Question

我正在为我正在处理的网站制作一个安装脚本，我在制作生成 config.php 文件的脚本时遇到了麻烦。以下是我为配置文件生成字符串的方法：

<?php
if (file_exists("config.php")) {
    header("Location: index.php");
} else {
    if (isset($_POST['smtp_password'])) {
        if (!($configFile = fopen("config.php", "c"))) {
            print("ERROR: Cannot write in this directory!");
            exit();
        }
$config = <<<EOT
<?php
    $_AMCFG['login_dir']        = '{$_POST['login_dir']}'; /* LINE 12 */
    $_AMCFG['server_key']       = '{$_POST['server_key']}';
    $_AMCFG['host']             = '{$_POST['host']}';
    $_AMCFG['database']         = '{$_POST['database']}';
    $_AMCFG['user']             = '{$_POST['user']}';
    $_AMCFG['password']         = '{$_POST['password']}';
    $_AMCFG['smtp_name']        = '{$_POST['smtp_name']}';
    $_AMCFG['smtp_mail']        = '{$_POST['smtp_mail']}';
    $_AMCFG['smtp_host']        = '{$_POST['smtp_host']}';
    $_AMCFG['smtp_port']        = {$_POST['smtp_port']};
    $_AMCFG['smtp_user']        = '{$_POST['smtp_user']}';
    $_AMCFG['smtp_password']    = '{$_POST['smtp_password']}';
?>
EOT;

        fwrite($configFile, $config);
        $db = mysqli_connect($_POST['host'], $_POST['user'], $_POST['password']);
        mysqli_select_db($db, $_POST['database']);
        $sqlFile = file_get_contents("install.sql");
        mysqli_multi_query($sqlFile);
        mysqli_query($db, "INSERT INTO admins (steamid, name, mail, disabled, superadmin) VALUES (\"".escape($_POST['admin_steamid'])."\", \"".escape($_POST['admin_name'])."\", \"".escape($_POST['admin_email'])."\", 0, 1)");
    }
}
?>

这是我得到的错误:(第12行）

Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING)

我做错了什么？

Answer 1

以下是我对此问题的贡献，这是否是预期结果，请参阅下面的结果。

<?php
$config = "<?php\n\n";
$config .= '$_AMCFG[\'login_dir\']'     . " = " . '$_POST[\'login_dir\']' . ";\n";
$config .= '$_AMCFG[\'server_key\']'    . " = " . '$_POST[\'server_key\']' . ";\n\n";
$config .= '$_AMCFG[\'host\']'          . " = " . '$_POST[\'host\']' . ";\n";
$config .= '$_AMCFG[\'database\']'      . " = " . '$_POST[\'database\']' . ";\n";
$config .= '$_AMCFG[\'user\']'          . " = " . '$_POST[\'user\']' . ";\n";
$config .= '$_AMCFG[\'password\']'      . " = " . '$_POST[\'password\']' . ";\n\n";
$config .= '$_AMCFG[\'smtp_name\']'     . " = " . '$_POST[\'smtp_name\']' . ";\n";
$config .= '$_AMCFG[\'smtp_mail\']'     . " = " . '$_POST[\'smtp_mail\']' . ";\n";
$config .= '$_AMCFG[\'smtp_host\']'     . " = " . '$_POST[\'smtp_host\']' . ";\n";
$config .= '$_AMCFG[\'smtp_port\']'     . " = " . '$_POST[\'smtp_port\']' . ";\n";
$config .= '$_AMCFG[\'smtp_user\']'     . " = " . '$_POST[\'smtp_user\']' . ";\n";
$config .= '$_AMCFG[\'smtp_password\']' . " = " . '$_POST[\'smtp_password\']' . ";\n\n";
$config .= "?>";

echo $config;

在HTML源代码中回显：

<?php

$_AMCFG['login_dir'] = $_POST['login_dir'];
$_AMCFG['server_key'] = $_POST['server_key'];

$_AMCFG['host'] = $_POST['host'];
$_AMCFG['database'] = $_POST['database'];
$_AMCFG['user'] = $_POST['user'];
$_AMCFG['password'] = $_POST['password'];

$_AMCFG['smtp_name'] = $_POST['smtp_name'];
$_AMCFG['smtp_mail'] = $_POST['smtp_mail'];
$_AMCFG['smtp_host'] = $_POST['smtp_host'];
$_AMCFG['smtp_port'] = $_POST['smtp_port'];
$_AMCFG['smtp_user'] = $_POST['smtp_user'];
$_AMCFG['smtp_password'] = $_POST['smtp_password'];

?>

Answer 2

$config .= "    $_AMCFG['smtp_port']        = ".$_POST['smtp_port'].";\n";

应该是：

$config .= "    $_AMCFG['smtp_port']        = '".$_POST['smtp_port'].";\n";

我认为这个代码在＆＃34; evaled＆＃34;时失败了。

Answer 3

我认为您的问题是在每行的第一部分使用双引号。你试图为$ _AMCFG [*]插入一个值，当它应该只是一个文字字符串时。

尝试在这部分周围使用单引号：

$config .= '    $_AMCFG['login_dir']        = \''.$_POST['login_dir']."';\n";

或者，我可以为此建议heredoc语法吗？那看起来像是：

$config = <<<EOT
<?php
    $_AMCFG['login_dir']        = '{$_POST['login_dir']}';
    $_AMCFG['server_key']       = '{$_POST['server_key']}';
    ...
    $_AMCFG['smtp_password']    = '{$_POST['smtp_password']}';
?>
EOT;

更清洁。实际上，您可以将该配置开发为单独的PHP文件，然后将其剪切/粘贴到位。

Answer 4

您的代码中有两个<?php个实例。你只需要第一个。

第二个用于输出文件，需要进行转义。

$config = <<<EOT
\x3C?php

带有两个实例的旧版本让php解析器感到困惑。

字符串解析错误（多行和变量）

4 个答案: