警告:字符串偏移量为XXX

当您尝试使用方括号语法访问数组元素时，就会发生这种情况，但您是在字符串上执行此操作，而不是在数组上执行此操作，因此该操作显然没有意义。

例子:

$var = "test";
echo $var["a_key"];

如果你认为变量应该是一个数组，看看它来自哪里并修复那里的问题。

解析错误:语法错误，意外的T_ENCAPSED_AND_WHITESPACE

在PHP 8.0及以上版本中，该消息改为:

语法错误，意外的字符串内容“”，期望“-”或标识符或变量或数字

当试图在双引号字符串中引用带有引号键的数组值时，当整个复杂变量结构没有包含在{}中时，最常遇到此错误。

错误情况:

这将导致意外的T_ENCAPSED_AND_WHITESPACE:

echo "This is a double-quoted string with a quoted array key in $array['key']";
//---------------------------------------------------------------------^^^^^

可能的修复:

在双引号字符串中，PHP将允许数组关键字字符串不带引号使用，并且不会发出E_NOTICE。所以上面可以写成:

echo "This is a double-quoted string with an un-quoted array key in $array[key]";
//------------------------------------------------------------------------^^^^^

整个复杂数组变量和键可以被括在{}中，在这种情况下，它们应该被引用以避免E_NOTICE。PHP文档对于复杂变量推荐使用这种语法。

echo "This is a double-quoted string with a quoted array key in {$array['key']}";
//--------------------------------------------------------------^^^^^^^^^^^^^^^
// Or a complex array property of an object:
echo "This is a a double-quoted string with a complex {$object->property->array['key']}";

当然，上述任何一种方法的替代方法都是将数组变量连接到，而不是插值它:

echo "This is a double-quoted string with an array variable". $array['key'] . " concatenated inside.";
//----------------------------------------------------------^^^^^^^^^^^^^^^^^^^^^

如需参考，请参阅PHP字符串手册页中的变量解析部分

致命错误:调用成员函数…在一个非物体上

发生在类似于xyz->method()的代码中，其中xyz不是对象，因此不能调用该方法。

这是一个致命错误，将停止脚本(向前兼容性注意:它将从PHP 7开始成为一个可捕获的错误)。

大多数情况下，这表明代码缺少对错误条件的检查。在调用一个对象的方法之前，验证它确实是一个对象。

一个典型的例子是

// ... some code using PDO
$statement = $pdo->prepare('invalid query', ...);
$statement->execute(...);

在上面的例子中，查询不能被准备，并且prepare()会将false赋值给$statement。尝试调用execute()方法将导致致命错误，因为false是一个“非对象”，因为该值是一个布尔值。

弄清楚为什么函数返回一个布尔值而不是一个对象。例如，检查$pdo对象中最近发生的错误。关于如何调试的细节将取决于如何处理特定函数/对象/类的错误。

如果->准备失败，那么你的$pdo数据库句柄对象没有被传递到当前作用域。找到它被定义的地方。然后将其作为参数传递，将其作为属性存储，或通过全局作用域共享。

另一个问题可能是有条件地创建一个对象，然后尝试在该条件块之外调用一个方法。例如

if ($someCondition) {
    $myObj = new MyObj();
}
// ...
$myObj->someMethod();

如果试图在条件块之外执行方法，则可能无法定义对象。

相关问题:

调用非对象上的成员函数 列出所有PHP“致命错误:调用成员函数…”关于Stackoverflow的问题

解析错误:语法错误，意外的T_VARIABLE

可能的场景

我似乎找不到我的代码出了什么问题。以下是我的全部错误:

解析错误:语法错误，意外的T_VARIABLE在x行

我正在尝试的

$sql = 'SELECT * FROM dealer WHERE id="'$id.'"';

回答

解析错误:程序的语法问题，例如在语句的末尾遗漏了分号，或者像上面的情况一样，遗漏了。操作符。解释器在遇到解析错误时停止运行程序。

简单地说，这是一个语法错误，这意味着您的代码中有一些东西阻止了它被正确解析，从而无法运行。

您应该做的是仔细检查错误所在的行周围是否有任何简单的错误。

该错误消息意味着，在文件的第x行，PHP解释器期望看到一个开括号，但相反，它遇到了名为T_VARIABLE的东西。T_VARIABLE这个东西叫做令牌。它是PHP解释器表达程序不同基本部分的方式。当解释器读入程序时，它将您所编写的内容转换为令牌列表。无论你在程序中放入一个变量，解释器的列表中都会有一个aT_VARIABLE令牌。

不错的阅读:解析器令牌列表

因此，请确保在php.ini中至少启用了E_PARSE。产品脚本中不应该存在解析错误。

我总是建议在编码时添加以下语句:

error_reporting(E_ALL);

PHP错误报告

另外，使用IDE是个好主意，它可以让你在输入时知道解析错误。你可以使用:

NetBeans(一个漂亮的免费软件)(在我看来是最好的) PhpStorm(戈登叔叔喜欢这个:P，付费计划，包含专有和免费软件) Eclipse(美女与野兽，免费软件)

相关问题:

参考:PHP语法错误;以及如何解决这些问题?

