我连接到数据库并想要选择tweetid在我的数组中的行但它在响应中给了我警告。这是我的代码。如果有人帮助找出我的错误,将会感激不尽
$con=mysqli_connect("localhost","root","6245","twitr");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();}
$ids = join(',',$tweetids);
$result = mysqli_query($con,"SELECT * FROM tweet WHERE tweetid in $ids");
while($row = mysqli_fetch_array($result)) {}
mysqli_close($con);
我的警告是:
mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in
答案 0 :(得分:3)
IN是一个函数,您需要将值设为
select * from table where col in (1,2,3);
http://dev.mysql.com/doc/refman/5.6/en/comparison-operators.html#function_in
答案 1 :(得分:0)
尝试使用FIND_IN_SET
$result = mysqli_query($con,"SELECT * FROM tweet WHERE find_in_set(tweetid,$ids)");