PHP解析错误：语法错误

这个问题在这里已经有了答案：

PHP解析/语法错误; 以及如何解决它们？ 13个答案

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将WHERE'age'！=“”更改为WHERE'age'！=''

使用

$query = mysqli_query($con, "SELECT a.* FROM (SELECT `id` as `id`, `age` as `age` FROM `register` WHERE `age` !='') as a INNER JOIN (SELECT `one` as `f1` FROM `friends` WHERE `two`='".$my_id."' UNION SELECT `two` as `f2` FROM `friends` WHERE `one` = '".$my_id."') as b ON a.id=b.f1");

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你应该在这里避免双引号

WHERE `age` != ""

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