PHP 意外的“var”（T_VARIABLE）
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PHP Unexpected 'var' (T_VARIABLE)
提问by ZephireNZ
For whatever reason, I have a small bit of PHP code that no matter where I put var $blah;, it always gives this error in the logs: PHP?Parse?error:??syntax?error,?unexpected?'var'?(T_VAR)?in?/path/to/file.php on line xx
无论出于何种原因，我都有一小段 PHP 代码，无论我把var $blah;它放在哪里，它总是在日志中给出这个错误：PHP?Parse?error:??syntax?error,?unexpected?'var'?(T_VAR)?in?/path/to/file.php on line xx
I have no idea why it wouldn't accept this. A class that is included (which creates the $proverbSitevariable in another php section) uses plenty of 'var $blah', with no problems. I also realise this is probably just an embarassingly simple mistake.
我不知道为什么它不接受这一点。包含的类（$proverbSite在另一个 php 部分创建变量）使用了大量的“var $blah”，没有问题。我也意识到这可能只是一个非常简单的错误。
<?php
$proverbSite->dbConnect();
$result = $proverbSite->dbQuery("randProverb");
if($result != null) {
$row = $result->fetch_assoc();
echo $row['proverb'];
echo "<br>";
}
?>
回答by Lex
Keyword varis only used in classes (in PHP). In the plain scope variables are automatically declared as you mention them. Just erase it, and it should work.
关键字var仅在类中使用（在 PHP 中）。在普通范围中，变量会在您提到它们时自动声明。只需擦除它，它应该可以工作。
回答by PHPman
Check the line before xxbecause you may have forgotten a ;and this may cause PHP to interpret it incorrectly
检查之前的行，xx因为您可能忘记了 a;这可能会导致 PHP 错误地解释它
回答by stephenspann
Without seeing all the sources involved, I'm guessing you're combining JavaScript and PHP.
没有看到所有涉及的来源，我猜你正在结合 JavaScript 和 PHP。
JavaScript variable declarations begin with 'var' and PHP does not... If PHP were to encounter the code 'var' before a variable, it would give the error message you listed. T_VAR usually indicates PHP is trying to interpret a constant, which would be the case with a JavaScript 'var'.
JavaScript 变量声明以“var”开头，而 PHP 不...如果 PHP 在变量之前遇到代码“var”，它会给出您列出的错误消息。T_VAR 通常表示 PHP 正在尝试解释一个常量，JavaScript 'var' 就是这种情况。
Now, as for the plugin/library your are using, it may be echoing javascript, but including a PHP variable, ex:
现在，对于您正在使用的插件/库，它可能会回显 javascript，但包括一个 PHP 变量，例如：
echo "var myJavasScriptVar = '$phpVar'";
In PHP, '$' in a string means that it will be replaced by a variable.
在 PHP 中，字符串中的 '$' 表示它将被变量替换。
Hope this helps!
希望这可以帮助！
回答by Mirza Sisic
I know that in newer versions of Laravel it works if you leave out the varkeyword, for example:
我知道在较新版本的 Laravel 中，如果您省略var关键字，它会起作用，例如：
Route::get('/read', function(){
$results = DB::select('select * from posts where id = ?', [1]);
return dump($results);
});
