如何在php中编写JSON_表查询?
我正在phpmyadmin(mysql 8.0.13)中运行此查询: 它可以像预期的那样工作,但是当我在php中尝试相同的查询时,什么都不起作用如何在php中编写JSON_表查询?,php,json,mysqli,Php,Json,Mysqli,我正在phpmyadmin(mysql 8.0.13)中运行此查询: 它可以像预期的那样工作,但是当我在php中尝试相同的查询时,什么都不起作用 $result = mysqli_query($conn, "SELECT people.* FROM product, JSON_TABLE(attributes, '$.people[*]' COLUMNS (firstname VARCHAR(40) PATH '$."firstname"')) people"); while($row = my
$result = mysqli_query($conn, "SELECT people.* FROM product, JSON_TABLE(attributes, '$.people[*]' COLUMNS (firstname VARCHAR(40) PATH '$."firstname"')) people");
while($row = mysqli_fetch_assoc($result))
{
$firstname = $row['firstname'];
}
有人能告诉我我做错了什么吗
$result = mysqli_query($conn, "SELECT people.* FROM product, JSON_TABLE(attributes, '$.people[*]' COLUMNS (firstname VARCHAR(40) PATH '$.firstname')) people");
您的查询中有语法错误。您在firstname中混合了双引号和单引号。尝试一下上面的查询,让我知道它是否有效
您的查询中有语法错误。您在firstname中混合了双引号和单引号。请尝试上面的查询,并让我知道它是否有效。也许$symbol会被解释为PHP变量?尝试将SQL放在单引号中。也许$symbol会被解释为PHP变量?尝试将SQL放在单引号中。
$result = mysqli_query($conn, "SELECT people.* FROM product, JSON_TABLE(attributes, '$.people[*]' COLUMNS (firstname VARCHAR(40) PATH '$.firstname')) people");