>";
?>

Here is PHP code that I received for testing purposes. I copied the code word for word, but for some reason I get the following errors:

Notice: Undefined index: $num in C:\xampp\htdocs\test.php on line 18

Notice: Array to string conversion in C:\xampp\htdocs\test.php on line 22

Notice: Undefined variable: hello in C:\xampp\htdocs\test.php on line 24

This was a question given on a midterm where I was supposed to interpret the output. It is possible that these errors were made on purpose, but I don't understand why. Could someone please explain why these errors are taking place?

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4条回答 默认 最新

  • dongtong7990 2014-04-11 23:09
    关注

    Notice: Undefined index: $num in C:\xampp\htdocs\test.php on line 18

    The single quotes makes it a string literal and the variable is not interpolated:

    echo $data[$num].'is a 
     number';
    

    Notice: Array to string conversion in C:\xampp\htdocs\test.php on line 22

    $data is an array, not a string. So you can't echo it out like a string:

    print_r($data);
    

    Notice: Undefined variable: hello in C:\xampp\htdocs\test.php on line 24

    You never declare $hello so it obviously doesn't exist and you can't use it. Either declare it as an empty string or remove the code that attempts to use it.

    $hello = '';
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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