mysql_connect(): user 'name'@'host'拒绝访问

当您连接到无效或缺少凭据(用户名/密码)的MySQL/MariaDB服务器时，会出现此警告。所以这通常不是代码问题，而是服务器配置问题。

See the manual page on mysql_connect("localhost", "user", "pw") for examples. Check that you actually used a $username and $password. It's uncommon that you gain access using no password - which is what happened when the Warning: said (using password: NO). Only the local test server usually allows to connect with username root, no password, and the test database name. You can test if they're really correct using the command line client: mysql --user="username" --password="password" testdb Username and password are case-sensitive and whitespace is not ignored. If your password contains meta characters like $, escape them, or put the password in single quotes. Most shared hosting providers predeclare mysql accounts in relation to the unix user account (sometimes just prefixes or extra numeric suffixes). See the docs for a pattern or documentation, and CPanel or whatever interface for setting a password. See the MySQL manual on Adding user accounts using the command line. When connected as admin user you can issue a query like: CREATE USER 'username'@'localhost' IDENTIFIED BY 'newpassword'; Or use Adminer or WorkBench or any other graphical tool to create, check or correct account details. If you can't fix your credentials, then asking the internet to "please help" will have no effect. Only you and your hosting provider have permissions and sufficient access to diagnose and fix things. Verify that you could reach the database server, using the host name given by your provider: ping dbserver.hoster.example.net Check this from a SSH console directly on your webserver. Testing from your local development client to your shared hosting server is rarely meaningful. Often you just want the server name to be "localhost", which normally utilizes a local named socket when available. Othertimes you can try "127.0.0.1" as fallback. Should your MySQL/MariaDB server listen on a different port, then use "servername:3306". If that fails, then there's a perhaps a firewall issue. (Off-topic, not a programming question. No remote guess-helping possible.) When using constants like e.g. DB_USER or DB_PASSWORD, check that they're actually defined. If you get a "Warning: Access defined for 'DB_USER'@'host'" and a "Notice: use of undefined constant 'DB_PASS'", then that's your problem. Verify that your e.g. xy/db-config.php was actually included and whatelse. Check for correctly set GRANT permissions. It's not sufficient to have a username+password pair. Each MySQL/MariaDB account can have an attached set of permissions. Those can restrict which databases you are allowed to connect to, from which client/server the connection may originate from, and which queries are permitted. The "Access denied" warning thus may as well show up for mysql_query calls, if you don't have permissions to SELECT from a specific table, or INSERT/UPDATE, and more commonly DELETE anything. You can adapt account permissions when connected per command line client using the admin account with a query like: GRANT ALL ON yourdb.* TO 'username'@'localhost'; If the warning shows up first with Warning: mysql_query(): Access denied for user ''@'localhost' then you may have a php.ini-preconfigured account/password pair. Check that mysql.default_user= and mysql.default_password= have meaningful values. Oftentimes this is a provider-configuration. So contact their support for mismatches. Find the documentation of your shared hosting provider: e.g. HostGator, GoDaddy, 1and1, DigitalOcean, BlueHost, DreamHost, MediaTemple, ixWebhosting, lunarhosting, or just google yours´. Else consult your webhosting provider through their support channels. Note that you may also have depleted the available connection pool. You'll get access denied warnings for too many concurrent connections. (You have to investigate the setup. That's an off-topic server configuration issue, not a programming question.) Your libmysql client version may not be compatible with the database server. Normally MySQL and MariaDB servers can be reached with PHPs compiled in driver. If you have a custom setup, or an outdated PHP version, and a much newer database server, or significantly outdated one - then the version mismatch may prevent connections. (No, you have to investigate yourself. Nobody can guess your setup).

更多的引用:

mysql服务器错误:mysql用户无法访问root用户 mysql_connect():拒绝访问 mysql_select_db()拒绝访问用户" @'localhost'(使用密码:NO) PHPMyAdmin拒绝用户“root”@“localhost”的访问

顺便说一句，你可能不想再使用mysql_*函数了。新来者经常迁移到mysqli，然而这也一样乏味。相反，阅读PDO和准备好的语句。 $db =新PDO("mysql:主机=localhost;dbname=testdb"， "用户名"，"密码");

警告:mysql_fetch_array()期望参数1是resource, boolean给定

首先，也是最重要的:

请不要在新代码中使用mysql_*函数。它们不再被维护，并且已被正式弃用。看到红框了吗?转而学习准备语句，并使用PDO或MySQLi——本文将帮助您选择哪一种。如果您选择PDO，这里有一个很好的教程。

当您试图从mysql_query的结果中获取数据但查询失败时，就会发生这种情况。

这是一个警告，不会停止脚本，但会使您的程序出错。

您需要检查mysql_query返回的结果by

$res = mysql_query($sql);
if (!$res) {
   trigger_error(mysql_error(),E_USER_ERROR);
}
// after checking, do the fetch

相关问题:

Mysql_fetch_array()期望参数1是resource，在select中给出的布尔值 所有“mysql_fetch_array()期望参数1是资源，布尔给定”的问题

相关的错误:

警告:[function]期望参数1是给定的布尔值

其他mysql*函数也期望mysql结果资源作为参数将出于同样的原因产生相同的错误。