Php prompts Warning: mysql

My height code is as follows: include (conn. php); if (! Empty ($ _ GET [id]) {$ sqlselect * fromnewswhere 'id '. $ _ GET [id].; $ querymysql_query ($ SQL); $ rsmysql_fetch_array ($ query);} the error Warning: mysql_fetch_array () expectsparameter1 is returned.

My height code is as follows: include ("conn. php"); if (! Empty ($ _ GET ['id']) {$ SQL = "select * from news where 'id' = '". $ _ GET ['id']. "'"; $ query = mysql_query ($ SQL); $ rs = mysql_fetch_array ($ query);} the error Warning: mysql_fetch_array () expects parameter 1 is returned.

My height code is as follows:

Include ("conn. php ");

If (! Empty ($ _ GET ['id']) {
$ SQL = "select * from news where 'id' = '". $ _ GET ['id']. "'";
$ Query = mysql_query ($ SQL );
$ Rs = mysql_fetch_array ($ query );
}

Error Message

Warning: mysql_fetch_array () expects parameter 1 to be resource, boolean given in F: xmappwww.111cn. netmylibrarysearch_result.php on line 32

Then I upgraded the code.

If (! Empty ($ _ GET ['id']) {
$ SQL = "select * from news where 'id' = '". $ _ GET ['id']. "'";
$ Query = mysql_query ($ SQL );
If (mysql_num_rows ($ query)
{
$ Rs = mysql_fetch_array ($ query );
}
Else
{
Mysql_error ();
}
}

In this way, the mysql error message is displayed. The result is that the SQL statement is faulty. Modify the SQL statement,

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