我试图在实际服务器上运行此代码,但是它会给出语法错误,而相同的查询在我的localhost上完美运行。我尝试了几种可能但没有运气。谁能告诉我这是什么问题?
<?php
$connection = new mysqli("localhost", "username", "password", "database");
$first_id = $connection->query("SELECT MIN(id) AS first_id FROM sample")->fetch_array(MYSQLI_ASSOC)['first_id'];
echo $first_id;
?>
这是我得到的语法错误。
解析错误:语法错误,意外&#39; [&#39;在第5行
答案 0 :(得分:5)
您正在执行array dereferencing,仅适用于PHP 5.4+。你没有运行PHP 5.4 +。
变化
$first_id = $connection->query("SELECT MIN(id) AS first_id FROM sample")->fetch_array(MYSQLI_ASSOC)['first_id'];
为:
$first = $connection->query("SELECT MIN(id) AS first_id FROM sample")->fetch_array(MYSQLI_ASSOC);
$first_id = $first['first_id'];