使用php文件替换另一个文件中一行的几个部分

$dbhost = "DBHOST";                          /*      "localhost"      */
$dbuname = "DBUSERNAME";                     /* "database user name"  */
$dbpass = "DBPASSWORD";                      /* "database password"   */
$dbname = "DBNAME";                          /*    "database name"    */

<form action="edit.php">
        DB Server: <input type="text" name="host" placeholder="localhost"><br>
        DB Username: <input type="text" name="dbuser" placeholder="admin"><br>
        DB Password: <input type="password" name="dbpassword" placeholder="********"><br>
        DB Name: <input type="text" name="db_name" placeholder="pnc"><br>
        <button type="submit">Next Step</button>
</form>

假设你有这样一个文件:

/home/网站/example.com/conf.php

<?php
$dbhost = "DBHOST";                          /*      "localhost"      */
$dbuname = "DBUSERNAME";                     /* "database user name"  */
$dbpass = "DBPASSWORD";                      /* "database password"   */
$dbname = "DBNAME"; 

要替换这些占位符值，您所需要的只是在变量中获取页面源并进行一些字符串替换，然后将其写回文件。

$confpath = '/home/websites/example.com/conf.php';
$confphp = file_get_contents($confphp);
// Note the POST values need to be CLEANED and validated first.
// This especially means whitelisting what characters are allowed
// in the POST-received variables, encoding and removing " and '.
// You have to be VERY careful to authentic the user, and verify
// that user has permissions to do what they're doing, with the 
// databases/tables they're working with, unless this is part of
// an install script, which would need to be removed immediately
// upon finishing the install.
$confphp = str_replace(
    array('DBHOST','DBUSERNAME','DBPASSWORD','DBNAME'),
    array(
        addslashes($post_cleaned_dbhost), 
        addslashes($post_cleaned_dbusername), 
        addslashes($post_cleaned_dbpassword), 
        addslashes($post_cleaned_db_name)
    ),
    $confphp
);
file_put_contents($confpath, $confphp);

注意，要替换的值必须是文件文本内容的唯一值(例如，在同一文件中没有$DBUSERNAME)，否则它也会替换这些值。如果文件没有写入，它返回false，所以您可以测试file_put_contents()的返回来验证它是否工作。

http://www.php.net/manual/en/function.file-get-contents.php

http://www.php.net/manual/en/function.file-put-contents.php

创建一个带有控制器的表单

HTML中的表单代码

<form name="dbSet" id="dbSet" action="form_set_db.php" method="POST" />
    <label>Host</label> <input type="text" name="host" id="host" /><br />
    <label>Username</label> <input type="text" name="username" id="username" /><br />
    <label>Password</label> <input type="password" name="password" id="password" /><br />
    <label>Database</label> <input type="database" name="database" id="database" />
</form>

form_set_db.php

require_once 'config.php';
if(isset($_POST["host"], $_POST["username"], $_POST["password"], $_POST["database"]))
{
    $dbhost = $_POST["host"];
    $dbuname = $_POST["username"];
    $dbpass = $_POST["password"];
    $dbname = $_POST["database"];
}

这将用您在表单中输入的值替换$dbhost $dbuname $dbpass $dbname的值。

编辑

注意，这只是暂时替换它。如果您想要永久存储配置文件(例如，如果您正在构建安装程序)，我会动态创建config.php文件，而不是从头开始创建它。然后在form_set_db.php

内做这样的操作

if(isset($_POST["host"], $_POST["username"], $_POST["password"], $_POST["database"]))
{
    $file = fopen("config.php", "w");
    if(!is_resource($file)) { return false; }
    fwrite($file, "$dbhost = '" . $_POST["host"] . "';'n");
    fwrite($file, "$dbuname = '" . $_POST["username"] . "';'n");
    fwrite($file, "$dbpass = '" . $_POST["password"] . "';'n");
    fwrite($file, "$dbname = '" . $_POST["database"] . "';");
    fclose($file);
}

您想创建一个接受用户输入并更改文件中的行的表单吗?我不建议直接从表单设置数据库连接。但如果有必要，请尝试使用常规格式存储数据，如JSON

 $data = json_decode(file_get_contents($dataFileName));

写

 file_put_contents(json_encode($data, $dataFileName));

参考查阅http://www.php.net//manual/en/book.json.php

如果您的文件是典型的数据格式，那么它将清楚地表明您的文件包含数据。你可以给这个数据库命名。Json for instance

创建名为database的文件。Json包含以下内容:

{
     "username": "...",
     "password": "...",
     "database": "...",
     "host": "..."
}

然后使用

读取这些数据

$data = json_decode(file_get_contents('database.json'));
$data['username'] = $newValue;
$username = $data['username'];

一旦你设置了新的值

file_put_contents(json_encode('database.json', $data));

从用户提交的数据连接到数据库存在问题，例如切换到不同的数据库并在您的页面中显示可用于XSS的恶意数据http://en.wikipedia.org/wiki/Cross-site_scripting