the code is like this
$sql="SELECT tahun FROM tahunajar ";
i want the value of $sql can write in this query
$sql1= "SELECT * FROM soal where idSoal LIKE '-->HERE<--' ";
can someone tell me how do it?
Ofcourse, you can write it via like this
$sql1= "SELECT * FROM soal where idSoal IN (".$sql.") ";
Simply use the variable in double quoted string. It will work. Following is the example:
$sql1= "SELECT * FROM soal where idSoal LIKE $sql";
You can learn more from HERE.