Question

我写了以下脚本。我的目的是做以下两件事：

让用户查看$score。
让用户按“分数”将$score增加1按钮。

$score的值存储在数据库中。

只有一个问题 - 当我点击“分数！”时按钮，$score的值不会增加1 - 无论原始值是什么，它都会重置为零。

<?php
    $page_title = "";
    ?>
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
        "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
    <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=utf-8"/>
        <title><?php print($page_title) ?></title>
    </head>
    <body>

    <?php // Script number 1.2, filename show_score.php

    // error handling
    ini_set('display errors',1);  // Let me learn from my mistakes!
    error_reporting(E_ALL|E_STRICT); // Show all possible problems! 

    // Set page title. 
    $page_title = "Game Score";

    // Connect to the database:
    $link = mysql_connect('localhost','username','password');
    mysql_select_db('game_scores',$link);

    // Create database query that gets the value of the score_counter cell
    $sql = 'SELECT score_counter FROM scores';
    $result = mysql_query($sql,$link);

    // Create a variable, $score_counter, that holds the array that 
    //is the result of the previous SQL query

    $score_counter = mysql_fetch_array($result);

    // You don't really want the whole array, just score_counter[0],
    // so assign score_counter[0] its own variable, $score

    $scores = $score_counter[0];

    // Now that you've retrieved the current number of scores from the 
    // database, and extracted that number from an array and 
    // put it in its own variable, print it out on the screen.

    echo 'The current number of scores is ' . $scores . ' scores.';


    // Now let users add to the number of scores by clicking the "score" button. 

    if(isset($_POST['score'])){

        // increment the number of scores:
        $scores++;

        // create the SQL query:
        $query='UPDATE scores SET score_counter="$scores" WHERE score_id=1';

        // run the SQL query:
        mysql_query($query) or die("Cannot update");
        }

    ?>

    <H1>Score Counter</H1>

    <p>Click to add one point.</p>

    <form action ="show_score.php" method ="post">
    <input type ="submit" name ="score" value="score">
    </form>

    </body>
    </html>

Answer 1

因为您的查询是单引号，这使得您的变量$scores成为字面词$ score。因此，您的数字数据类型（prob int）将其转换为零。

将您的查询更改为：

$query='UPDATE scores SET score_counter="'.$scores.'" WHERE score_id=1';

我使用连接来确保在查询中使用$scores的值而不是文字$ score。

Answer 2

要在PHP字符串中使用变量插值，您需要使用双引号：

$query="UPDATE scores SET score_counter=$scores WHERE score_id=1";

Answer 3

替换

$query='UPDATE scores SET score_counter="$scores" WHERE score_id=1';

与

$query='UPDATE scores SET score_counter="' .$scores. '" WHERE score_id=1';

它将$得分作为单词，这在DB

中不存在

无法将数据写入数据库

3 个答案: